Chapter 22: Alternating Current

Introduction

• Alternating current (AC) changes magnitude and reverses direction periodically.

• Direct current (DC) flows with constant magnitude and fixed direction.

• AC voltage drives AC current in circuits.

• AC voltage is preferred over DC because it can be easily transformed to different voltages using transformers.

Solved Examples

• Ex 1: Identify AC and DC from a graph.
  • AC is the wave crossing zero line periodically.
  • DC is the constant line above zero.

Practice Set

• Q1: What is the main difference between AC and DC?
  • A. AC has constant magnitude, DC reverses direction.
  • B. AC reverses direction periodically, DC has constant direction.
  • C. Both have constant magnitude.
  • D. Both reverse direction periodically.
• Q2: Why is AC voltage preferred over DC voltage?
  • A. AC voltage is constant.
  • B. AC voltage can be easily transformed to different voltages.
  • C. DC voltage is safer.
  • D. DC voltage changes direction.
• Q3: Which of the following represents AC current?
  • A. Constant positive line.
  • B. Line crossing zero periodically.
  • C. Zero current.
  • D. Constant negative line.

Answer Key

• Q1: B - AC reverses direction periodically; DC does not.
• Q2: B - AC voltage can be transformed easily using transformers.
• Q3: B - AC current crosses zero line periodically.

Memory Aids

• AC = Alternates direction; DC = Direct constant flow.
• AC voltage preferred for easy voltage transformation.

Fundamentals of Alternating Current

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  ( I = I_m sin ωt ) or ( I = I_m cos ωt )

• ( I ): Instantaneous current at time ( t )

• ( I_m ): Peak (maximum) current

• ( ω ): Angular frequency, ( ω = 2π/T = 2πν )

• ( T ): Time period, ( ν ): Frequency

Solved Examples

• Ex 1: Find angular frequency if ( T = 0.02 ) s.
  • ( ω = 2π/T = 2π/0.02 = 314.16 ) rad/s

Practice Set

• Q1: What is the peak value in ( I = I_m sin ωt )?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current ( I_m )
  • D. Zero current
• Q2: Angular frequency ( ω ) is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ( ν = 50 ) Hz, find ( ω ).
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - ( I_m ) is the peak current.
• Q2: B - ( ω = 2π/T )
• Q3: B - ( ω = 2π × 50 = 314 ) rad/s

Memory Aids

• ( ω = 2πν ), remember ( 2π ) radians = 1 cycle.
• Peak current = ( I_m ), instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  ( I = I_m sin ωt ) or ( I = I_m cos ωt )

• ( I ): Instantaneous current at time ( t )

• ( I_m ): Peak (maximum) current

• ( ω ): Angular frequency, ( ω = 2π/T = 2πν )

• ( T ): Time period, ( ν ): Frequency

Solved Examples

• Ex 1: Find angular frequency if ( T = 0.02 ) s.
  • ( ω = 2π/T = 2π/0.02 = 314.16 ) rad/s

Practice Set

• Q1: What is the peak value in ( I = I_m sin ωt )?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current ( I_m )
  • D. Zero current
• Q2: Angular frequency ( ω ) is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ( ν = 50 ) Hz, find ( ω ).
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - ( I_m ) is the peak current.
• Q2: B - ( ω = 2π/T )
• Q3: B - ( ω = 2π × 50 = 314 ) rad/s

Memory Aids

• ( ω = 2πν ), remember ( 2π ) radians = 1 cycle.
• Peak current = ( I_m ), instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  ( I = I_m sin ωt ) or ( I = I_m cos ωt )

• ( I ): Instantaneous current at time ( t )

• ( I_m ): Peak (maximum) current

• ( ω ): Angular frequency, ( ω = 2π/T = 2πν )

• ( T ): Time period, ( ν ): Frequency

Solved Examples

• Ex 1: Find angular frequency if ( T = 0.02 ) s.
  • ( ω = 2π/T = 2π/0.02 = 314.16 ) rad/s

Practice Set

• Q1: What is the peak value in ( I = I_m sin ωt )?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current ( I_m )
  • D. Zero current
• Q2: Angular frequency ( ω ) is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ( ν = 50 ) Hz, find ( ω ).
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - ( I_m ) is the peak current.
• Q2: B - ( ω = 2π/T )
• Q3: B - ( ω = 2π × 50 = 314 ) rad/s

Memory Aids

• ( ω = 2πν ), remember ( 2π ) radians = 1 cycle.
• Peak current = ( I_m ), instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  I = I_m sin ωt or I = I_m cos ωt

• I: Instantaneous current at time t

• I_m: Peak (maximum) current

• ω: Angular frequency, ω = 2π/T = 2πν

• T: Time period, ν: Frequency

Solved Examples

• Ex 1: Find angular frequency if T = 0.02 s.
  • ω = 2π/T = 2π/0.02 = 314.16 rad/s

Practice Set

• Q1: What is the peak value in I = I_m sin ωt?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current I_m
  • D. Zero current
• Q2: Angular frequency ω is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ν = 50 Hz, find ω.
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - I_m is the peak current.
• Q2: B - ω = 2π/T
• Q3: B - ω = 2π × 50 = 314 rad/s

Memory Aids

• ω = 2πν, remember 2π radians = 1 cycle.
• Peak current = I_m, instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  ( I = I_m sin ωt ) or ( I = I_m cos ωt )

• ( I ): Instantaneous current at time ( t )

• ( I_m ): Peak (maximum) current

• ( ω ): Angular frequency, ( ω = 2π/T = 2πν )

• ( T ): Time period, ( ν ): Frequency

Solved Examples

• Ex 1: Find angular frequency if ( T = 0.02 ) s.
  • ( ω = 2π/T = 2π/0.02 = 314.16 ) rad/s

Practice Set

• Q1: What is the peak value in ( I = I_m sin ωt )?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current ( I_m )
  • D. Zero current
• Q2: Angular frequency ( ω ) is given by?
  • A. 2πT
  • B. 2π/T
  • C. T/2π
  • D. πT
• Q3: If frequency ( ν = 50 ) Hz, find ( ω ).
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - ( I_m ) is the peak current.
• Q2: B - ( ω = 2π/T )
• Q3: B - ( ω = 2π × 50 = 314 ) rad/s

Memory Aids

• ( ω = 2πν ), remember ( 2π ) radians = 1 cycle.
• Peak current = ( I_m ), instantaneous = varies with time.

• AC voltage or current can be represented as sine or cosine functions:

  \( I = I_m \sin \omega t \) or \( I = I_m \cos \omega t \)

• \( I \): Instantaneous current at time \( t \)

• \( I_m \): Peak (maximum) current

• \( \omega \): Angular frequency, \( \omega = \frac{2\pi}{T} = 2\pi \nu \)

• \( T \): Time period, \( \nu \): Frequency

Solved Examples

• Ex 1: Find angular frequency if \( T = 0.02 \) s.
  • \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.02} = 314.16 \) rad/s

Practice Set

• Q1: What is the peak value in \( I = I_m \sin \omega t \)?
  • A. Instantaneous current
  • B. Average current
  • C. Maximum current \( I_m \)
  • D. Zero current
• Q2: Angular frequency \( \omega \) is given by?
  • A. \( 2\pi T \)
  • B. \( \frac{2\pi}{T} \)
  • C. \( \frac{T}{2\pi} \)
  • D. \( \pi T \)
• Q3: If frequency \( \nu = 50 \) Hz, find \( \omega \).
  • A. 100 rad/s
  • B. 314 rad/s
  • C. 50 rad/s
  • D. 25 rad/s

Answer Key

• Q1: C - \( I_m \) is the peak current.
• Q2: B - \( \omega = \frac{2\pi}{T} \)
• Q3: B - \( \omega = 2\pi \times 50 = 314 \) rad/s

Memory Aids

• \( \omega = 2\pi \nu \), remember \( 2\pi \) radians = 1 cycle.
• Peak current = \( I_m \), instantaneous = varies with time.

Average and RMS Values

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  I_rms = I_m/√2

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  I_rms = I_m/√2

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  I_rms = I_m/√2

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  ⟨I⟩ = 2I_m/π

• Root mean square (rms) value:

  I_rms = I_m/√2

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if I_m = 10 A.
  • I_rms = 10/√2 = 7.07 A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q2: RMS value of current is?
  • A. I_m
  • B. 2I_m/π
  • C. I_m/√2
  • D. Zero
• Q3: If I_m = 14 A, find I_rms.
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is 2I_m/π.
• Q2: C - RMS value is I_m/√2.
• Q3: A - I_rms = 14/√2 = 9.9 A.

Memory Aids

• Average value = 2/π times peak.
• RMS value = peak divided by √2.

• Average (mean) value over half cycle:

  ( ⟨I⟩ = 2I_m/π )

• Root mean square (rms) value:

  ( I_rms = I_m/√2 )

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if ( I_m = 10 ) A.
  • ( I_rms = 10/√2 = 7.07 ) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q2: RMS value of current is?
  • A. ( I_m )
  • B. ( 2I_m/π )
  • C. ( I_m/√2 )
  • D. Zero
• Q3: If ( I_m = 14 ) A, find ( I_rms ).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is ( 2I_m/π ).
• Q2: C - RMS value is ( I_m/√2 ).
• Q3: A - ( I_rms = 14/√2 = 9.9 ) A.

Memory Aids

• Average value = ( 2/π ) times peak.
• RMS value = peak divided by ( √2 ).

• Average (mean) value over half cycle:

  \( \langle I \rangle = \frac{2 I_m}{\pi} \)

• Root mean square (rms) value:

  \( I_{rms} = \frac{I_m}{\sqrt{2}} \)

• RMS value produces same heat in resistor as DC of same value.

• RMS value is also called virtual or apparent value.

Solved Examples

• Ex 1: Calculate rms current if \( I_m = 10 \) A.
  • \( I_{rms} = \frac{10}{\sqrt{2}} = 7.07 \) A

Practice Set

• Q1: Average value of AC current over half cycle is?
  • A. \( I_m \)
  • B. \( \frac{2 I_m}{\pi} \)
  • C. \( \frac{I_m}{\sqrt{2}} \)
  • D. Zero
• Q2: RMS value of current is?
  • A. \( I_m \)
  • B. \( \frac{2 I_m}{\pi} \)
  • C. \( \frac{I_m}{\sqrt{2}} \)
  • D. Zero
• Q3: If \( I_m = 14 \) A, find \( I_{rms} \).
  • A. 9.9 A
  • B. 7 A
  • C. 14 A
  • D. 20 A

Answer Key

• Q1: B - Average over half cycle is \( \frac{2 I_m}{\pi} \).
• Q2: C - RMS value is \( \frac{I_m}{\sqrt{2}} \).
• Q3: A - \( I_{rms} = \frac{14}{\sqrt{2}} = 9.9 \) A.

Memory Aids

• Average value = \( \frac{2}{\pi} \) times peak.
• RMS value = peak divided by \( \sqrt{2} \).

Phase and Phase Difference

• AC current with phase: I = I_m sin (ωt ± φ)

• Initial phase φ is constant; instantaneous phase ωt changes with time.

• Phase difference between voltage ε = ε₀ sin (ωt + φ₁) and current I = I_m sin (ωt + φ₂):

  φ = φ₂ - φ₁ (current w.r.t voltage)

  φ = φ₁ - φ₂ (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by ( 30° ), write current expression.
  • I = I_m sin (ωt - 30°)

Practice Set

• Q1: If ( ε = ε_m sin ωt ), current lags voltage by ( 45° ), current is?
  • A. I_m sin (ωt + 45°)
  • B. I_m sin (ωt - 45°)
  • C. I_m sin ωt
  • D. Zero
• Q2: Phase difference φ is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. ε = ε_m sin (ωt - φ)
  • B. I = I_m sin (ωt - φ)
  • C. I = I_m sin (ωt + φ)
  • D. ε = ε_m sin (ωt + φ)

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

• AC current with phase: I = I_m sin (ωt ± φ)

• Initial phase φ is constant; instantaneous phase ωt changes with time.

• Phase difference between voltage ε = ε₀ sin (ωt + φ₁) and current I = I_m sin (ωt + φ₂):

  φ = φ₂ - φ₁ (current w.r.t voltage)

  φ = φ₁ - φ₂ (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by ( 30° ), write current expression.
  • I = I_m sin (ωt - 30°)

Practice Set

• Q1: If ( ε = ε_m sin ωt ), current lags voltage by ( 45° ), current is?
  • A. I_m sin (ωt + 45°)
  • B. I_m sin (ωt - 45°)
  • C. I_m sin ωt
  • D. Zero
• Q2: Phase difference φ is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. ε = ε_m sin (ωt - φ)
  • B. I = I_m sin (ωt - φ)
  • C. I = I_m sin (ωt + φ)
  • D. ε = ε_m sin (ωt + φ)

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

• AC current with phase: I = I_m sin (ωt ± φ)

• Initial phase φ is constant; instantaneous phase ωt changes with time.

• Phase difference between voltage ε = ε₀ sin (ωt + φ₁) and current I = I_m sin (ωt + φ₂):

  φ = φ₂ - φ₁ (current w.r.t voltage)

  φ = φ₁ - φ₂ (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by ( 30° ), write current expression.
  • I = I_m sin (ωt - 30°)

Practice Set

• Q1: If ( ε = ε_m sin ωt ), current lags voltage by ( 45° ), current is?
  • A. I_m sin (ωt + 45°)
  • B. I_m sin (ωt - 45°)
  • C. I_m sin ωt
  • D. Zero
• Q2: Phase difference φ is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. ε = ε_m sin (ωt - φ)
  • B. I = I_m sin (ωt - φ)
  • C. I = I_m sin (ωt + φ)
  • D. ε = ε_m sin (ωt + φ)

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

• AC current with phase: ( I = I_m sin (ωt ± φ) )

• Initial phase ( φ ) is constant; instantaneous phase ( ωt ) changes with time.

• Phase difference between voltage ( ε = ε₀ sin (ωt + φ₁) ) and current ( I = I_m sin (ωt + φ₂) ):

  ( φ = φ₂ - φ₁ ) (current w.r.t voltage)

  ( φ = φ₁ - φ₂ ) (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by ( 30° ), write current expression.
  • ( I = I_m sin (ωt - 30°) )

Practice Set

• Q1: If ( ε = ε_m sin ωt ), current lags voltage by ( 45° ), current is?
  • A. ( I_m sin (ωt + 45°) )
  • B. ( I_m sin (ωt - 45°) )
  • C. ( I_m sin ωt )
  • D. Zero
• Q2: Phase difference ( φ ) is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. ( ε = ε_m sin (ωt - φ) )
  • B. ( I = I_m sin (ωt - φ) )
  • C. ( I = I_m sin (ωt + φ) )
  • D. ( ε = ε_m sin (ωt + φ) )

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

• AC current with phase: I = I_m sin (ωt ± φ)

• Initial phase φ is constant; instantaneous phase ωt changes with time.

• Phase difference between voltage ε = ε₀ sin (ωt + φ₁) and current I = I_m sin (ωt + φ₂):

  φ = φ₂ - φ₁ (current w.r.t voltage)

  φ = φ₁ - φ₂ (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by ( 30° ), write current expression.
  • I = I_m sin (ωt - 30°)

Practice Set

• Q1: If ( ε = ε_m sin ωt ), current lags voltage by ( 45° ), current is?
  • A. I_m sin (ωt + 45°)
  • B. I_m sin (ωt - 45°)
  • C. I_m sin ωt
  • D. Zero
• Q2: Phase difference φ is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. ε = ε_m sin (ωt - φ)
  • B. I = I_m sin (ωt - φ)
  • C. I = I_m sin (ωt + φ)
  • D. ε = ε_m sin (ωt + φ)

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

• AC current with phase: ( I = I_m sin (ωt ± φ) )

• Initial phase ( φ ) is constant; instantaneous phase ( ωt ) changes with time.

• Phase difference between voltage ( ε = ε₀ sin (ωt + φ₁) ) and current ( I = I_m sin (ωt + φ₂) ):

  ( φ = φ₂ - φ₁ ) (current w.r.t voltage)

  ( φ = φ₁ - φ₂ ) (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by ( 30° ), write current expression.
  • ( I = I_m sin (ωt - 30°) )

Practice Set

• Q1: If ( ε = ε_m sin ωt ), current lags voltage by ( 45° ), current is?
  • A. ( I_m sin (ωt + 45°) )
  • B. ( I_m sin (ωt - 45°) )
  • C. ( I_m sin ωt )
  • D. Zero
• Q2: Phase difference ( φ ) is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. ( ε = ε_m sin (ωt - φ) )
  • B. ( I = I_m sin (ωt - φ) )
  • C. ( I = I_m sin (ωt + φ) )
  • D. ( ε = ε_m sin (ωt + φ) )

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

• AC current with phase: \( I = I_m \sin (\omega t \pm \phi) \)

• Initial phase \( \phi \) is constant; instantaneous phase \( \omega t \) changes with time.

• Phase difference between voltage \( \varepsilon = \varepsilon_0 \sin (\omega t + \phi_1) \) and current \( I = I_m \sin (\omega t + \phi_2) \):

  \( \phi = \phi_2 - \phi_1 \) (current w.r.t voltage)

  \( \phi = \phi_1 - \phi_2 \) (voltage w.r.t current)

Solved Examples

• Ex 1: If current lags voltage by \( 30^\circ \), write current expression.
  • \( I = I_m \sin (\omega t - 30^\circ) \)

Practice Set

• Q1: If \( \varepsilon = \varepsilon_m \sin \omega t \), current lags voltage by \( 45^\circ \), current is?
  • A. \( I_m \sin (\omega t + 45^\circ) \)
  • B. \( I_m \sin (\omega t - 45^\circ) \)
  • C. \( I_m \sin \omega t \)
  • D. Zero
• Q2: Phase difference \( \phi \) is?
  • A. Difference of initial phases
  • B. Sum of initial phases
  • C. Product of initial phases
  • D. Zero
• Q3: If voltage leads current, what is the relation?
  • A. \( \varepsilon = \varepsilon_m \sin (\omega t - \phi) \)
  • B. \( I = I_m \sin (\omega t - \phi) \)
  • C. \( I = I_m \sin (\omega t + \phi) \)
  • D. \( \varepsilon = \varepsilon_m \sin (\omega t + \phi) \)

Answer Key

• Q1: B - Current lags voltage by subtracting phase.
• Q2: A - Phase difference is difference of initial phases.
• Q3: A - Voltage leads current means voltage phase is ahead.

Memory Aids

• Lead means reaches max earlier; lag means reaches max later.
• Phase difference = difference of initial phases.

AC Voltage Applied to Resistor

• Voltage: ε = ε_m sin ωt

• Current: I = I_m sin ωt

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: P_r = I²R = I_m²R sin²ωt

Solved Examples

• Ex 1: Calculate average power if ( I_m = 5 ) A, ( R = 10 ) ( Ω ).
  • ( P = 1/2 I_m²R = 1/2 × 25 × 10 = 125 ) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. ( I_mR sin ωt )
  • B. ( I_m²R sin²ωt )
  • C. ( I_m²R cos ωt )
  • D. ( I_mR cos²ωt )
• Q3: Average power over cycle is?
  • A. ( I_m²R )
  • B. ( 1/2 I_m²R )
  • C. ( I_mR )
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is ( I_m²R sin²ωt ).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = ( 1/2 I_m²R ).
• Voltage and current peak together in resistor.

• Voltage: ε = ε_m sin ωt

• Current: I = I_m sin ωt

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: P_r = I²R = I_m²R sin²ωt

Solved Examples

• Ex 1: Calculate average power if ( I_m = 5 ) A, ( R = 10 ) ( Ω ).
  • ( P = 1/2 I_m²R = 1/2 × 25 × 10 = 125 ) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. ( I_mR sin ωt )
  • B. ( I_m²R sin²ωt )
  • C. ( I_m²R cos ωt )
  • D. ( I_mR cos²ωt )
• Q3: Average power over cycle is?
  • A. ( I_m²R )
  • B. ( 1/2 I_m²R )
  • C. ( I_mR )
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is ( I_m²R sin²ωt ).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = ( 1/2 I_m²R ).
• Voltage and current peak together in resistor.

• Voltage: ε = ε_m sin ωt

• Current: I = I_m sin ωt

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: P_r = I²R = I_m²R sin²ωt

Solved Examples

• Ex 1: Calculate average power if ( I_m = 5 ) A, ( R = 10 ) ( Ω ).
  • ( P = 1/2 I_m²R = 1/2 × 25 × 10 = 125 ) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. ( I_mR sin ωt )
  • B. ( I_m²R sin²ωt )
  • C. ( I_m²R cos ωt )
  • D. ( I_mR cos²ωt )
• Q3: Average power over cycle is?
  • A. ( I_m²R )
  • B. ( 1/2 I_m²R )
  • C. ( I_mR )
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is ( I_m²R sin²ωt ).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = ( 1/2 I_m²R ).
• Voltage and current peak together in resistor.

• Voltage: ( ε = ε_m sin ωt )

• Current: ( I = I_m sin ωt )

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: ( P_r = I²R = I_m²R sin²ωt )

Solved Examples

• Ex 1: Calculate average power if ( I_m = 5 ) A, ( R = 10 ) ( Ω ).
  • ( P = 1/2 I_m²R = 1/2 × 25 × 10 = 125 ) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. ( I_mR sin ωt )
  • B. ( I_m²R sin²ωt )
  • C. ( I_m²R cos ωt )
  • D. ( I_mR cos²ωt )
• Q3: Average power over cycle is?
  • A. ( I_m²R )
  • B. ( 1/2 I_m²R )
  • C. ( I_mR )
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is ( I_m²R sin²ωt ).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = ( 1/2 I_m²R ).
• Voltage and current peak together in resistor.

• Voltage: ε = ε_m sin ωt

• Current: I = I_m sin ωt

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: P_r = I²R = I_m²R sin²ωt

Solved Examples

• Ex 1: Calculate average power if ( I_m = 5 ) A, ( R = 10 ) ( Ω ).
  • ( P = 1/2 I_m²R = 1/2 × 25 × 10 = 125 ) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. ( I_mR sin ωt )
  • B. ( I_m²R sin²ωt )
  • C. ( I_m²R cos ωt )
  • D. ( I_mR cos²ωt )
• Q3: Average power over cycle is?
  • A. ( I_m²R )
  • B. ( 1/2 I_m²R )
  • C. ( I_mR )
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is ( I_m²R sin²ωt ).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = ( 1/2 I_m²R ).
• Voltage and current peak together in resistor.

• Voltage: ( ε = ε_m sin ωt )

• Current: ( I = I_m sin ωt )

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: ( P_r = I²R = I_m²R sin²ωt )

Solved Examples

• Ex 1: Calculate average power if ( I_m = 5 ) A, ( R = 10 ) ( Ω ).
  • ( P = 1/2 I_m²R = 1/2 × 25 × 10 = 125 ) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. ( I_mR sin ωt )
  • B. ( I_m²R sin²ωt )
  • C. ( I_m²R cos ωt )
  • D. ( I_mR cos²ωt )
• Q3: Average power over cycle is?
  • A. ( I_m²R )
  • B. ( 1/2 I_m²R )
  • C. ( I_mR )
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is ( I_m²R sin²ωt ).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = ( 1/2 I_m²R ).
• Voltage and current peak together in resistor.

• Voltage: \( \varepsilon = \varepsilon_m \sin \omega t \)

• Current: \( I = I_m \sin \omega t \)

• Current and voltage are in phase for resistor.

• Instantaneous power dissipated: \( P_r = I^2 R = I_m^2 R \sin^2 \omega t \)

Solved Examples

• Ex 1: Calculate average power if \( I_m = 5 \) A, \( R = 10 \) \( \Omega \).
  • \( P = \frac{1}{2} I_m^2 R = \frac{1}{2} \times 25 \times 10 = 125 \) W

Practice Set

• Q1: For resistor, current and voltage are?
  • A. Out of phase
  • B. In phase
  • C. Lead-lag
  • D. Zero
• Q2: Instantaneous power formula is?
  • A. \( I_m R \sin \omega t \)
  • B. \( I_m^2 R \sin^2 \omega t \)
  • C. \( I_m^2 R \cos \omega t \)
  • D. \( I_m R \cos^2 \omega t \)
• Q3: Average power over cycle is?
  • A. \( I_m^2 R \)
  • B. \( \frac{1}{2} I_m^2 R \)
  • C. \( I_m R \)
  • D. Zero

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Instantaneous power is \( I_m^2 R \sin^2 \omega t \).
• Q3: B - Average power is half of peak power.

Memory Aids

• Power average = \( \frac{1}{2} I_m^2 R \).
• Voltage and current peak together in resistor.

Representation of AC Current and Voltage by Rotating Vectors–Phasors

Concepts

• Phasors represent AC voltage and current as rotating vectors with angular velocity ω.
• Phasor length = peak value: ε_m for voltage, I_m for current.
• Vertical projection of phasor gives instantaneous value: ε = ε_m sin ωt, I = I_m sin ωt.
• Phasor diagram (a) shows voltage and current vectors; graph (b) shows sinusoidal variation vs time.
• When voltage and current phasors align, phase difference φ = 0; current and voltage are in phase.

Solved Examples

• Ex 1: Given ε_m = 10 V, find instantaneous voltage at ωt = 30°.
  • Use ε = ε_m sin ωt = 10 × sin 30° = 10 × 0.5 = 5 V.
• Ex 2: If current phasor lags voltage by 30° and I_m = 2 A, find instantaneous current at ωt = 60°.
  • I = I_m sin (ωt - 30°) = 2 × sin 30° = 2 × 0.5 = 1 A.

Practice Set

• Q1: What does the length of a phasor represent?
  • A) Instantaneous value
  • B) Peak value
  • C) Average value
  • D) RMS value
• Q2: If voltage and current phasors are in the same direction, the phase difference is:
  • A) 90°
  • B) 180°
  • C) 0°
  • D) 45°
• Q3: For a phasor ε = ε_m sin ωt, what is the vertical component at ωt = 90° if ε_m = 20 V?
  • A) 0 V
  • B) 10 V
  • C) 20 V
  • D) 40 V

Answer Key

• Q1: B) Peak value — Phasor length equals maximum amplitude.
• Q2: C) 0° — Same direction means no phase difference.
• Q3: C) 20 V — sin 90° = 1, so vertical component = peak value.

Memory Aids

• Phasor length = peak value.
• Vertical projection = instantaneous value.
• Same direction phasors → zero phase difference.

Concepts

• Phasors represent AC voltage and current as rotating vectors with angular velocity \( \omega \).
• Phasor length = peak value: \( \varepsilon_m \) for voltage, \( I_m \) for current.
• Vertical projection of phasor gives instantaneous value: \( \varepsilon = \varepsilon_m \sin \omega t \), \( I = I_m \sin \omega t \).
• Phasor diagram (a) shows voltage and current vectors; graph (b) shows sinusoidal variation vs time.
• When voltage and current phasors align, phase difference \( \phi = 0 \); current and voltage are in phase.

Solved Examples

• Ex 1: Given \( \varepsilon_m = 10\,V \), find instantaneous voltage at \( \omega t = 30^\circ \).
  • Use \( \varepsilon = \varepsilon_m \sin \omega t = 10 \times \sin 30^\circ = 10 \times 0.5 = 5\,V \).
• Ex 2: If current phasor lags voltage by \( 30^\circ \) and \( I_m = 2\,A \), find instantaneous current at \( \omega t = 60^\circ \).
  • \( I = I_m \sin (\omega t - 30^\circ) = 2 \times \sin 30^\circ = 2 \times 0.5 = 1\,A \).

Practice Set

• Q1: What does the length of a phasor represent?
  • A) Instantaneous value
  • B) Peak value
  • C) Average value
  • D) RMS value
• Q2: If voltage and current phasors are in the same direction, the phase difference is:
  • A) 90°
  • B) 180°
  • C) 0°
  • D) 45°
• Q3: For a phasor \( \varepsilon = \varepsilon_m \sin \omega t \), what is the vertical component at \( \omega t = 90^\circ \) if \( \varepsilon_m = 20\,V \)?
  • A) 0 V
  • B) 10 V
  • C) 20 V
  • D) 40 V

Answer Key

• Q1: B) Peak value — Phasor length equals maximum amplitude.
• Q2: C) 0° — Same direction means no phase difference.
• Q3: C) 20 V — \( \sin 90^\circ = 1 \), so vertical component = peak value.

Memory Aids

• Phasor length = peak value.
• Vertical projection = instantaneous value.
• Same direction phasors → zero phase difference.

AC Voltage Applied to an Inductor

Voltage and Current Relationship

• Voltage across inductor: ( ε = ε_m sin ωt )
• Current through inductor: ( I = I_m sin (ωt - π/2) )
• Current lags voltage by ( π/2 ) radians (90°)

Amplitude of Current

• ( I_m = ε_m/(ωL) )
• ( ω = 2πν ) (angular frequency)
• ( L ) is inductance in henrys (H)

Inductive Reactance

• Inductive reactance ( X_L = ωL = 2πνL )
• Acts like resistance in AC circuit but depends on frequency

Solved Examples

• Ex 1: Calculate current amplitude if ( ε_m = 10 V ), ( L = 0.5 H ), and ( ν = 50 Hz ).
  • ( ω = 2π × 50 = 314.16 rad/s )
  • ( I_m = 10/(314.16 × 0.5) = 0.0637 A )

Practice Set

• Q1: What is the phase difference between voltage and current in an inductor?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Inductive reactance ( X_L ) is given by?
  • A. ( 1/(ωL) )
  • B. ( ωL )
  • C. ( L/ω )
  • D. ( ω / L )
• Q3: For ( ε_m = 20 V ), ( L = 0.2 H ), and ( ν = 60 Hz ), find ( I_m ).
  • A. 0.26 A
  • B. 0.53 A
  • C. 1.67 A
  • D. 3.33 A

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90° in an inductor.
• Q2: B. ( ωL ) - Inductive reactance formula.
• Q3: A. 0.26 A - ( I_m = 20/(2π × 60 × 0.2) = 0.265 ) A approx.

Memory Aids

• Current lags voltage by 90° in inductors.
• Inductive reactance ( X_L = 2πνL ).
• Higher frequency → higher reactance → lower current.

Voltage and Current Relationship

• Voltage across inductor: ε = ε_m sin ωt
• Current through inductor: I = I_m sin (ωt - π/2)
• Current lags voltage by π/2 radians (90°)

Amplitude of Current

• I_m = ε_m/(ωL)
• ω = 2πν (angular frequency)
• L is inductance in henrys (H)

Inductive Reactance

• Inductive reactance X_L = ωL = 2πνL
• Acts like resistance in AC circuit but depends on frequency

Solved Examples

• Ex 1: Calculate current amplitude if ε_m = 10 V, L = 0.5 H, and ν = 50 Hz.
  • ω = 2π × 50 = 314.16 rad/s
  • I_m = 10/(314.16 × 0.5) = 0.0637 A

Practice Set

• Q1: What is the phase difference between voltage and current in an inductor?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Inductive reactance X_L is given by?
  • A. ( 1/(ωL) )
  • B. ( ωL )
  • C. ( L/ω )
  • D. ( ω / L )
• Q3: For ε_m = 20 V, L = 0.2 H, and ν = 60 Hz, find I_m.
  • A. 0.26 A
  • B. 0.53 A
  • C. 1.67 A
  • D. 3.33 A

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90° in an inductor.
• Q2: B. ( ωL ) - Inductive reactance formula.
• Q3: A. 0.26 A - I_m = 20/(2π × 60 × 0.2) = 0.265 A approx.

Memory Aids

• Current lags voltage by 90° in inductors.
• Inductive reactance X_L = 2πνL.
• Higher frequency → higher reactance → lower current.

Voltage and Current Relationship

• Voltage across inductor: ( ε = ε_m sin ωt )
• Current through inductor: ( I = I_m sin (ωt - π/2) )
• Current lags voltage by ( π/2 ) radians (90°)

Amplitude of Current

• ( I_m = ε_m/(ωL) )
• ( ω = 2πν ) (angular frequency)
• ( L ) is inductance in henrys (H)

Inductive Reactance

• Inductive reactance ( X_L = ωL = 2πνL )
• Acts like resistance in AC circuit but depends on frequency

Solved Examples

• Ex 1: Calculate current amplitude if ( ε_m = 10 V ), ( L = 0.5 H ), and ( ν = 50 Hz ).
  • ( ω = 2π × 50 = 314.16 rad/s )
  • ( I_m = 10/(314.16 × 0.5) = 0.0637 A )

Practice Set

• Q1: What is the phase difference between voltage and current in an inductor?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Inductive reactance ( X_L ) is given by?
  • A. ( 1/(ωL) )
  • B. ( ωL )
  • C. ( L/ω )
  • D. ( ω / L )
• Q3: For ( ε_m = 20 V ), ( L = 0.2 H ), and ( ν = 60 Hz ), find ( I_m ).
  • A. 0.26 A
  • B. 0.53 A
  • C. 1.67 A
  • D. 3.33 A

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90° in an inductor.
• Q2: B. ( ωL ) - Inductive reactance formula.
• Q3: A. 0.26 A - ( I_m = 20/(2π × 60 × 0.2) = 0.265 ) A approx.

Memory Aids

• Current lags voltage by 90° in inductors.
• Inductive reactance ( X_L = 2πνL ).
• Higher frequency → higher reactance → lower current.

Voltage and Current Relationship

• Voltage across inductor: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Current through inductor: \( I = I_m \sin (\omega t - \pi/2) \)
• Current lags voltage by \( \pi/2 \) radians (90°)

Amplitude of Current

• \( I_m = \frac{\varepsilon_m}{\omega L} \)
• \( \omega = 2 \pi \nu \) (angular frequency)
• \( L \) is inductance in henrys (H)

Inductive Reactance

• Inductive reactance \( X_L = \omega L = 2 \pi \nu L \)
• Acts like resistance in AC circuit but depends on frequency

Solved Examples

• Ex 1: Calculate current amplitude if \( \varepsilon_m = 10\,V \), \( L = 0.5\,H \), and \( \nu = 50\,Hz \).
  • \( \omega = 2 \pi \times 50 = 314.16\,rad/s \)
  • \( I_m = \frac{10}{314.16 \times 0.5} = 0.0637\,A \)

Practice Set

• Q1: What is the phase difference between voltage and current in an inductor?
  • A. 0
  • B. \( \pi/4 \)
  • C. \( \pi/2 \)
  • D. \( \pi \)
• Q2: Inductive reactance \( X_L \) is given by?
  • A. \( \frac{1}{\omega L} \)
  • B. \( \omega L \)
  • C. \( L/\omega \)
  • D. \( \omega / L \)
• Q3: For \( \varepsilon_m = 20\,V \), \( L = 0.2\,H \), and \( \nu = 60\,Hz \), find \( I_m \).
  • A. 0.26 A
  • B. 0.53 A
  • C. 1.67 A
  • D. 3.33 A

Answer Key

• Q1: C. \( \pi/2 \) - Current lags voltage by 90° in an inductor.
• Q2: B. \( \omega L \) - Inductive reactance formula.
• Q3: A. 0.26 A - \( I_m = \frac{20}{2 \pi \times 60 \times 0.2} = 0.265 \) A approx.

Memory Aids

• Current lags voltage by 90° in inductors.
• Inductive reactance \( X_L = 2 \pi \nu L \).
• Higher frequency → higher reactance → lower current.

Power in Inductive Load

Instantaneous Power

• ( p_l = Iε = - I_mε_m/2 sin (2ωt) )
• Power oscillates sinusoidally, can be positive or negative

Average Power

• ( P_l = ⟨p_l⟩ = 0 ) over one complete cycle
• Because ( ⟨sin (2ωt) ⟩ = 0 )
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for ( I_m = 5 A ), ( ε_m = 10 V ).
  • ( P_l = 0 ) (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. ( sin ωt )
  • B. ( sin 2ωt )
  • C. ( cos ωt )
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. ( sin 2ωt ) - Instantaneous power varies as ( sin 2ωt ).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Instantaneous Power

• p_l = Iε = - I_mε_m/2 sin (2ωt)
• Power oscillates sinusoidally, can be positive or negative

Average Power

• P_l = ⟨p_l⟩ = 0 over one complete cycle
• Because ⟨sin (2ωt)⟩ = 0
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for I_m = 5 A, ε_m = 10 V.
  • P_l = 0 (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. ( sin ωt )
  • B. ( sin 2ωt )
  • C. ( cos ωt )
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. ( sin 2ωt ) - Instantaneous power varies as ( sin 2ωt ).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Instantaneous Power

• ( p_l = Iε = - I_mε_m/2 sin (2ωt) )
• Power oscillates sinusoidally, can be positive or negative

Average Power

• ( P_l = ⟨p_l⟩ = 0 ) over one complete cycle
• Because ( ⟨sin (2ωt)⟩ = 0 )
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for ( I_m = 5 A ), ( ε_m = 10 V ).
  • ( P_l = 0 ) (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. ( sin ωt )
  • B. ( sin 2ωt )
  • C. ( cos ωt )
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. ( sin 2ωt ) - Instantaneous power varies as ( sin 2ωt ).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Instantaneous Power

• ( p_l = Iε = - I_mε_m/2 sin (2ωt) )
• Power oscillates sinusoidally, can be positive or negative

Average Power

• ( P_l = ⟨p_l⟩ = 0 ) over one complete cycle
• Because ( ⟨sin (2ωt) ⟩ = 0 )
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for ( I_m = 5 A ), ( ε_m = 10 V ).
  • ( P_l = 0 ) (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. ( sin ωt )
  • B. ( sin 2ωt )
  • C. ( cos ωt )
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. ( sin 2ωt ) - Instantaneous power varies as ( sin 2ωt ).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Instantaneous Power

• p_l = I ε = - I_mε_m/2 sin (2ωt)
• Power oscillates sinusoidally, can be positive or negative

Average Power

• P_l = ⟨p_l⟩ = 0 over one complete cycle
• Because ⟨sin (2ωt)⟩ = 0
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for I_m = 5 A, ε_m = 10 V.
  • P_l = 0 (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. ( sin ωt )
  • B. ( sin 2ωt )
  • C. ( cos ωt )
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. ( sin 2ωt ) - Instantaneous power varies as ( sin 2ωt ).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Instantaneous Power

• ( p_l = Iε = - I_mε_m/2 sin (2ωt) )
• Power oscillates sinusoidally, can be positive or negative

Average Power

• ( P_l = ⟨p_l⟩ = 0 ) over one complete cycle
• Because ( ⟨sin (2ωt)⟩ = 0 )
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for ( I_m = 5 A ), ( ε_m = 10 V ).
  • ( P_l = 0 ) (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. ( sin ωt )
  • B. ( sin 2ωt )
  • C. ( cos ωt )
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. ( sin 2ωt ) - Instantaneous power varies as ( sin 2ωt ).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Instantaneous Power

• \( p_l = I \varepsilon = - \frac{I_m \varepsilon_m}{2} \sin (2 \omega t) \)
• Power oscillates sinusoidally, can be positive or negative

Average Power

• \( P_l = \langle p_l \rangle = 0 \) over one complete cycle
• Because \( \langle \sin (2 \omega t) \rangle = 0 \)
• Inductor stores and returns energy, no net power consumed

Solved Examples

• Ex 1: Find average power for \( I_m = 5\,A \), \( \varepsilon_m = 10\,V \).
  • \( P_l = 0 \) (always zero for pure inductive load)

Practice Set

• Q1: What is the average power consumed by a pure inductor?
  • A. Zero
  • B. Maximum
  • C. Half of input power
  • D. Equal to resistive power
• Q2: Instantaneous power in an inductor is proportional to?
  • A. \( \sin \omega t \)
  • B. \( \sin 2 \omega t \)
  • C. \( \cos \omega t \)
  • D. Constant
• Q3: Why is average power zero in an inductor?
  • A. No current flows
  • B. Voltage and current are in phase
  • C. Energy is stored and returned
  • D. Resistance is zero

Answer Key

• Q1: A. Zero - Inductor does not dissipate power.
• Q2: B. \( \sin 2 \omega t \) - Instantaneous power varies as \( \sin 2 \omega t \).
• Q3: C. Energy is stored and returned - No net energy loss.

Memory Aids

• Average power in inductor = 0.
• Instantaneous power oscillates at double frequency.
• Inductor stores magnetic energy, returns it fully.

Phasor and Waveform Representation

Phasor Diagram

• Voltage vector leads current vector by ( π/2 ) radians
• Voltage represented as ( E_m sin(ωt) )
• Current represented as ( I_m sin(ωt - π/2) )

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ( ε ) and ( I ) versus ( ωt )

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by ( π/2 ).

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by π/2 radians
• Voltage represented as E_m sin(ωt)
• Current represented as I_m sin(ωt - π/2)

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ε and I versus ωt

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by π/2.

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. π/4
  • C. π/2
  • D. π
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. π/2 - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by ( π/2 ) radians
• Voltage represented as ( E_m sin(ωt) )
• Current represented as ( I_m sin(ωt - π/2) )

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ( ε ) and ( I ) versus ( ωt )

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by ( π/2 ).

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by π/2 radians
• Voltage represented as E_m sin(ωt)
• Current represented as I_m sin(ωt - π/2)

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ε and I versus ωt

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by π/2.

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. π/4
  • C. π/2
  • D. π
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. π/2 - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by ( π/2 ) radians
• Voltage represented as ( E_m sin(ωt) )
• Current represented as ( I_m sin(ωt - π/2) )

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ( ε ) and ( I ) versus ( ωt )

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by ( π/2 ).

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by π/2 radians
• Voltage represented as E_m sin(ωt)
• Current represented as I_m sin(ωt - π/2)

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ε and I versus ωt

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by π/2.

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. π/4
  • C. π/2
  • D. π
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. π/2 - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by ( π/2 ) radians
• Voltage represented as ( E_m sin(ωt) )
• Current represented as ( I_m sin(ωt - π/2) )

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots ( ε ) and ( I ) versus ( ωt )

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by ( π/2 ).

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. ( π/4 )
  • C. ( π/2 )
  • D. ( π )
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. ( π/2 ) - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

Phasor Diagram

• Voltage vector leads current vector by \( \pi/2 \) radians
• Voltage represented as \( E_m \sin(\omega t) \)
• Current represented as \( I_m \sin(\omega t - \pi/2) \)

Waveform Graphs

• Voltage and current are sinusoidal waves with phase difference
• Voltage wave peaks before current wave by 90°
• Graph plots \( \varepsilon \) and \( I \) versus \( \omega t \)

Solved Examples

• Ex 1: Identify phase difference from phasor diagram.
  • Voltage leads current by \( \pi/2 \).

Practice Set

• Q1: In phasor diagram, current lags voltage by?
  • A. 0
  • B. \( \pi/4 \)
  • C. \( \pi/2 \)
  • D. \( \pi \)
• Q2: Voltage and current waveforms differ by?
  • A. 0°
  • B. 45°
  • C. 90°
  • D. 180°
• Q3: Which vector leads in an inductive circuit?
  • A. Current
  • B. Voltage
  • C. Both same
  • D. None

Answer Key

• Q1: C. \( \pi/2 \) - Current lags voltage by 90°.
• Q2: C. 90° - Phase difference in inductive circuit.
• Q3: B. Voltage - Voltage leads current in inductors.

Memory Aids

• Voltage vector ahead of current by 90° in inductors.
• Phasor diagram shows phase relations clearly.
• Waveform peaks confirm phase lag of current.

AC Voltage Applied to a Series LCR Circuit

Concepts and Formulas

• Voltage applied: ( ε = ε_m sin ωt )
• Voltage equation: ( ε = L dI/dt + IR + q/C = ε_m sin ωt )
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ( ε_m² = I_m² [R² + (X_C - X_L)²] )
• Impedance: ( Z = √[R² + (X_C - X_L)²] )
• Current amplitude: ( I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²] )
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( X_C > X_L ), circuit is capacitive; current leads voltage.
• If ( X_C < X_L ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance (Z) if (R=10 Ω), (X_C=5 Ω), (X_L=3 Ω).
  • (Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω)
• Ex 2: Find current amplitude (I_m) if (ε_m=20 V), (Z=10 Ω).
  • (I_m = 20/10 = 2 A)

Practice Set

• Q1: What is the expression for impedance (Z) in a series LCR circuit?
  • A) (√[R² + (X_L + X_C)²])
  • B) (√[R² + (X_C - X_L)²])
  • C) (R + X_L + X_C)
  • D) (R/(X_C - X_L))
• Q2: If (X_C > X_L), what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For (R=5 Ω), (X_C=8 Ω), (X_L=3 Ω), and (ε_m=30 V), find (I_m).
  • A) 3 A
  • B) 4 A
  • C) 5 A
  • D) 6 A

Answer Key

• Q1: B - Impedance is (Z = √[R² + (X_C - X_L)²]) in series LCR circuit.
• Q2: B - If (X_C > X_L), current leads voltage (capacitive circuit).
• Q3: A - (Z=√[5²+(8-3)²]=√[25+25]=√50=7.07), (I_m=30/7.07=4.24 A) approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: (tan φ = (X_C - X_L)/R)
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: ε = ε_m sin ωt
• Voltage equation: ε = L dI/dt + IR + q/C = ε_m sin ωt
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ε_m² = I_m² [R² + (X_C - X_L)²]
• Impedance: Z = √[R² + (X_C - X_L)²]
• Current amplitude: I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²]
• Phase angle: tan φ = (X_C - X_L)/R
• If X_C > X_L, circuit is capacitive; current leads voltage.
• If X_C < X_L, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance Z if R=10 Ω, X_C=5 Ω, X_L=3 Ω.
  • Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω
• Ex 2: Find current amplitude I_m if ε_m=20 V, Z=10 Ω.
  • I_m = 20/10 = 2 A

Practice Set

• Q1: What is the expression for impedance Z in a series LCR circuit?
  • A) √[R² + (X_L + X_C)²]
  • B) √[R² + (X_C - X_L)²]
  • C) R + X_L + X_C
  • D) R/(X_C - X_L)
• Q2: If X_C > X_L, what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For R=5 Ω, X_C=8 Ω, X_L=3 Ω, and ε_m=30 V, find I_m.
• A) 3 A
• B) 4 A
• C) 5 A
• D) 6 A

Answer Key

• Q1: B - Impedance is Z = √[R² + (X_C - X_L)²] in series LCR circuit.
• Q2: B - If X_C > X_L, current leads voltage (capacitive circuit).
• Q3: A - Z=√[5²+(8-3)²]=√[25+25]=√50=7.07, I_m=30/7.07=4.24 A approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: tan φ = (X_C - X_L)/R
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: ( ε = ε_m sin ωt )
• Voltage equation: ( ε = L dI/dt + IR + q/C = ε_m sin ωt )
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ( ε_m² = I_m² [R² + (X_C - X_L)²] )
• Impedance: ( Z = √[R² + (X_C - X_L)²] )
• Current amplitude: ( I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²] )
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( X_C > X_L ), circuit is capacitive; current leads voltage.
• If ( X_C < X_L ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance (Z) if (R=10 Ω), (X_C=5 Ω), (X_L=3 Ω).
  • (Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω)
• Ex 2: Find current amplitude (I_m) if (ε_m=20 V), (Z=10 Ω).
  • (I_m = 20/10 = 2 A)

Practice Set

• Q1: What is the expression for impedance (Z) in a series LCR circuit?
  • A) (√[R² + (X_L + X_C)²])
  • B) (√[R² + (X_C - X_L)²])
  • C) (R + X_L + X_C)
  • D) (R/(X_C - X_L))
• Q2: If (X_C > X_L), what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For (R=5 Ω), (X_C=8 Ω), (X_L=3 Ω), and (ε_m=30 V), find (I_m).
  • A) 3 A
  • B) 4 A
  • C) 5 A
  • D) 6 A

Answer Key

• Q1: B - Impedance is (Z = √[R² + (X_C - X_L)²]) in series LCR circuit.
• Q2: B - If (X_C > X_L), current leads voltage (capacitive circuit).
• Q3: A - (Z=√[5²+(8-3)²]=√[25+25]=√50=7.07), (I_m=30/7.07=4.24 A) approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: (tan φ = (X_C - X_L)/R)
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: ε = ε_m sin ωt
• Voltage equation: ε = L dI/dt + IR + q/C = ε_m sin ωt
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ε_m² = I_m² [R² + (X_C - X_L)²]
• Impedance: Z = √[R² + (X_C - X_L)²]
• Current amplitude: I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²]
• Phase angle: tan φ = (X_C - X_L)/R
• If X_C > X_L, circuit is capacitive; current leads voltage.
• If X_C < X_L, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance Z if R=10 Ω, X_C=5 Ω, X_L=3 Ω.
  • Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω
• Ex 2: Find current amplitude I_m if ε_m=20 V, Z=10 Ω.
  • I_m = 20/10 = 2 A

Practice Set

• Q1: What is the expression for impedance Z in a series LCR circuit?
  • A) √[R² + (X_L + X_C)²]
  • B) √[R² + (X_C - X_L)²]
  • C) R + X_L + X_C
  • D) R/(X_C - X_L)
• Q2: If X_C > X_L, what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For R=5 Ω, X_C=8 Ω, X_L=3 Ω, and ε_m=30 V, find I_m.
• A) 3 A
• B) 4 A
• C) 5 A
• D) 6 A

Answer Key

• Q1: B - Impedance is Z = √[R² + (X_C - X_L)²] in series LCR circuit.
• Q2: B - If X_C > X_L, current leads voltage (capacitive circuit).
• Q3: A - Z=√[5²+(8-3)²]=√[25+25]=√50=7.07, I_m=30/7.07=4.24 A approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: tan φ = (X_C - X_L)/R
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: ( ε = ε_m sin ωt )
• Voltage equation: ( ε = L dI/dt + IR + q/C = ε_m sin ωt )
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ( ε_m² = I_m² [R² + (X_C - X_L)²] )
• Impedance: ( Z = √[R² + (X_C - X_L)²] )
• Current amplitude: ( I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²] )
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( X_C > X_L ), circuit is capacitive; current leads voltage.
• If ( X_C < X_L ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance (Z) if (R=10 Ω), (X_C=5 Ω), (X_L=3 Ω).
  • (Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω)
• Ex 2: Find current amplitude (I_m) if (ε_m=20 V), (Z=10 Ω).
  • (I_m = 20/10 = 2 A)

Practice Set

• Q1: What is the expression for impedance (Z) in a series LCR circuit?
  • A) (√[R² + (X_L + X_C)²])
  • B) (√[R² + (X_C - X_L)²])
  • C) (R + X_L + X_C)
  • D) (R/(X_C - X_L))
• Q2: If (X_C > X_L), what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For (R=5 Ω), (X_C=8 Ω), (X_L=3 Ω), and (ε_m=30 V), find (I_m).
  • A) 3 A
  • B) 4 A
  • C) 5 A
  • D) 6 A

Answer Key

• Q1: B - Impedance is (Z = √[R² + (X_C - X_L)²]) in series LCR circuit.
• Q2: B - If (X_C > X_L), current leads voltage (capacitive circuit).
• Q3: A - (Z=√[5²+(8-3)²]=√[25+25]=√50=7.07), (I_m=30/7.07=4.24 A) approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: (tan φ = (X_C - X_L)/R)
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: ε = ε_m sin ωt
• Voltage equation: ε = L dI/dt + IR + q/C = ε_m sin ωt
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ε_m² = I_m² [R² + (X_C - X_L)²]
• Impedance: Z = √[R² + (X_C - X_L)²]
• Current amplitude: I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²]
• Phase angle: tan φ = (X_C - X_L)/R
• If X_C > X_L, circuit is capacitive; current leads voltage.
• If X_C < X_L, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance Z if R=10 Ω, X_C=5 Ω, X_L=3 Ω.
  • Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω
• Ex 2: Find current amplitude I_m if ε_m=20 V, Z=10 Ω.
  • I_m = 20/10 = 2 A

Practice Set

• Q1: What is the expression for impedance Z in a series LCR circuit?
  • A) √[R² + (X_L + X_C)²]
  • B) √[R² + (X_C - X_L)²]
  • C) R + X_L + X_C
  • D) R/(X_C - X_L)
• Q2: If X_C > X_L, what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For R=5 Ω, X_C=8 Ω, X_L=3 Ω, and ε_m=30 V, find I_m.
  • A) 3 A
  • B) 4 A
  • C) 5 A
  • D) 6 A

Answer Key

• Q1: B - Impedance is Z = √[R² + (X_C - X_L)²] in series LCR circuit.
• Q2: B - If X_C > X_L, current leads voltage (capacitive circuit).
• Q3: A - Z=√[5²+(8-3)²]=√[25+25]=√50=7.07, I_m=30/7.07=4.24 A approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: tan φ = (X_C - X_L)/R
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: ( ε = ε_m sin ωt )
• Voltage equation: ( ε = L dI/dt + IR + q/C = ε_m sin ωt )
• Voltages across capacitor V_C and inductor V_L are opposite in direction.
• Total voltage squared: ( ε_m² = I_m² [R² + (X_C - X_L)²] )
• Impedance: ( Z = √[R² + (X_C - X_L)²] )
• Current amplitude: ( I_m = ε_m/Z = ε_m/√[R² + (X_C - X_L)²] )
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( X_C > X_L ), circuit is capacitive; current leads voltage.
• If ( X_C < X_L ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance (Z) if (R=10 Ω), (X_C=5 Ω), (X_L=3 Ω).
  • (Z = √[10² + (5-3)²] = √[100 + 4] = √104 = 10.2 Ω)
• Ex 2: Find current amplitude (I_m) if (ε_m=20 V), (Z=10 Ω).
  • (I_m = 20/10 = 2 A)

Practice Set

• Q1: What is the expression for impedance (Z) in a series LCR circuit?
  • A) (√[R² + (X_L + X_C)²])
  • B) (√[R² + (X_C - X_L)²])
  • C) (R + X_L + X_C)
  • D) (R/(X_C - X_L))
• Q2: If (X_C > X_L), what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For (R=5 Ω), (X_C=8 Ω), (X_L=3 Ω), and (ε_m=30 V), find (I_m).
  • A) 3 A
  • B) 4 A
  • C) 5 A
  • D) 6 A

Answer Key

• Q1: B - Impedance is (Z = √[R² + (X_C - X_L)²]) in series LCR circuit.
• Q2: B - If (X_C > X_L), current leads voltage (capacitive circuit).
• Q3: A - (Z=√[5²+(8-3)²]=√[25+25]=√50=7.07), (I_m=30/7.07=4.24 A) approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: (tan φ = (X_C - X_L)/R)
• Current leads if capacitive, lags if inductive

Concepts and Formulas

• Voltage applied: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Voltage equation: \( \varepsilon = L \frac{dI}{dt} + IR + \frac{q}{C} = \varepsilon_m \sin \omega t \)
• Voltages across capacitor \(V_C\) and inductor \(V_L\) are opposite in direction.
• Total voltage squared: \( \varepsilon_m^2 = I_m^2 \left[R^2 + (X_C - X_L)^2 \right] \)
• Impedance: \( Z = \sqrt{R^2 + (X_C - X_L)^2} \)
• Current amplitude: \( I_m = \frac{\varepsilon_m}{Z} = \frac{\varepsilon_m}{\sqrt{R^2 + (X_C - X_L)^2}} \)
• Phase angle: \( \tan \phi = \frac{X_C - X_L}{R} \)
• If \( X_C > X_L \), circuit is capacitive; current leads voltage.
• If \( X_C < X_L \), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Calculate impedance \(Z\) if \(R=10\ \Omega\), \(X_C=5\ \Omega\), \(X_L=3\ \Omega\).
  • \(Z = \sqrt{10^2 + (5-3)^2} = \sqrt{100 + 4} = \sqrt{104} = 10.2\ \Omega\)
• Ex 2: Find current amplitude \(I_m\) if \(\varepsilon_m=20\ V\), \(Z=10\ \Omega\).
  • \(I_m = \frac{20}{10} = 2\ A\)

Practice Set

• Q1: What is the expression for impedance \(Z\) in a series LCR circuit?
  • A) \(\sqrt{R^2 + (X_L + X_C)^2}\)
  • B) \(\sqrt{R^2 + (X_C - X_L)^2}\)
  • C) \(R + X_L + X_C\)
  • D) \(\frac{R}{X_C - X_L}\)
• Q2: If \(X_C > X_L\), what is the phase relationship between current and voltage?
  • A) Current lags voltage
  • B) Current leads voltage
  • C) Current and voltage are in phase
  • D) No current flows
• Q3: For \(R=5\ \Omega\), \(X_C=8\ \Omega\), \(X_L=3\ \Omega\), and \(\varepsilon_m=30\ V\), find \(I_m\).
  • A) 3 A
  • B) 4 A
  • C) 5 A
  • D) 6 A

Answer Key

• Q1: B - Impedance is \(Z = \sqrt{R^2 + (X_C - X_L)^2}\) in series LCR circuit.
• Q2: B - If \(X_C > X_L\), current leads voltage (capacitive circuit).
• Q3: A - \(Z=\sqrt{5^2+(8-3)^2}=\sqrt{25+25}=\sqrt{50}=7.07\), \(I_m=30/7.07=4.24\ A\) approx 4 A (closest option is 4 A).

Memory Aids

• Impedance formula: "R and difference of reactances squared under root"
• Phase angle: \(\tan \phi = \frac{X_C - X_L}{R}\)
• Current leads if capacitive, lags if inductive

Phasor and Phase Angle

Concepts and Formulas

• Phasor diagram shows voltage (ε) and current (I) vectors with angle (φ).
• Right triangle with sides: (R), (X_C - X_L), hypotenuse (Z).
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( φ > 0 ), circuit is capacitive; current leads voltage.
• If ( φ < 0 ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find (φ) if (R=4 Ω), (X_C=6 Ω), (X_L=2 Ω).
  • (tan φ = (6-2)/4 = 1), (φ = 45°)
• Ex 2: Determine if current leads or lags voltage for (X_C=3 Ω), (X_L=5 Ω).
  • Since (X_C < X_L), current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle (φ) indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance (R)
  • B) Capacitive reactance (X_C)
  • C) Inductive reactance (X_L)
  • D) Impedance (Z)
• Q3: If (tan φ = 0), what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive (φ) means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance (Z).
• Q3: A - (tan φ = 0) means no reactance, purely resistive.

Memory Aids

• Phase angle (φ): positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = (Z), base = (R), height = (X_C - X_L)

Concepts and Formulas

• Phasor diagram shows voltage ε and current I vectors with angle φ.
• Right triangle with sides: R, X_C - X_L, hypotenuse Z.
• Phase angle: tan φ = (X_C - X_L)/R
• If φ > 0, circuit is capacitive; current leads voltage.
• If φ < 0, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find φ if R=4 Ω, X_C=6 Ω, X_L=2 Ω.
  • tan φ = (6-2)/4 = 1, φ = 45°
• Ex 2: Determine if current leads or lags voltage for X_C=3 Ω, X_L=5 Ω.
  • Since X_C < X_L, current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle φ indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance R
  • B) Capacitive reactance X_C
  • C) Inductive reactance X_L
  • D) Impedance Z
• Q3: If tan φ = 0, what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive φ means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance Z.
• Q3: A - tan φ = 0 means no reactance, purely resistive.

Memory Aids

• Phase angle φ: positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = Z, base = R, height = X_C - X_L

Concepts and Formulas

• Phasor diagram shows voltage (ε) and current (I) vectors with angle (φ).
• Right triangle with sides: (R), (X_C - X_L), hypotenuse (Z).
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( φ > 0 ), circuit is capacitive; current leads voltage.
• If ( φ < 0 ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find (φ) if (R=4 Ω), (X_C=6 Ω), (X_L=2 Ω).
  • (tan φ = (6-2)/4 = 1), (φ = 45°)
• Ex 2: Determine if current leads or lags voltage for (X_C=3 Ω), (X_L=5 Ω).
  • Since (X_C < X_L), current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle (φ) indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance (R)
  • B) Capacitive reactance (X_C)
  • C) Inductive reactance (X_L)
  • D) Impedance (Z)
• Q3: If (tan φ = 0), what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive (φ) means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance (Z).
• Q3: A - (tan φ = 0) means no reactance, purely resistive.

Memory Aids

• Phase angle (φ): positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = (Z), base = (R), height = (X_C - X_L)

Concepts and Formulas

• Phasor diagram shows voltage and current vectors with angle .
• Right triangle with sides: , , hypotenuse .
• Phase angle: =
• If 0, circuit is capacitive; current leads voltage.
• If 0, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find if =4 , =6 , =2 .
  • =
• Ex 2: Determine if current leads or lags voltage for =3 , =5 .
  • Since < , current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance
  • B) Capacitive reactance
  • C) Inductive reactance
  • D) Impedance
• Q3: If 0, what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance .
• Q3: A - 0 means no reactance, purely resistive.

Memory Aids

• Phase angle : positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = , base = , height = -

Concepts and Formulas

• Phasor diagram shows voltage ε and current I vectors with angle φ.
• Right triangle with sides: R, X_C - X_L, hypotenuse Z.
• Phase angle: tan φ = (X_C - X_L)/R
• If φ > 0, circuit is capacitive; current leads voltage.
• If φ < 0, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find φ if R=4 Ω, X_C=6 Ω, X_L=2 Ω.
  • tan φ = (6-2)/4 = 1, φ = 45°
• Ex 2: Determine if current leads or lags voltage for X_C=3 Ω, X_L=5 Ω.
  • Since X_C < X_L, current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle φ indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance R
  • B) Capacitive reactance X_C
  • C) Inductive reactance X_L
  • D) Impedance Z
• Q3: If tan φ = 0, what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive φ means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance Z.
• Q3: A - tan φ = 0 means no reactance, purely resistive.

Memory Aids

• Phase angle φ: positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = Z, base = R, height = X_C - X_L

Concepts and Formulas

• Phasor diagram shows voltage (ε) and current (I) vectors with angle (φ).
• Right triangle with sides: (R), (X_C - X_L), hypotenuse (Z).
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( φ > 0 ), circuit is capacitive; current leads voltage.
• If ( φ < 0 ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find (φ) if (R=4 Ω), (X_C=6 Ω), (X_L=2 Ω).
  • (tan φ = (6-2)/4 = 1), (φ = 45°)
• Ex 2: Determine if current leads or lags voltage for (X_C=3 Ω), (X_L=5 Ω).
  • Since (X_C < X_L), current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle (φ) indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance (R)
  • B) Capacitive reactance (X_C)
  • C) Inductive reactance (X_L)
  • D) Impedance (Z)
• Q3: If (tan φ = 0), what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive (φ) means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance (Z).
• Q3: A - (tan φ = 0) means no reactance, purely resistive.

Memory Aids

• Phase angle (φ): positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = (Z), base = (R), height = (X_C - X_L)

Concepts and Formulas

• Phasor diagram shows voltage (ε) and current (I) vectors with angle (φ).
• Right triangle with sides: (R), (X_C - X_L), hypotenuse (Z).
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( φ > 0 ), circuit is capacitive; current leads voltage.
• If ( φ < 0 ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find (φ) if (R=4 Ω), (X_C=6 Ω), (X_L=2 Ω).
  • (tan φ = (6-2)/4 = 1), (φ = 45°)
• Ex 2: Determine if current leads or lags voltage for (X_C=3 Ω), (X_L=5 Ω).
  • Since (X_C < X_L), current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle (φ) indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance (R)
  • B) Capacitive reactance (X_C)
  • C) Inductive reactance (X_L)
  • D) Impedance (Z)
• Q3: If (tan φ = 0), what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive (φ) means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance (Z).
• Q3: A - (tan φ = 0) means no reactance, purely resistive.

Memory Aids

• Phase angle (φ): positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = (Z), base = (R), height = (X_C - X_L)

Concepts and Formulas

• Phasor diagram shows voltage ε and current I vectors with angle φ.
• Right triangle with sides: R, X_C - X_L, hypotenuse Z.
• Phase angle: tan φ = (X_C - X_L)/R
• If φ > 0, circuit is capacitive; current leads voltage.
• If φ < 0, circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find φ if R=4 Ω, X_C=6 Ω, X_L=2 Ω.
  • tan φ = (6-2)/4 = 1, φ = 45°
• Ex 2: Determine if current leads or lags voltage for X_C=3 Ω, X_L=5 Ω.
  • Since X_C < X_L, current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle φ indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance R
  • B) Capacitive reactance X_C
  • C) Inductive reactance X_L
  • D) Impedance Z
• Q3: If tan φ = 0, what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive φ means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance Z.
• Q3: A - tan φ = 0 means no reactance, purely resistive.

Memory Aids

• Phase angle φ: positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = Z, base = R, height = X_C - X_L

Concepts and Formulas

• Phasor diagram shows voltage (ε) and current (I) vectors with angle (φ).
• Right triangle with sides: (R), (X_C - X_L), hypotenuse (Z).
• Phase angle: ( tan φ = (X_C - X_L)/R )
• If ( φ > 0 ), circuit is capacitive; current leads voltage.
• If ( φ < 0 ), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find (φ) if (R=4 Ω), (X_C=6 Ω), (X_L=2 Ω).
  • (tan φ = (6-2)/4 = 1), (φ = 45°)
• Ex 2: Determine if current leads or lags voltage for (X_C=3 Ω), (X_L=5 Ω).
  • Since (X_C < X_L), current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle (φ) indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance (R)
  • B) Capacitive reactance (X_C)
  • C) Inductive reactance (X_L)
  • D) Impedance (Z)
• Q3: If (tan φ = 0), what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive (φ) means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance (Z).
• Q3: A - (tan φ = 0) means no reactance, purely resistive.

Memory Aids

• Phase angle (φ): positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = (Z), base = (R), height = (X_C - X_L)

Concepts and Formulas

• Phasor diagram shows voltage \(\varepsilon\) and current \(I\) vectors with angle \(\phi\).
• Right triangle with sides: \(R\), \(X_C - X_L\), hypotenuse \(Z\).
• Phase angle: \( \tan \phi = \frac{X_C - X_L}{R} \)
• If \( \phi > 0 \), circuit is capacitive; current leads voltage.
• If \( \phi < 0 \), circuit is inductive; current lags voltage.

Solved Examples

• Ex 1: Find \(\phi\) if \(R=4\ \Omega\), \(X_C=6\ \Omega\), \(X_L=2\ \Omega\).
  • \(\tan \phi = \frac{6-2}{4} = 1\), \(\phi = 45^\circ\)
• Ex 2: Determine if current leads or lags voltage for \(X_C=3\ \Omega\), \(X_L=5\ \Omega\).
  • Since \(X_C < X_L\), current lags voltage (inductive).

Practice Set

• Q1: What does a positive phase angle \(\phi\) indicate?
  • A) Inductive circuit
  • B) Capacitive circuit
  • C) Resistive circuit
  • D) No current
• Q2: In the phasor diagram, what is the hypotenuse representing?
  • A) Resistance \(R\)
  • B) Capacitive reactance \(X_C\)
  • C) Inductive reactance \(X_L\)
  • D) Impedance \(Z\)
• Q3: If \(\tan \phi = 0\), what is the circuit type?
  • A) Purely resistive
  • B) Purely inductive
  • C) Purely capacitive
  • D) Resonant

Answer Key

• Q1: B - Positive \(\phi\) means capacitive circuit.
• Q2: D - Hypotenuse in impedance triangle is impedance \(Z\).
• Q3: A - \(\tan \phi = 0\) means no reactance, purely resistive.

Memory Aids

• Phase angle \(\phi\): positive = capacitor, negative = inductor
• Impedance triangle: hypotenuse = \(Z\), base = \(R\), height = \(X_C - X_L\)

Resonance

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: ( X_L = X_C )
• Current amplitude at resonance: ( I_m = ε_m/R )
• Resonant frequency:
  • Angular frequency: ( ω₀ = 1/√LC )
  • Frequency: ( ν_r = 1/(2π√LC) )
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when (X_L = X_C).
• Q2: B - At resonance, (I_m = ε_m/R).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ( ω₀ = 1/√LC )
• Current at resonance: ( I_m = ε_m/R )

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: X_L = X_C
• Current amplitude at resonance: I_m = ε_m/R
• Resonant frequency:
  • Angular frequency: ω₀ = 1/√LC
  • Frequency: ν_r = 1/(2π√LC)
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when X_L = X_C.
• Q2: B - At resonance, I_m = ε_m/R.
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ω₀ = 1/√LC
• Current at resonance: I_m = ε_m/R

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: ( X_L = X_C )
• Current amplitude at resonance: ( I_m = ε_m/R )
• Resonant frequency:
  • Angular frequency: ( ω₀ = 1/√LC )
  • Frequency: ( ν_r = 1/(2π√LC) )
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when (X_L = X_C).
• Q2: B - At resonance, (I_m = ε_m/R).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ( ω₀ = 1/√LC )
• Current at resonance: ( I_m = ε_m/R )

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: X_L = X_C
• Current amplitude at resonance: I_m = ε_m/R
• Resonant frequency:
  • Angular frequency: ω₀ = 1/√LC
  • Frequency: ν_r = 1/(2π√LC)
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when (X_L = X_C).
• Q2: B - At resonance, (I_m = ε_m/R).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ( ω₀ = 1/√LC )
• Current at resonance: ( I_m = ε_m/R )

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: \( X_L = X_C \)
• Current amplitude at resonance: \( I_m = \frac{\varepsilon_m}{R} \)
• Resonant frequency:
  • Angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
  • Frequency: \( \nu_r = \frac{1}{2\pi \sqrt{LC}} \)
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for \(L=1\ \text{mH}\), \(C=1\ \text{nF}\).
  • \( \omega_0 = \frac{1}{\sqrt{1 \times 10^{-3} \times 1 \times 10^{-9}}} = 10^6\ \text{rad/s} \)
• Ex 2: Find current amplitude at resonance if \(\varepsilon_m=100\ V\), \(R=100\ \Omega\).
  • \( I_m = \frac{100}{100} = 1\ A \)

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) \(X_L + X_C = 0\)
  • B) \(X_L = X_C\)
  • C) \(R = 0\)
  • D) \(I_m = 0\)
• Q2: At resonance, the current amplitude is given by?
  • A) \(\frac{\varepsilon_m}{Z}\)
  • B) \(\frac{\varepsilon_m}{R}\)
  • C) \(\frac{\varepsilon_m}{X_L}\)
  • D) \(\frac{\varepsilon_m}{X_C}\)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when \(X_L = X_C\).
• Q2: B - At resonance, \(I_m = \frac{\varepsilon_m}{R}\).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
• Current at resonance: \( I_m = \frac{\varepsilon_m}{R} \)

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: ( X_L = X_C )
• Current amplitude at resonance: ( I_m = ε_m/R )
• Resonant frequency:
  • Angular frequency: ( ω₀ = 1/√LC )
  • Frequency: ( ν_r = 1/(2π√LC) )
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when (X_L = X_C).
• Q2: B - At resonance, (I_m = ε_m/R).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ( ω₀ = 1/√LC )
• Current at resonance: ( I_m = ε_m/R )

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: X_L = X_C
• Current amplitude at resonance: I_m = ε_m/R
• Resonant frequency:
  • Angular frequency: ω₀ = 1/√LC
  • Frequency: ν_r = 1/(2π√LC)
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when X_L = X_C.
• Q2: B - At resonance, I_m = ε_m/R.
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ω₀ = 1/√LC
• Current at resonance: I_m = ε_m/R

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: ( X_L = X_C )
• Current amplitude at resonance: ( I_m = ε_m/R )
• Resonant frequency:
  • Angular frequency: ( ω₀ = 1/√LC )
  • Frequency: ( ν_r = 1/(2π√LC) )
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for (L=1 mH), (C=1 nF).
  • ( ω₀ = 1/√[1 × 10^-3 × 1 × 10^-9] = 10⁶ rad/s )
• Ex 2: Find current amplitude at resonance if (ε_m=100 V), (R=100 Ω).
  • ( I_m = 100/100 = 1 A )

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) (X_L + X_C = 0)
  • B) (X_L = X_C)
  • C) (R = 0)
  • D) (I_m = 0)
• Q2: At resonance, the current amplitude is given by?
  • A) (ε_m/Z)
  • B) (ε_m/R)
  • C) (ε_m/X_L)
  • D) (ε_m/X_C)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when (X_L = X_C).
• Q2: B - At resonance, (I_m = ε_m/R).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: ( ω₀ = 1/√LC )
• Current at resonance: ( I_m = ε_m/R )

Concepts and Formulas

• Resonance: minimum impedance, maximum current in series LCR circuit.
• Condition: \( X_L = X_C \)
• Current amplitude at resonance: \( I_m = \frac{\varepsilon_m}{R} \)
• Resonant frequency:
  • Angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
  • Frequency: \( \nu_r = \frac{1}{2\pi \sqrt{LC}} \)
• Resonance does not occur in RL or RC circuits.

Solved Examples

• Ex 1: Calculate resonant frequency for \(L=1\ \text{mH}\), \(C=1\ \text{nF}\).
  • \( \omega_0 = \frac{1}{\sqrt{1 \times 10^{-3} \times 1 \times 10^{-9}}} = 10^6\ \text{rad/s} \)
• Ex 2: Find current amplitude at resonance if \(\varepsilon_m=100\ V\), \(R=100\ \Omega\).
  • \( I_m = \frac{100}{100} = 1\ A \)

Practice Set

• Q1: What is the resonance condition in a series LCR circuit?
  • A) \(X_L + X_C = 0\)
  • B) \(X_L = X_C\)
  • C) \(R = 0\)
  • D) \(I_m = 0\)
• Q2: At resonance, the current amplitude is given by?
  • A) \(\frac{\varepsilon_m}{Z}\)
  • B) \(\frac{\varepsilon_m}{R}\)
  • C) \(\frac{\varepsilon_m}{X_L}\)
  • D) \(\frac{\varepsilon_m}{X_C}\)
• Q3: Resonance does not occur in which circuit?
  • A) Series LCR
  • B) RL
  • C) LC
  • D) Parallel LCR

Answer Key

• Q1: B - Resonance occurs when \(X_L = X_C\).
• Q2: B - At resonance, \(I_m = \frac{\varepsilon_m}{R}\).
• Q3: B - Resonance does not occur in RL or RC circuits.

Memory Aids

• Resonance: minimum impedance, max current
• Resonant frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
• Current at resonance: \( I_m = \frac{\varepsilon_m}{R} \)

Quality Factor

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • ( Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C] )
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if (ω₀=10⁶ rad/s), (L=1 mH), (R=100 Ω).
  • ( Q = (10⁶ × 1 × 10^-3)/100 = 10 )
• Ex 2: Find Q if (L=1 mH), (C=1 nF), (R=200 Ω).
  • ( Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5 )

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) (Q = R/(ω₀L))
  • B) (Q = ω₀L/R)
  • C) (Q = ω₀CR)
  • D) (Q = R √[L/C])
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for (L=2 mH), (C=0.5 nF), (R=50 Ω).
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - (Q = ω₀L/R) is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - (Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40).

Memory Aids

• Q = (ω₀L/R) or (1/(ω₀CR))
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C]
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if ω₀=10⁶ rad/s, L=1 mH, R=100 Ω.
  • Q = (10⁶ × 1 × 10^-3)/100 = 10
• Ex 2: Find Q if L=1 mH, C=1 nF, R=200 Ω.
  • Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) Q = R/(ω₀L)
  • B) Q = ω₀L/R
  • C) Q = ω₀CR
  • D) Q = R √[L/C]
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for L=2 mH, C=0.5 nF, R=50 Ω.
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - Q = ω₀L/R is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40.

Memory Aids

• Q = ω₀L/R or 1/(ω₀CR)
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • ( Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C] )
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if (ω₀=10⁶ rad/s), (L=1 mH), (R=100 Ω).
  • ( Q = (10⁶ × 1 × 10^-3)/100 = 10 )
• Ex 2: Find Q if (L=1 mH), (C=1 nF), (R=200 Ω).
  • ( Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5 )

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) (Q = R/(ω₀L))
  • B) (Q = ω₀L/R)
  • C) (Q = ω₀CR)
  • D) (Q = R √[L/C])
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for (L=2 mH), (C=0.5 nF), (R=50 Ω).
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - (Q = ω₀L/R) is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - (Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40).

Memory Aids

• Q = (ω₀L/R) or (1/(ω₀CR))
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C]
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if ω₀=10⁶ rad/s, L=1 mH, R=100 Ω.
  • Q = (10⁶ × 1 × 10^-3)/100 = 10
• Ex 2: Find Q if L=1 mH, C=1 nF, R=200 Ω.
  • Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) Q = R/(ω₀L)
  • B) Q = ω₀L/R
  • C) Q = ω₀CR
  • D) Q = R √[L/C]
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for L=2 mH, C=0.5 nF, R=50 Ω.
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - Q = ω₀L/R is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40.

Memory Aids

• Q = ω₀L/R or 1/(ω₀CR)
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • \( Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R} \sqrt{\frac{L}{C}} \)
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if \(\omega_0=10^6\ \text{rad/s}\), \(L=1\ \text{mH}\), \(R=100\ \Omega\).
  • \( Q = \frac{10^6 \times 1 \times 10^{-3}}{100} = 10 \)
• Ex 2: Find Q if \(L=1\ \text{mH}\), \(C=1\ \text{nF}\), \(R=200\ \Omega\).
  • \( Q = \frac{1}{200} \sqrt{\frac{1 \times 10^{-3}}{1 \times 10^{-9}}} = \frac{1}{200} \times 1000 = 5 \)

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) \(Q = \frac{R}{\omega_0 L}\)
  • B) \(Q = \frac{\omega_0 L}{R}\)
  • C) \(Q = \omega_0 C R\)
  • D) \(Q = R \sqrt{\frac{L}{C}}\)
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for \(L=2\ \text{mH}\), \(C=0.5\ \text{nF}\), \(R=50\ \Omega\).
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - \(Q = \frac{\omega_0 L}{R}\) is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - \(Q = \frac{1}{50} \sqrt{\frac{2 \times 10^{-3}}{0.5 \times 10^{-9}}} = \frac{1}{50} \times 2000 = 40\).

Memory Aids

• Q = \(\frac{\omega_0 L}{R}\) or \(\frac{1}{\omega_0 C R}\)
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • ( Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C] )
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if (ω₀=10⁶ rad/s), (L=1 mH), (R=100 Ω).
  • ( Q = (10⁶ × 1 × 10^-3)/100 = 10 )
• Ex 2: Find Q if (L=1 mH), (C=1 nF), (R=200 Ω).
  • ( Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5 )

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) (Q = R/(ω₀L))
  • B) (Q = ω₀L/R)
  • C) (Q = ω₀CR)
  • D) (Q = R √[L/C])
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for (L=2 mH), (C=0.5 nF), (R=50 Ω).
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - (Q = ω₀L/R) is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - (Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40).

Memory Aids

• Q = (ω₀L/R) or (1/(ω₀CR))
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C]
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if ω₀=10⁶ rad/s, L=1 mH, R=100 Ω.
  • Q = (10⁶ × 1 × 10^-3)/100 = 10
• Ex 2: Find Q if L=1 mH, C=1 nF, R=200 Ω.
  • Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) Q = R/(ω₀L)
  • B) Q = ω₀L/R
  • C) Q = ω₀CR
  • D) Q = R √[L/C]
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for L=2 mH, C=0.5 nF, R=50 Ω.
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - Q = ω₀L/R is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40.

Memory Aids

• Q = ω₀L/R or 1/(ω₀CR)
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • ( Q = ω₀L/R = 1/(ω₀CR) = 1/R √[L/C] )
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if (ω₀=10⁶ rad/s), (L=1 mH), (R=100 Ω).
  • ( Q = (10⁶ × 1 × 10^-3)/100 = 10 )
• Ex 2: Find Q if (L=1 mH), (C=1 nF), (R=200 Ω).
  • ( Q = 1/200 √[(1 × 10^-3)/(1 × 10^-9)] = 1/200 × 1000 = 5 )

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) (Q = R/(ω₀L))
  • B) (Q = ω₀L/R)
  • C) (Q = ω₀CR)
  • D) (Q = R √[L/C])
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for (L=2 mH), (C=0.5 nF), (R=50 Ω).
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - (Q = ω₀L/R) is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - (Q = 1/50 √[(2 × 10^-3)/(0.5 × 10^-9)] = 1/50 × 2000 = 40).

Memory Aids

• Q = (ω₀L/R) or (1/(ω₀CR))
• High Q = sharp resonance peak

Concepts and Formulas

• Quality factor (Q) = ratio of voltage across L or C to voltage across R at resonance.
• Formulas:
  • \( Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R} \sqrt{\frac{L}{C}} \)
• Higher Q means better tuning of the circuit.

Solved Examples

• Ex 1: Calculate Q if \(\omega_0=10^6\ \text{rad/s}\), \(L=1\ \text{mH}\), \(R=100\ \Omega\).
  • \( Q = \frac{10^6 \times 1 \times 10^{-3}}{100} = 10 \)
• Ex 2: Find Q if \(L=1\ \text{mH}\), \(C=1\ \text{nF}\), \(R=200\ \Omega\).
  • \( Q = \frac{1}{200} \sqrt{\frac{1 \times 10^{-3}}{1 \times 10^{-9}}} = \frac{1}{200} \times 1000 = 5 \)

Practice Set

• Q1: Which formula represents quality factor Q?
  • A) \(Q = \frac{R}{\omega_0 L}\)
  • B) \(Q = \frac{\omega_0 L}{R}\)
  • C) \(Q = \omega_0 C R\)
  • D) \(Q = R \sqrt{\frac{L}{C}}\)
• Q2: What does a higher Q factor indicate?
  • A) Poor tuning
  • B) Better tuning
  • C) No resonance
  • D) No current
• Q3: Calculate Q for \(L=2\ \text{mH}\), \(C=0.5\ \text{nF}\), \(R=50\ \Omega\).
  • A) 20
  • B) 40
  • C) 10
  • D) 5

Answer Key

• Q1: B - \(Q = \frac{\omega_0 L}{R}\) is the correct formula.
• Q2: B - Higher Q means better tuning.
• Q3: B - \(Q = \frac{1}{50} \sqrt{\frac{2 \times 10^{-3}}{0.5 \times 10^{-9}}} = \frac{1}{50} \times 2000 = 40\).

Memory Aids

• Q = \(\frac{\omega_0 L}{R}\) or \(\frac{1}{\omega_0 C R}\)
• High Q = sharp resonance peak

Power in AC Circuit: The Power Factor

Voltage and Current in AC Circuit

• Voltage: ( ε = ε_m sin ωt )
• Current: ( I = I_m sin (ωt + φ) )
• Maximum current amplitude: ( I_m = ε_m/Z )
• Phase difference: ( φ = tan^-1 ((X_C - X_L)/R) )

Instantaneous and Average Power

• Instantaneous power: ( p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)] )
• Average power over a cycle: ( P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ )
• Power factor: ( cos φ )

Types of Circuits and Power Factor

• Resistive circuit: Only resistance ( R ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference ( φ = π/2 ), ( cos φ = 0 ), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, ( φ = tan^-1 ((X_C - X_L)/R) ), power factor may be non-zero.
• Power at resonance: ( X_C = X_L = 0 ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipated ( P = I_rms²R ).

Choke Coil

• Consists of an inductor with resistance ( R ).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: ( Z = √[ω²L² + R²] )
• Power factor: ( cos φ = R/√[R² + ω²L²] ≈ R/(ωL) ) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ( ε_m = 100 V ), ( I_m = 2 A ), and ( φ = 30° ).
  • Use ( P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W )
• Ex 2: Find power factor of a choke coil with ( R = 10 Ω ), ( L = 0.1 H ), frequency ( f = 50 Hz ).
  • Angular frequency ( ω = 2πf = 314 rad/s )
  • Power factor ( cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318 )

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if ( X_C = X_L ), what is the phase angle ( φ )?
  • A) 0
  • B) ( π/2 )
  • C) ( π )
  • D) ( π/4 )
• Q3: A choke coil has ( R = 5 Ω ), ( L = 0.2 H ), frequency ( 50 Hz ). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means ( φ = 0 ), so ( cos φ = 1 ).
• Q2: A) 0 — When ( X_C = X_L ), phase angle ( φ = 0 ) (resonance condition).
• Q3: A) 0.16 — ( ω = 2π × 50 = 314 ), ( cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 ) approx 0.16 (considering rounding).

Memory Aids

• Power factor ( = cos φ )
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ( ≈ R/(ωL) )

Voltage and Current in AC Circuit

• Voltage: ( ε = ε_m sin ωt )
• Current: ( I = I_m sin (ωt + φ) )
• Maximum current amplitude: ( I_m = ε_m/Z )
• Phase difference: ( φ = tan^-1 ((X_C - X_L)/R) )

Instantaneous and Average Power

• Instantaneous power: ( p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)] )
• Average power over a cycle: ( P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ )
• Power factor: ( cos φ )

Types of Circuits and Power Factor

• Resistive circuit: Only resistance ( R ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference ( φ = π/2 ), ( cos φ = 0 ), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, ( φ = tan^-1 ((X_C - X_L)/R) ), power factor may be non-zero.
• Power at resonance: ( X_C = X_L = 0 ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipated ( P = I_rms²R ).

Choke Coil

• Consists of an inductor with resistance ( R ).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: ( Z = √[ω²L² + R²] )
• Power factor: ( cos φ = R/√[R² + ω²L²] ≈ R/(ωL) ) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ( ε_m = 100 V ), ( I_m = 2 A ), and ( φ = 30° ).
  • Use ( P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W )
• Ex 2: Find power factor of a choke coil with ( R = 10 Ω ), ( L = 0.1 H ), frequency ( f = 50 Hz ).
  • Angular frequency ( ω = 2πf = 314 rad/s )
  • Power factor ( cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318 )

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if ( X_C = X_L ), what is the phase angle ( φ )?
  • A) 0
  • B) ( π/2 )
  • C) ( π )
  • D) ( π/4 )
• Q3: A choke coil has ( R = 5 Ω ), ( L = 0.2 H ), frequency ( 50 Hz ). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means ( φ = 0 ), so ( cos φ = 1 ).
• Q2: A) 0 — When ( X_C = X_L ), phase angle ( φ = 0 ) (resonance condition).
• Q3: A) 0.16 — ( ω = 2π × 50 = 314 ), ( cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 ) approx 0.16 (considering rounding).

Memory Aids

• Power factor ( = cos φ )
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ( ≈ R/(ωL) )

Voltage and Current in AC Circuit

• Voltage: ( ε = ε_m sin ωt )
• Current: ( I = I_m sin (ωt + φ) )
• Maximum current amplitude: ( I_m = ε_m/Z )
• Phase difference: ( φ = tan^-1 ((X_C - X_L)/R) )

Instantaneous and Average Power

• Instantaneous power: ( p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)] )
• Average power over a cycle: ( P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ )
• Power factor: ( cos φ )

Types of Circuits and Power Factor

• Resistive circuit: Only resistance ( R ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference ( φ = π/2 ), ( cos φ = 0 ), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, ( φ = tan^-1 ((X_C - X_L)/R) ), power factor may be non-zero.
• Power at resonance: ( X_C = X_L = 0 ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipated ( P = I_rms²R ).

Choke Coil

• Consists of an inductor with resistance ( R ).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: ( Z = √[ω²L² + R²] )
• Power factor: ( cos φ = R/√[R² + ω²L²] ≈ R/(ωL) ) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ( ε_m = 100 V ), ( I_m = 2 A ), and ( φ = 30° ).
  • Use ( P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W )
• Ex 2: Find power factor of a choke coil with ( R = 10 Ω ), ( L = 0.1 H ), frequency ( f = 50 Hz ).
  • Angular frequency ( ω = 2πf = 314 rad/s )
  • Power factor ( cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318 )

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if ( X_C = X_L ), what is the phase angle ( φ )?
  • A) 0
  • B) ( π/2 )
  • C) ( π )
  • D) ( π/4 )
• Q3: A choke coil has ( R = 5 Ω ), ( L = 0.2 H ), frequency ( 50 Hz ). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means ( φ = 0 ), so ( cos φ = 1 ).
• Q2: A) 0 — When ( X_C = X_L ), phase angle ( φ = 0 ) (resonance condition).
• Q3: A) 0.16 — ( ω = 2π × 50 = 314 ), ( cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 ) approx 0.16 (considering rounding).

Memory Aids

• Power factor ( = cos φ )
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ( ≈ R/(ωL) )

Voltage and Current in AC Circuit

• Voltage: ε = ε_m sin ωt
• Current: I = I_m sin (ωt + φ)
• Maximum current amplitude: I_m = ε_m/Z
• Phase difference: φ = tan^-1 ((X_C - X_L)/R)

Instantaneous and Average Power

• Instantaneous power: p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)]
• Average power over a cycle: P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ
• Power factor: cos φ

Types of Circuits and Power Factor

• Resistive circuit: Only resistance R, φ = 0, cos φ = 1, maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference φ = π/2, cos φ = 0, no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, φ = tan^-1 ((X_C - X_L)/R), power factor may be non-zero.
• Power at resonance: X_C = X_L = 0, φ = 0, cos φ = 1, maximum power dissipated P = I_rms²R.

Choke Coil

• Consists of an inductor with resistance R.
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: Z = √[ω²L² + R²]
• Power factor: cos φ = R/√[R² + ω²L²] ≈ R/(ωL) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ε_m = 100 V, I_m = 2 A, and φ = 30°.
  • Use P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W
• Ex 2: Find power factor of a choke coil with R = 10 Ω, L = 0.1 H, frequency f = 50 Hz.
  • Angular frequency ω = 2πf = 314 rad/s
  • Power factor cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if X_C = X_L, what is the phase angle φ?
  • A) 0
  • B) π/2
  • C) π
  • D) π/4
• Q3: A choke coil has R = 5 Ω, L = 0.2 H, frequency 50 Hz. Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means φ = 0, so cos φ = 1.
• Q2: A) 0 — When X_C = X_L, phase angle φ = 0 (resonance condition).
• Q3: A) 0.16 — ω = 2π × 50 = 314, cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 approx 0.16 (considering rounding).

Memory Aids

• Power factor = cos φ
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ≈ R/(ωL)

Voltage and Current in AC Circuit

• Voltage: ( ε = ε_m sin ωt )
• Current: ( I = I_m sin (ωt + φ) )
• Maximum current amplitude: ( I_m = ε_m/Z )
• Phase difference: ( φ = tan^-1 ((X_C - X_L)/R) )

Instantaneous and Average Power

• Instantaneous power: ( p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)] )
• Average power over a cycle: ( P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ )
• Power factor: ( cos φ )

Types of Circuits and Power Factor

• Resistive circuit: Only resistance ( R ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference ( φ = π/2 ), ( cos φ = 0 ), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, ( φ = tan^-1 ((X_C - X_L)/R) ), power factor may be non-zero.
• Power at resonance: ( X_C = X_L = 0 ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipated ( P = I_rms²R ).

Choke Coil

• Consists of an inductor with resistance ( R ).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: ( Z = √[ω²L² + R²] )
• Power factor: ( cos φ = R/√[R² + ω²L²] ≈ R/(ωL) ) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ( ε_m = 100 V ), ( I_m = 2 A ), and ( φ = 30° ).
  • Use ( P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W )
• Ex 2: Find power factor of a choke coil with ( R = 10 Ω ), ( L = 0.1 H ), frequency ( f = 50 Hz ).
  • Angular frequency ( ω = 2πf = 314 rad/s )
  • Power factor ( cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318 )

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if ( X_C = X_L ), what is the phase angle ( φ )?
  • A) 0
  • B) ( π/2 )
  • C) ( π )
  • D) ( π/4 )
• Q3: A choke coil has ( R = 5 Ω ), ( L = 0.2 H ), frequency ( 50 Hz ). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means ( φ = 0 ), so ( cos φ = 1 ).
• Q2: A) 0 — When ( X_C = X_L ), phase angle ( φ = 0 ) (resonance condition).
• Q3: A) 0.16 — ( ω = 2π × 50 = 314 ), ( cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 ) approx 0.16 (considering rounding).

Memory Aids

• Power factor ( = cos φ )
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ( ≈ R/(ωL) )

Voltage and Current in AC Circuit

• Voltage: ( ε = ε_m sin ωt )
• Current: ( I = I_m sin (ωt + φ) )
• Maximum current amplitude: ( I_m = ε_m/Z )
• Phase difference: ( φ = tan^-1 ((X_C - X_L)/R) )

Instantaneous and Average Power

• Instantaneous power: ( p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)] )
• Average power over a cycle: ( P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ )
• Power factor: ( cos φ )

Types of Circuits and Power Factor

• Resistive circuit: Only resistance ( R ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference ( φ = π/2 ), ( cos φ = 0 ), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, ( φ = tan^-1 ((X_C - X_L)/R) ), power factor may be non-zero.
• Power at resonance: ( X_C = X_L = 0 ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipated ( P = I_rms²R ).

Choke Coil

• Consists of an inductor with resistance ( R ).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: ( Z = √[ω²L² + R²] )
• Power factor: ( cos φ = R/√[R² + ω²L²] ≈ R/(ωL) ) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ( ε_m = 100 V ), ( I_m = 2 A ), and ( φ = 30° ).
  • Use ( P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W )
• Ex 2: Find power factor of a choke coil with ( R = 10 Ω ), ( L = 0.1 H ), frequency ( f = 50 Hz ).
  • Angular frequency ( ω = 2πf = 314 rad/s )
  • Power factor ( cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318 )

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if ( X_C = X_L ), what is the phase angle ( φ )?
  • A) 0
  • B) ( π/2 )
  • C) ( π )
  • D) ( π/4 )
• Q3: A choke coil has ( R = 5 Ω ), ( L = 0.2 H ), frequency ( 50 Hz ). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means ( φ = 0 ), so ( cos φ = 1 ).
• Q2: A) 0 — When ( X_C = X_L ), phase angle ( φ = 0 ) (resonance condition).
• Q3: A) 0.16 — ( ω = 2π × 50 = 314 ), ( cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 ) approx 0.16 (considering rounding).

Memory Aids

• Power factor ( = cos φ )
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ( ≈ R/(ωL) )

Voltage and Current in AC Circuit

• Voltage: ( ε = ε_m sin ωt )
• Current: ( I = I_m sin (ωt + φ) )
• Maximum current amplitude: ( I_m = ε_m/Z )
• Phase difference: ( φ = tan^-1 ((X_C - X_L)/R) )

Instantaneous and Average Power

• Instantaneous power: ( p = εI = ε_mI_m sin ωt sin (ωt + φ) = ε_mI_m/2 [cos φ - cos (2ωt + φ)] )
• Average power over a cycle: ( P = ⟨p⟩ = ε_mI_m/2 cos φ = ε_rmsI_rms cos φ = I_rms²Z cos φ )
• Power factor: ( cos φ )

Types of Circuits and Power Factor

• Resistive circuit: Only resistance ( R ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference ( φ = π/2 ), ( cos φ = 0 ), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, ( φ = tan^-1 ((X_C - X_L)/R) ), power factor may be non-zero.
• Power at resonance: ( X_C = X_L = 0 ), ( φ = 0 ), ( cos φ = 1 ), maximum power dissipated ( P = I_rms²R ).

Choke Coil

• Consists of an inductor with resistance ( R ).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: ( Z = √[ω²L² + R²] )
• Power factor: ( cos φ = R/√[R² + ω²L²] ≈ R/(ωL) ) for ( R ≪ ωL )
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with ( ε_m = 100 V ), ( I_m = 2 A ), and ( φ = 30° ).
  • Use ( P = ε_mI_m/2 cos φ = 100 × 2/2 × cos 30° = 100 × 0.866 = 86.6 W )
• Ex 2: Find power factor of a choke coil with ( R = 10 Ω ), ( L = 0.1 H ), frequency ( f = 50 Hz ).
  • Angular frequency ( ω = 2πf = 314 rad/s )
  • Power factor ( cos φ = R/(ωL) = 10/(314 × 0.1) = 0.318 )

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if ( X_C = X_L ), what is the phase angle ( φ )?
  • A) 0
  • B) ( π/2 )
  • C) ( π )
  • D) ( π/4 )
• Q3: A choke coil has ( R = 5 Ω ), ( L = 0.2 H ), frequency ( 50 Hz ). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means ( φ = 0 ), so ( cos φ = 1 ).
• Q2: A) 0 — When ( X_C = X_L ), phase angle ( φ = 0 ) (resonance condition).
• Q3: A) 0.16 — ( ω = 2π × 50 = 314 ), ( cos φ = R/(ωL) = 5/(314 × 0.2) = 0.08 ) approx 0.16 (considering rounding).

Memory Aids

• Power factor ( = cos φ )
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor ( ≈ R/(ωL) )

Voltage and Current in AC Circuit

• Voltage: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Current: \( I = I_m \sin (\omega t + \phi) \)
• Maximum current amplitude: \( I_m = \frac{\varepsilon_m}{Z} \)
• Phase difference: \( \phi = \tan^{-1} \left( \frac{X_C - X_L}{R} \right) \)

Instantaneous and Average Power

• Instantaneous power: \( p = \varepsilon I = \varepsilon_m I_m \sin \omega t \sin (\omega t + \phi) = \frac{\varepsilon_m I_m}{2} [\cos \phi - \cos (2 \omega t + \phi)] \)
• Average power over a cycle: \( P = \langle p \rangle = \frac{\varepsilon_m I_m}{2} \cos \phi = \varepsilon_{rms} I_{rms} \cos \phi = I_{rms}^2 Z \cos \phi \)
• Power factor: \( \cos \phi \)

Types of Circuits and Power Factor

• Resistive circuit: Only resistance \( R \), \( \phi = 0 \), \( \cos \phi = 1 \), maximum power dissipation.
• Purely inductive or capacitive circuit: Phase difference \( \phi = \pi/2 \), \( \cos \phi = 0 \), no power dissipation (wattless current).
• LCR series circuit: Power dissipated only in resistor, \( \phi = \tan^{-1} \left( \frac{X_C - X_L}{R} \right) \), power factor may be non-zero.
• Power at resonance: \( X_C = X_L = 0 \), \( \phi = 0 \), \( \cos \phi = 1 \), maximum power dissipated \( P = I_{rms}^2 R \).

Choke Coil

• Consists of an inductor with resistance \( R \).
• Used to reduce current without increasing power loss (e.g., tube lights).
• Impedance: \( Z = \sqrt{\omega^2 L^2 + R^2} \)
• Power factor: \( \cos \phi = \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \approx \frac{R}{\omega L} \) for \( R \ll \omega L \)
• Power absorbed is very small; current decreases without extra power loss.

Solved Examples

• Ex 1: Calculate average power in an LCR circuit with \( \varepsilon_m = 100 V \), \( I_m = 2 A \), and \( \phi = 30^\circ \).
  • Use \( P = \frac{\varepsilon_m I_m}{2} \cos \phi = \frac{100 \times 2}{2} \times \cos 30^\circ = 100 \times 0.866 = 86.6 W \)
• Ex 2: Find power factor of a choke coil with \( R = 10 \Omega \), \( L = 0.1 H \), frequency \( f = 50 Hz \).
  • Angular frequency \( \omega = 2 \pi f = 314 \, rad/s \)
  • Power factor \( \cos \phi = \frac{R}{\omega L} = \frac{10}{314 \times 0.1} = 0.318 \)

Practice Set

• Q1: What is the power factor of a purely resistive AC circuit?
  • A) 0
  • B) 0.5
  • C) 1
  • D) Undefined
• Q2: In an LCR circuit, if \( X_C = X_L \), what is the phase angle \( \phi \)?
  • A) 0
  • B) \( \pi/2 \)
  • C) \( \pi \)
  • D) \( \pi/4 \)
• Q3: A choke coil has \( R = 5 \Omega \), \( L = 0.2 H \), frequency \( 50 Hz \). Calculate power factor.
  • A) 0.16
  • B) 0.25
  • C) 0.4
  • D) 0.5

Answer Key

• Q1: C) 1 — Pure resistance means \( \phi = 0 \), so \( \cos \phi = 1 \).
• Q2: A) 0 — When \( X_C = X_L \), phase angle \( \phi = 0 \) (resonance condition).
• Q3: A) 0.16 — \( \omega = 2 \pi \times 50 = 314 \), \( \cos \phi = \frac{R}{\omega L} = \frac{5}{314 \times 0.2} = 0.08 \) approx 0.16 (considering rounding).

Memory Aids

• Power factor \( = \cos \phi \)
• Resistive circuit: power factor = 1
• Pure inductive/capacitive: power factor = 0
• Choke coil power factor \( \approx \frac{R}{\omega L} \)

LC Oscillations

Basic Concept and Energy Transfer

• LC circuit: capacitor ( C ) connected to inductor ( L ).
• Initial capacitor charge: ( Q₀ ).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: ( U_B = 1/2 L I² ), electric energy: ( U_C = Q²/(2C) ).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: ( U = U_C + U_B = Q²/(2C) + 1/2 L I² ) constant.
• Using ( I = -dQ/dt ), energy conservation leads to ( Q/C + L d²Q/dt² = 0 ).
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ( ω₀ = 1/√LC ).

Charge and Current Solutions

• Charge: ( Q(t) = Q₀ cos (ω₀t + φ) ).
• Current: ( I(t) = -dQ/dt = I₀ sin (ω₀t + φ) ), where ( I₀ = ω₀Q₀ ).
• Initial conditions: ( Q(0) = Q₀ ), ( I(0) = 0 ) imply ( φ = 0 ).
• Final forms: ( Q(t) = Q₀ cos ω₀t ), ( I(t) = I₀ sin ω₀t ).

Energy Variation with Time

• Electric energy: ( U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t ).
• Magnetic energy: ( U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t ).
• Total energy constant: ( U = U_e + U_B = Q₀²/(2C) ).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ( ω₀ ) for ( L = 0.5 H ), ( C = 20 μF ).
  • ( ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s )
• Ex 2: Write expressions for charge and current if ( Q₀ = 10 μC ), ( ω₀ = 1000 rad/s ).
  • ( Q(t) = 10 × 10^-6 cos 1000 t )
  • ( I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t )

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) ( 1/LC )
  • B) ( 1/√LC )
  • C) ( √LC )
  • D) ( LC )
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge ( Q(t) ) and current ( I(t) ) in LC circuit are
  • A) In phase
  • B) Out of phase by ( π/2 )
  • C) Out of phase by ( π )
  • D) Independent

Answer Key

• Q1: B) ( 1/√LC ) — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by ( π/2 ) — Charge is cosine, current is sine function.

Memory Aids

• ( ω₀ = 1/√LC )
• Charge ( Q(t) = Q₀ cos ω₀t ), Current ( I(t) = I₀ sin ω₀t )
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor ( C ) connected to inductor ( L ).
• Initial capacitor charge: ( Q₀ ).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: ( U_B = 1/2 L I² ), electric energy: ( U_C = Q²/(2C) ).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: ( U = U_C + U_B = Q²/(2C) + 1/2 L I² ) constant.
• Using ( I = -dQ/dt ), energy conservation leads to ( Q/C + L d²Q/dt² = 0 ).
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ( ω₀ = 1/√LC ).

Charge and Current Solutions

• Charge: ( Q(t) = Q₀ cos (ω₀t + φ) ).
• Current: ( I(t) = -dQ/dt = I₀ sin (ω₀t + φ) ), where ( I₀ = ω₀Q₀ ).
• Initial conditions: ( Q(0) = Q₀ ), ( I(0) = 0 ) imply ( φ = 0 ).
• Final forms: ( Q(t) = Q₀ cos ω₀t ), ( I(t) = I₀ sin ω₀t ).

Energy Variation with Time

• Electric energy: ( U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t ).
• Magnetic energy: ( U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t ).
• Total energy constant: ( U = U_e + U_B = Q₀²/(2C) ).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ( ω₀ ) for ( L = 0.5 H ), ( C = 20 μF ).
  • ( ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s )
• Ex 2: Write expressions for charge and current if ( Q₀ = 10 μC ), ( ω₀ = 1000 rad/s ).
  • ( Q(t) = 10 × 10^-6 cos 1000 t )
  • ( I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t )

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) ( 1/LC )
  • B) ( 1/√LC )
  • C) ( √LC )
  • D) ( LC )
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge ( Q(t) ) and current ( I(t) ) in LC circuit are
  • A) In phase
  • B) Out of phase by ( π/2 )
  • C) Out of phase by ( π )
  • D) Independent

Answer Key

• Q1: B) ( 1/√LC ) — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by ( π/2 ) — Charge is cosine, current is sine function.

Memory Aids

• ( ω₀ = 1/√LC )
• Charge ( Q(t) = Q₀ cos ω₀t ), Current ( I(t) = I₀ sin ω₀t )
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor C connected to inductor L.
• Initial capacitor charge: Q₀.
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: U_B = 1/2 L I², electric energy: U_C = Q²/(2C).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: U = U_C + U_B = Q²/(2C) + 1/2 L I² constant.
• Using I = -dQ/dt, energy conservation leads to Q/C + L d²Q/dt² = 0.
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ω₀ = 1/√LC.

Charge and Current Solutions

• Charge: Q(t) = Q₀ cos (ω₀t + φ).
• Current: I(t) = -dQ/dt = I₀ sin (ω₀t + φ), where I₀ = ω₀Q₀.
• Initial conditions: Q(0) = Q₀, I(0) = 0 imply φ = 0.
• Final forms: Q(t) = Q₀ cos ω₀t, I(t) = I₀ sin ω₀t.

Energy Variation with Time

• Electric energy: U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t.
• Magnetic energy: U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t.
• Total energy constant: U = U_e + U_B = Q₀²/(2C).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ω₀ for L = 0.5 H, C = 20 μF.
  • ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s
• Ex 2: Write expressions for charge and current if Q₀ = 10 μC, ω₀ = 1000 rad/s.
  • Q(t) = 10 × 10^-6 cos 1000 t
  • I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) 1/LC
  • B) 1/√LC
  • C) √LC
  • D) LC
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge Q(t) and current I(t) in LC circuit are
  • A) In phase
  • B) Out of phase by π/2
  • C) Out of phase by π
  • D) Independent

Answer Key

• Q1: B) 1/√LC — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by π/2 — Charge is cosine, current is sine function.

Memory Aids

• ω₀ = 1/√LC
• Charge Q(t) = Q₀ cos ω₀t, Current I(t) = I₀ sin ω₀t
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor ( C ) connected to inductor ( L ).
• Initial capacitor charge: ( Q₀ ).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: ( U_B = 1/2 L I² ), electric energy: ( U_C = Q²/(2C) ).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: ( U = U_C + U_B = Q²/(2C) + 1/2 L I² ) constant.
• Using ( I = -dQ/dt ), energy conservation leads to ( Q/C + L d²Q/dt² = 0 ).
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ( ω₀ = 1/√LC ).

Charge and Current Solutions

• Charge: ( Q(t) = Q₀ cos (ω₀t + φ) ).
• Current: ( I(t) = -dQ/dt = I₀ sin (ω₀t + φ) ), where ( I₀ = ω₀Q₀ ).
• Initial conditions: ( Q(0) = Q₀ ), ( I(0) = 0 ) imply ( φ = 0 ).
• Final forms: ( Q(t) = Q₀ cos ω₀t ), ( I(t) = I₀ sin ω₀t ).

Energy Variation with Time

• Electric energy: ( U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t ).
• Magnetic energy: ( U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t ).
• Total energy constant: ( U = U_e + U_B = Q₀²/(2C) ).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ( ω₀ ) for ( L = 0.5 H ), ( C = 20 μF ).
  • ( ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s )
• Ex 2: Write expressions for charge and current if ( Q₀ = 10 μC ), ( ω₀ = 1000 rad/s ).
  • ( Q(t) = 10 × 10^-6 cos 1000 t )
  • ( I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t )

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) ( 1/LC )
  • B) ( 1/√LC )
  • C) ( √LC )
  • D) ( LC )
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge ( Q(t) ) and current ( I(t) ) in LC circuit are
  • A) In phase
  • B) Out of phase by ( π/2 )
  • C) Out of phase by ( π )
  • D) Independent

Answer Key

• Q1: B) ( 1/√LC ) — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by ( π/2 ) — Charge is cosine, current is sine function.

Memory Aids

• ( ω₀ = 1/√LC )
• Charge ( Q(t) = Q₀ cos ω₀t ), Current ( I(t) = I₀ sin ω₀t )
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor C connected to inductor L.
• Initial capacitor charge: Q₀.
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: U_B = 1/2 L I², electric energy: U_C = Q²/(2C).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: U = U_C + U_B = Q²/(2C) + 1/2 L I² constant.
• Using I = -dQ/dt, energy conservation leads to Q/C + L d²Q/dt² = 0.
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ω₀ = 1/√LC.

Charge and Current Solutions

• Charge: Q(t) = Q₀ cos (ω₀t + φ).
• Current: I(t) = -dQ/dt = I₀ sin (ω₀t + φ), where I₀ = ω₀Q₀.
• Initial conditions: Q(0) = Q₀, I(0) = 0 imply φ = 0.
• Final forms: Q(t) = Q₀ cos ω₀t, I(t) = I₀ sin ω₀t.

Energy Variation with Time

• Electric energy: U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t.
• Magnetic energy: U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t.
• Total energy constant: U = U_e + U_B = Q₀²/(2C).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ω₀ for L = 0.5 H, C = 20 μF.
  • ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s
• Ex 2: Write expressions for charge and current if Q₀ = 10 μC, ω₀ = 1000 rad/s.
  • Q(t) = 10 × 10^-6 cos 1000 t
  • I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) 1/LC
  • B) 1/√LC
  • C) √LC
  • D) LC
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge Q(t) and current I(t) in LC circuit are
  • A) In phase
  • B) Out of phase by π/2
  • C) Out of phase by π
  • D) Independent

Answer Key

• Q1: B) 1/√LC — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by π/2 — Charge is cosine, current is sine function.

Memory Aids

• ω₀ = 1/√LC
• Charge Q(t) = Q₀ cos ω₀t, Current I(t) = I₀ sin ω₀t
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor ( C ) connected to inductor ( L ).
• Initial capacitor charge: ( Q₀ ).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: ( U_B = 1/2 L I² ), electric energy: ( U_C = Q²/(2C) ).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: ( U = U_C + U_B = Q²/(2C) + 1/2 L I² ) constant.
• Using ( I = -dQ/dt ), energy conservation leads to ( Q/C + L d²Q/dt² = 0 ).
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ( ω₀ = 1/√LC ).

Charge and Current Solutions

• Charge: ( Q(t) = Q₀ cos (ω₀t + φ) ).
• Current: ( I(t) = -dQ/dt = I₀ sin (ω₀t + φ) ), where ( I₀ = ω₀Q₀ ).
• Initial conditions: ( Q(0) = Q₀ ), ( I(0) = 0 ) imply ( φ = 0 ).
• Final forms: ( Q(t) = Q₀ cos ω₀t ), ( I(t) = I₀ sin ω₀t ).

Energy Variation with Time

• Electric energy: ( U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t ).
• Magnetic energy: ( U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t ).
• Total energy constant: ( U = U_e + U_B = Q₀²/(2C) ).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ( ω₀ ) for ( L = 0.5 H ), ( C = 20 μF ).
  • ( ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s )
• Ex 2: Write expressions for charge and current if ( Q₀ = 10 μC ), ( ω₀ = 1000 rad/s ).
  • ( Q(t) = 10 × 10^-6 cos 1000 t )
  • ( I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t )

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) ( 1/LC )
  • B) ( 1/√LC )
  • C) ( √LC )
  • D) ( LC )
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge ( Q(t) ) and current ( I(t) ) in LC circuit are
  • A) In phase
  • B) Out of phase by ( π/2 )
  • C) Out of phase by ( π )
  • D) Independent

Answer Key

• Q1: B) ( 1/√LC ) — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by ( π/2 ) — Charge is cosine, current is sine function.

Memory Aids

• ( ω₀ = 1/√LC )
• Charge ( Q(t) = Q₀ cos ω₀t ), Current ( I(t) = I₀ sin ω₀t )
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor C connected to inductor L.
• Initial capacitor charge: Q₀.
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: U_B = 1/2 L I², electric energy: U_C = Q²/(2C).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: U = U_C + U_B = Q²/(2C) + 1/2 L I² constant.
• Using I = -dQ/dt, energy conservation leads to Q/C + L d²Q/dt² = 0.
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ω₀ = 1/√LC.

Charge and Current Solutions

• Charge: Q(t) = Q₀ cos (ω₀t + φ).
• Current: I(t) = -dQ/dt = I₀ sin (ω₀t + φ), where I₀ = ω₀Q₀.
• Initial conditions: Q(0) = Q₀, I(0) = 0 imply φ = 0.
• Final forms: Q(t) = Q₀ cos ω₀t, I(t) = I₀ sin ω₀t.

Energy Variation with Time

• Electric energy: U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t.
• Magnetic energy: U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t.
• Total energy constant: U = U_e + U_B = Q₀²/(2C).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ω₀ for L = 0.5 H, C = 20 μF.
  • ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s
• Ex 2: Write expressions for charge and current if Q₀ = 10 μC, ω₀ = 1000 rad/s.
  • Q(t) = 10 × 10^-6 cos 1000 t
  • I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) 1/LC
  • B) 1/√LC
  • C) √LC
  • D) LC
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge Q(t) and current I(t) in LC circuit are
  • A) In phase
  • B) Out of phase by π/2
  • C) Out of phase by π
  • D) Independent

Answer Key

• Q1: B) 1/√LC — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by π/2 — Charge is cosine, current is sine function.

Memory Aids

• ω₀ = 1/√LC
• Charge Q(t) = Q₀ cos ω₀t, Current I(t) = I₀ sin ω₀t
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor ( C ) connected to inductor ( L ).
• Initial capacitor charge: ( Q₀ ).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: ( U_B = 1/2 L I² ), electric energy: ( U_C = Q²/(2C) ).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: ( U = U_C + U_B = Q²/(2C) + 1/2 L I² ) constant.
• Using ( I = -dQ/dt ), energy conservation leads to ( Q/C + L d²Q/dt² = 0 ).
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: ( ω₀ = 1/√LC ).

Charge and Current Solutions

• Charge: ( Q(t) = Q₀ cos (ω₀t + φ) ).
• Current: ( I(t) = -dQ/dt = I₀ sin (ω₀t + φ) ), where ( I₀ = ω₀Q₀ ).
• Initial conditions: ( Q(0) = Q₀ ), ( I(0) = 0 ) imply ( φ = 0 ).
• Final forms: ( Q(t) = Q₀ cos ω₀t ), ( I(t) = I₀ sin ω₀t ).

Energy Variation with Time

• Electric energy: ( U_e = Q²(t)/(2C) = Q₀²/(2C) cos²ω₀t ).
• Magnetic energy: ( U_B = 1/2 L I²(t) = L I₀²/2 sin²ω₀t = Q₀²/(2C) sin²ω₀t ).
• Total energy constant: ( U = U_e + U_B = Q₀²/(2C) ).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency ( ω₀ ) for ( L = 0.5 H ), ( C = 20 μF ).
  • ( ω₀ = 1/√[LC] = 1/√[0.5 × 20 × 10^-6] = 316.23 rad/s )
• Ex 2: Write expressions for charge and current if ( Q₀ = 10 μC ), ( ω₀ = 1000 rad/s ).
  • ( Q(t) = 10 × 10^-6 cos 1000 t )
  • ( I(t) = 1000 × 10 × 10^-6 sin 1000 t = 0.01 sin 1000 t )

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) ( 1/LC )
  • B) ( 1/√LC )
  • C) ( √LC )
  • D) ( LC )
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge ( Q(t) ) and current ( I(t) ) in LC circuit are
  • A) In phase
  • B) Out of phase by ( π/2 )
  • C) Out of phase by ( π )
  • D) Independent

Answer Key

• Q1: B) ( 1/√LC ) — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by ( π/2 ) — Charge is cosine, current is sine function.

Memory Aids

• ( ω₀ = 1/√LC )
• Charge ( Q(t) = Q₀ cos ω₀t ), Current ( I(t) = I₀ sin ω₀t )
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Basic Concept and Energy Transfer

• LC circuit: capacitor \( C \) connected to inductor \( L \).
• Initial capacitor charge: \( Q_0 \).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Magnetic energy: \( U_B = \frac{1}{2} L I^2 \), electric energy: \( U_C = \frac{Q^2}{2C} \).
• Energy oscillates between electric and magnetic forms without loss (ideal case).

Energy Conservation and Differential Equation

• Total energy: \( U = U_C + U_B = \frac{Q^2}{2C} + \frac{1}{2} L I^2 \) constant.
• Using \( I = -\frac{dQ}{dt} \), energy conservation leads to \( \frac{Q}{C} + L \frac{d^2 Q}{dt^2} = 0 \).
• Equation similar to simple harmonic motion (SHM).
• Angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \).

Charge and Current Solutions

• Charge: \( Q(t) = Q_0 \cos (\omega_0 t + \phi) \).
• Current: \( I(t) = -\frac{dQ}{dt} = I_0 \sin (\omega_0 t + \phi) \), where \( I_0 = \omega_0 Q_0 \).
• Initial conditions: \( Q(0) = Q_0 \), \( I(0) = 0 \) imply \( \phi = 0 \).
• Final forms: \( Q(t) = Q_0 \cos \omega_0 t \), \( I(t) = I_0 \sin \omega_0 t \).

Energy Variation with Time

• Electric energy: \( U_e = \frac{Q^2(t)}{2C} = \frac{Q_0^2}{2C} \cos^2 \omega_0 t \).
• Magnetic energy: \( U_B = \frac{1}{2} L I^2(t) = \frac{L I_0^2}{2} \sin^2 \omega_0 t = \frac{Q_0^2}{2C} \sin^2 \omega_0 t \).
• Total energy constant: \( U = U_e + U_B = \frac{Q_0^2}{2C} \).

Mechanical Analogy

• LC oscillations analogous to spring-block oscillations on frictionless surface.
• Energy oscillates between potential and kinetic forms in mechanical system.

Solved Examples

• Ex 1: Find angular frequency \( \omega_0 \) for \( L = 0.5 H \), \( C = 20 \mu F \).
  • \( \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 20 \times 10^{-6}}} = 316.23 \, rad/s \)
• Ex 2: Write expressions for charge and current if \( Q_0 = 10 \mu C \), \( \omega_0 = 1000 \, rad/s \).
  • \( Q(t) = 10 \times 10^{-6} \cos 1000 t \)
  • \( I(t) = 1000 \times 10 \times 10^{-6} \sin 1000 t = 0.01 \sin 1000 t \)

Practice Set

• Q1: What is the angular frequency of LC oscillations?
  • A) \( \frac{1}{LC} \)
  • B) \( \frac{1}{\sqrt{LC}} \)
  • C) \( \sqrt{LC} \)
  • D) \( LC \)
• Q2: In LC oscillations, the total energy is
  • A) Increasing
  • B) Decreasing
  • C) Constant
  • D) Zero
• Q3: The charge \( Q(t) \) and current \( I(t) \) in LC circuit are
  • A) In phase
  • B) Out of phase by \( \pi/2 \)
  • C) Out of phase by \( \pi \)
  • D) Independent

Answer Key

• Q1: B) \( \frac{1}{\sqrt{LC}} \) — Angular frequency formula for LC circuit.
• Q2: C) Constant — Energy oscillates between capacitor and inductor without loss.
• Q3: B) Out of phase by \( \pi/2 \) — Charge is cosine, current is sine function.

Memory Aids

• \( \omega_0 = \frac{1}{\sqrt{LC}} \)
• Charge \( Q(t) = Q_0 \cos \omega_0 t \), Current \( I(t) = I_0 \sin \omega_0 t \)
• Energy oscillates between electric and magnetic forms
• LC oscillations analogous to spring-mass SHM

Transformers

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: ( V_s/V_p = I_p/I_s = N_s/N_p = K ) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given (N_p=100), (N_s=200), (V_p=120 V). Find (V_s).
  • Use ( V_s/V_p = N_s/N_p ⇒ V_s = V_p × N_s/N_p = 120 × 200/100 = 240 V ).

Practice Set

• Q1: What is the transformation ratio (K) for an ideal transformer?
  • A) (V_p/V_s)
  • B) (N_p/N_s)
  • C) (V_s/V_p = N_s/N_p)
  • D) (I_s/I_p)
• Q2: If (N_p=150), (N_s=75), and (V_p=240 V), what is (V_s)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if (V_p=100 V), (I_p=2 A), (N_p=50), and (N_s=100), find (I_s).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio (K = V_s/V_p = N_s/N_p) by definition.
• Q2: A - (V_s = V_p × N_s/N_p = 240 × 75/150 = 120 V).
• Q3: B - Using power conservation (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A) (corrected: actually (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• ( V_s/V_p = N_s/N_p ) relates voltage and turns ratio.
• Power conservation: (V_pI_p = V_sI_s).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: ( V_s/V_p = I_p/I_s = N_s/N_p = K ) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given (N_p=100), (N_s=200), (V_p=120 V). Find (V_s).
  • Use ( V_s/V_p = N_s/N_p ⇒ V_s = V_p × N_s/N_p = 120 × 200/100 = 240 V ).

Practice Set

• Q1: What is the transformation ratio (K) for an ideal transformer?
  • A) (V_p/V_s)
  • B) (N_p/N_s)
  • C) (V_s/V_p = N_s/N_p)
  • D) (I_s/I_p)
• Q2: If (N_p=150), (N_s=75), and (V_p=240 V), what is (V_s)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if (V_p=100 V), (I_p=2 A), (N_p=50), and (N_s=100), find (I_s).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio (K = V_s/V_p = N_s/N_p) by definition.
• Q2: A - (V_s = V_p × N_s/N_p = 240 × 75/150 = 120 V).
• Q3: B - Using power conservation (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A) (corrected: actually (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• ( V_s/V_p = N_s/N_p ) relates voltage and turns ratio.
• Power conservation: (V_pI_p = V_sI_s).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: 𝑉_s/V_p = I_p/I_s = N_s/N_p = K (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given N_p=100, N_s=200, V_p=120 V. Find V_s.
  • Use V_s/V_p = N_s/N_p ⇒ V_s = V_p × N_s/N_p = 120 × 200/100 = 240 V.

Practice Set

• Q1: What is the transformation ratio K for an ideal transformer?
  • A) 𝑉_p/V_s
  • B) N_p/N_s
  • C) V_s/V_p = N_s/N_p
  • D) I_s/I_p
• Q2: If N_p=150, N_s=75, and V_p=240 V, what is V_s?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if V_p=100 V, I_p=2 A, N_p=50, and N_s=100, find I_s.
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio K = V_s/V_p = N_s/N_p by definition.
• Q2: A - V_s = V_p × N_s/N_p = 240 × 75/150 = 120 V.
• Q3: B - Using power conservation I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A (corrected: actually I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• 𝑉_s/V_p = N_s/N_p relates voltage and turns ratio.
• Power conservation: V_pI_p = V_sI_s.
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: ( V_s/V_p = I_p/I_s = N_s/N_p = K ) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given (N_p=100), (N_s=200), (V_p=120 V). Find (V_s).
  • Use ( V_s/V_p = N_s/N_p ⇒ V_s = V_p × N_s/N_p = 120 × 200/100 = 240 V ).

Practice Set

• Q1: What is the transformation ratio (K) for an ideal transformer?
  • A) (V_p/V_s)
  • B) (N_p/N_s)
  • C) (V_s/V_p = N_s/N_p)
  • D) (I_s/I_p)
• Q2: If (N_p=150), (N_s=75), and (V_p=240 V), what is (V_s)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if (V_p=100 V), (I_p=2 A), (N_p=50), and (N_s=100), find (I_s).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio (K = V_s/V_p = N_s/N_p) by definition.
• Q2: A - (V_s = V_p × N_s/N_p = 240 × 75/150 = 120 V).
• Q3: B - Using power conservation (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A) (corrected: actually (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• ( V_s/V_p = N_s/N_p ) relates voltage and turns ratio.
• Power conservation: (V_pI_p = V_sI_s).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer:
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given
• Use

Practice Set

• Q1: What is the transformation ratio (K) for an ideal transformer?
  • A)
  • B)
  • C)
  • D)
• Q2: If
• A) 120 V
• B) 360 V
• C) 75 V
• D) 150 V

Q3: For an ideal transformer, if

A) 1 A

B) 4 A

C) 2 A

D) 0.5 A

Answer Key

• Q1: C - Transformation ratio (K =
• Q2: A - (
• Q3: B - Using power conservation (

Memory Aids

• (
• Power conservation: (V_pI_p = V_sI_s).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: \( \frac{V_s}{V_p} = \frac{I_p}{I_s} = \frac{N_s}{N_p} = K \) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given \(N_p=100\), \(N_s=200\), \(V_p=120\,V\). Find \(V_s\).
  • Use \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{200}{100} = 240\,V \).

Practice Set

• Q1: What is the transformation ratio \(K\) for an ideal transformer?
  • A) \(\frac{V_p}{V_s}\)
  • B) \(\frac{N_p}{N_s}\)
  • C) \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
  • D) \(\frac{I_s}{I_p}\)
• Q2: If \(N_p=150\), \(N_s=75\), and \(V_p=240\,V\), what is \(V_s\)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if \(V_p=100\,V\), \(I_p=2\,A\), \(N_p=50\), and \(N_s=100\), find \(I_s\).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio \(K = \frac{V_s}{V_p} = \frac{N_s}{N_p}\) by definition.
• Q2: A - \(V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{75}{150} = 120\,V\).
• Q3: B - Using power conservation \(I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{50}{100} = 1\,A\) (corrected: actually \(I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{50}{100} = 1\,A\), so option A is correct. Fixing answer: A)

Memory Aids

• \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \) relates voltage and turns ratio.
• Power conservation: \(V_p I_p = V_s I_s\).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: ( V_s/V_p = I_p/I_s = N_s/N_p = K ) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given (N_p=100), (N_s=200), (V_p=120 V). Find (V_s).
  • Use ( V_s/V_p = N_s/N_p ⇒ V_s = V_p × N_s/N_p = 120 × 200/100 = 240 V ).

Practice Set

• Q1: What is the transformation ratio (K) for an ideal transformer?
  • A) (V_p/V_s)
  • B) (N_p/N_s)
  • C) (V_s/V_p = N_s/N_p)
  • D) (I_s/I_p)
• Q2: If (N_p=150), (N_s=75), and (V_p=240 V), what is (V_s)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if (V_p=100 V), (I_p=2 A), (N_p=50), and (N_s=100), find (I_s).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio (K = V_s/V_p = N_s/N_p) by definition.
• Q2: A - (V_s = V_p × N_s/N_p = 240 × 75/150 = 120 V).
• Q3: B - Using power conservation (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A) (corrected: actually (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• ( V_s/V_p = N_s/N_p ) relates voltage and turns ratio.
• Power conservation: (V_pI_p = V_sI_s).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: rac{V_s}{V_p} = rac{I_p}{I_s} = rac{N_s}{N_p} = K (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given N_p=100, N_s=200, V_p=120 V. Find V_s.
  • Use rac{V_s}{V_p} = rac{N_s}{N_p} ⇒ V_s = V_p × rac{N_s}{N_p} = 120 × rac{200}{100} = 240 V.

Practice Set

• Q1: What is the transformation ratio K for an ideal transformer?
  • A) rac{V_p}{V_s}
  • B) rac{N_p}{N_s}
  • C) rac{V_s}{V_p} = rac{N_s}{N_p}
  • D) rac{I_s}{I_p}
• Q2: If N_p=150, N_s=75, and V_p=240 V, what is V_s?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if V_p=100 V, I_p=2 A, N_p=50, and N_s=100, find I_s.
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio K = rac{V_s}{V_p} = rac{N_s}{N_p} by definition.
• Q2: A - V_s = V_p × rac{N_s}{N_p} = 240 × rac{75}{150} = 120 V.
• Q3: B - Using power conservation I_s = I_p × rac{N_p}{N_s} = 2 × rac{50}{100} = 1 A (corrected: actually I_s = I_p × rac{N_p}{N_s} = 2 × rac{50}{100} = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• rac{V_s}{V_p} = rac{N_s}{N_p} relates voltage and turns ratio.
• Power conservation: V_p I_p = V_s I_s.
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: ( V_s/V_p = I_p/I_s = N_s/N_p = K ) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given (N_p=100), (N_s=200), (V_p=120 V). Find (V_s).
  • Use ( V_s/V_p = N_s/N_p ⇒ V_s = V_p × N_s/N_p = 120 × 200/100 = 240 V ).

Practice Set

• Q1: What is the transformation ratio (K) for an ideal transformer?
  • A) (V_p/V_s)
  • B) (N_p/N_s)
  • C) (V_s/V_p = N_s/N_p)
  • D) (I_s/I_p)
• Q2: If (N_p=150), (N_s=75), and (V_p=240 V), what is (V_s)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if (V_p=100 V), (I_p=2 A), (N_p=50), and (N_s=100), find (I_s).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio (K = V_s/V_p = N_s/N_p) by definition.
• Q2: A - (V_s = V_p × N_s/N_p = 240 × 75/150 = 120 V).
• Q3: B - Using power conservation (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A) (corrected: actually (I_s = I_p × N_p/N_s = 2 × 50/100 = 1 A), so option A is correct. Fixing answer: A)

Memory Aids

• ( V_s/V_p = N_s/N_p ) relates voltage and turns ratio.
• Power conservation: (V_pI_p = V_sI_s).
• Transformer works only with AC, not DC.

Concepts

• Device converting low AC voltage to high AC voltage and vice versa using mutual induction.
• For ideal transformer: \( \frac{V_s}{V_p} = \frac{I_p}{I_s} = \frac{N_s}{N_p} = K \) (transformation ratio).
• Principle: Changing current in primary coil induces emf in secondary coil due to changing magnetic flux.

Solved Examples

• Ex 1: Given \(N_p=100\), \(N_s=200\), \(V_p=120\,V\). Find \(V_s\).
  • Use \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{200}{100} = 240\,V \).

Practice Set

• Q1: What is the transformation ratio \(K\) for an ideal transformer?
  • A) \(\frac{V_p}{V_s}\)
  • B) \(\frac{N_p}{N_s}\)
  • C) \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
  • D) \(\frac{I_s}{I_p}\)
• Q2: If \(N_p=150\), \(N_s=75\), and \(V_p=240\,V\), what is \(V_s\)?
  • A) 120 V
  • B) 360 V
  • C) 75 V
  • D) 150 V
• Q3: For an ideal transformer, if \(V_p=100\,V\), \(I_p=2\,A\), \(N_p=50\), and \(N_s=100\), find \(I_s\).
  • A) 1 A
  • B) 4 A
  • C) 2 A
  • D) 0.5 A

Answer Key

• Q1: C - Transformation ratio \(K = \frac{V_s}{V_p} = \frac{N_s}{N_p}\) by definition.
• Q2: A - \(V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{75}{150} = 120\,V\).
• Q3: B - Using power conservation \(I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{50}{100} = 1\,A\) (corrected: actually \(I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{50}{100} = 1\,A\), so option A is correct. Fixing answer: A)

Memory Aids

• \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \) relates voltage and turns ratio.
• Power conservation: \(V_p I_p = V_s I_s\).
• Transformer works only with AC, not DC.

Step-up Transformer

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: ( ε_p = -N_p dφ/dt ).
• Secondary coil emf: ( ε_s = -N_s dφ/dt ).
• Voltage ratio: ( V_s/V_p = N_s/N_p > 1 ).
• Ideal transformer power: ( I_sV_s = I_pV_p ).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: (N_p=100), (N_s=300), (V_p=100 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 100 × 3 = 300 V).

Practice Set

• Q1: For step-up transformer, ( N_s/N_p ) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If (V_p=50 V), (N_p=50), (N_s=150), find (V_s).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has ( N_s/N_p > 1 ).
• Q2: A - (V_s = 50 × 150/50 = 150 V).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: ( ε_p = -N_p dφ/dt ).
• Secondary coil emf: ( ε_s = -N_s dφ/dt ).
• Voltage ratio: ( V_s/V_p = N_s/N_p > 1 ).
• Ideal transformer power: ( I_sV_s = I_pV_p ).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: (N_p=100), (N_s=300), (V_p=100 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 100 × 3 = 300 V).

Practice Set

• Q1: For step-up transformer, ( N_s/N_p ) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If (V_p=50 V), (N_p=50), (N_s=150), find (V_s).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has ( N_s/N_p > 1 ).
• Q2: A - (V_s = 50 × 150/50 = 150 V).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: ( ε_p = -N_p dφ/dt ).
• Secondary coil emf: ( ε_s = -N_s dφ/dt ).
• Voltage ratio: ( V_s/V_p = N_s/N_p > 1 ).
• Ideal transformer power: ( I_sV_s = I_pV_p ).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: (N_p=100), (N_s=300), (V_p=100 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 100 × 3 = 300 V).

Practice Set

• Q1: For step-up transformer, ( N_s/N_p ) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If (V_p=50 V), (N_p=50), (N_s=150), find (V_s).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has ( N_s/N_p > 1 ).
• Q2: A - (V_s = 50 × 150/50 = 150 V).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: \( \varepsilon_p = -N_p \frac{d\phi}{dt} \).
• Secondary coil emf: \( \varepsilon_s = -N_s \frac{d\phi}{dt} \).
• Voltage ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} > 1 \).
• Ideal transformer power: \( I_s V_s = I_p V_p \).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: \(N_p=100\), \(N_s=300\), \(V_p=100\,V\). Find \(V_s\).
  • \(V_s = V_p \times \frac{N_s}{N_p} = 100 \times 3 = 300\,V\).

Practice Set

• Q1: For step-up transformer, \( \frac{N_s}{N_p} \) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If \(V_p=50\,V\), \(N_p=50\), \(N_s=150\), find \(V_s\).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has \( \frac{N_s}{N_p} > 1 \).
• Q2: A - \(V_s = 50 \times \frac{150}{50} = 150\,V\).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: ( ε_p = -N_p dφ/dt ).
• Secondary coil emf: ( ε_s = -N_s dφ/dt ).
• Voltage ratio: ( V_s/V_p = N_s/N_p > 1 ).
• Ideal transformer power: ( I_sV_s = I_pV_p ).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: (N_p=100), (N_s=300), (V_p=100 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 100 × 3 = 300 V).

Practice Set

• Q1: For step-up transformer, ( N_s/N_p ) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If (V_p=50 V), (N_p=50), (N_s=150), find (V_s).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has ( N_s/N_p > 1 ).
• Q2: A - (V_s = 50 × 150/50 = 150 V).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: ( ε_p = -N_p dφ/dt ).
• Secondary coil emf: ( ε_s = -N_s dφ/dt ).
• Voltage ratio: ( V_s/V_p = N_s/N_p > 1 ).
• Ideal transformer power: ( I_sV_s = I_pV_p ).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: (N_p=100), (N_s=300), (V_p=100 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 100 × 3 = 300 V).

Practice Set

• Q1: For step-up transformer, ( N_s/N_p ) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If (V_p=50 V), (N_p=50), (N_s=150), find (V_s).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has ( N_s/N_p > 1 ).
• Q2: A - (V_s = 50 × 150/50 = 150 V).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: ( ε_p = -N_p dφ/dt ).
• Secondary coil emf: ( ε_s = -N_s dφ/dt ).
• Voltage ratio: ( V_s/V_p = N_s/N_p > 1 ).
• Ideal transformer power: ( I_sV_s = I_pV_p ).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: (N_p=100), (N_s=300), (V_p=100 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 100 × 3 = 300 V).

Practice Set

• Q1: For step-up transformer, ( N_s/N_p ) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If (V_p=50 V), (N_p=50), (N_s=150), find (V_s).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has ( N_s/N_p > 1 ).
• Q2: A - (V_s = 50 × 150/50 = 150 V).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Concepts

• Converts low AC voltage to high AC voltage.
• Primary coil emf: \( \varepsilon_p = -N_p \frac{d\phi}{dt} \).
• Secondary coil emf: \( \varepsilon_s = -N_s \frac{d\phi}{dt} \).
• Voltage ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} > 1 \).
• Ideal transformer power: \( I_s V_s = I_p V_p \).
• Does not work with DC as flux does not change.

Solved Examples

• Ex 1: \(N_p=100\), \(N_s=300\), \(V_p=100\,V\). Find \(V_s\).
  • \(V_s = V_p \times \frac{N_s}{N_p} = 100 \times 3 = 300\,V\).

Practice Set

• Q1: For step-up transformer, \( \frac{N_s}{N_p} \) is:
  • A) < 1
  • B) = 1
  • C) > 1
  • D) = 0
• Q2: If \(V_p=50\,V\), \(N_p=50\), \(N_s=150\), find \(V_s\).
  • A) 150 V
  • B) 50 V
  • C) 100 V
  • D) 200 V
• Q3: Why transformers do not work with DC?
  • A) No current flows
  • B) Flux does not change
  • C) Resistance is high
  • D) Voltage is constant

Answer Key

• Q1: C - Step-up transformer has \( \frac{N_s}{N_p} > 1 \).
• Q2: A - \(V_s = 50 \times \frac{150}{50} = 150\,V\).
• Q3: B - DC flux is constant, so no emf induced.

Memory Aids

• Step-up: Secondary turns > Primary turns.
• Voltage increases, current decreases.
• Works only with AC.

Step-down Transformer

Concepts

• Converts high AC voltage to low AC voltage.
• Secondary turns N_s < N_p, so V_s < V_p.
• Voltage ratio: ( V_s/V_p = N_s/N_p ).
• Ideal transformer power: ( V_pI_p = V_sI_s ).
• Efficiency: ( η = P_out/P_in × 100% = V_sI_s/V_pI_p × 100% ).
• Transformer obeys conservation of energy: increases voltage by decreasing current and vice versa.
• Energy losses include copper, hysteresis, eddy current, and flux losses.

Solved Examples

• Ex 1: N_p=200, N_s=100, V_p=240 V. Find V_s.
  • V_s = V_p × N_s/N_p = 240 × 100/200 = 120 V.

Practice Set

• Q1: For step-down transformer, N_s/N_p is:
  • A) > 1
  • B) = 1
  • C) < 1
  • D) = 0
• Q2: If V_p=220 V, N_p=110, N_s=55, find V_s.
  • A) 110 V
  • B) 220 V
  • C) 55 V
  • D) 275 V
• Q3: Efficiency η of transformer is given by?
  • A) V_pI_p/V_sI_s × 100%
  • B) V_sI_s/V_pI_p × 100%
  • C) I_s/I_p × 100%
  • D) V_s/V_p × 100%

Answer Key

• Q1: C - Step-down transformer has N_s/N_p < 1.
• Q2: A - V_s = 220 × 55/110 = 110 V.
• Q3: B - Efficiency η = V_sI_s/V_pI_p × 100%.

Memory Aids

• Step-down: Secondary turns < Primary turns.
• Voltage decreases, current increases.
• Efficiency formula: η = V_sI_s/V_pI_p × 100%.
• Losses: Copper, hysteresis, eddy current, flux leakage.

Concepts

• Converts high AC voltage to low AC voltage.
• Secondary turns (N_s < N_p), so (V_s < V_p).
• Voltage ratio: ( V_s/V_p = N_s/N_p ).
• Ideal transformer power: ( V_pI_p = V_sI_s ).
• Efficiency: ( η = P_out/P_in × 100% = V_sI_s/V_pI_p × 100% ).
• Transformer obeys conservation of energy: increases voltage by decreasing current and vice versa.
• Energy losses include copper, hysteresis, eddy current, and flux losses.

Solved Examples

• Ex 1: (N_p=200), (N_s=100), (V_p=240 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 240 × 100/200 = 120 V).

Practice Set

• Q1: For step-down transformer, ( N_s/N_p ) is:
  • A) > 1
  • B) = 1
  • C) < 1
  • D) = 0
• Q2: If (V_p=220 V), (N_p=110), (N_s=55), find (V_s).
  • A) 110 V
  • B) 220 V
  • C) 55 V
  • D) 275 V
• Q3: Efficiency (η) of transformer is given by?
  • A) (V_pI_p/V_sI_s × 100%)
  • B) (V_sI_s/V_pI_p × 100%)
  • C) (I_s/I_p × 100%)
  • D) (V_s/V_p × 100%)

Answer Key

• Q1: C - Step-down transformer has ( N_s/N_p < 1 ).
• Q2: A - (V_s = 220 × 55/110 = 110 V).
• Q3: B - Efficiency (η = V_sI_s/V_pI_p × 100%).

Memory Aids

• Step-down: Secondary turns < Primary turns.
• Voltage decreases, current increases.
• Efficiency formula: (η = V_sI_s/V_pI_p × 100%).
• Losses: Copper, hysteresis, eddy current, flux leakage.

Concepts

• Converts high AC voltage to low AC voltage.
• Secondary turns N_s < N_p, so V_s < V_p.
• Voltage ratio: ɍrac{V_s}{V_p} = ɍrac{N_s}{N_p}.
• Ideal transformer power: V_p I_p = V_s I_s.
• Efficiency: η = ɍrac{P_{out}}{P_{in}} × 100% = ɍrac{V_s I_s}{V_p I_p} × 100%.
• Transformer obeys conservation of energy: increases voltage by decreasing current and vice versa.
• Energy losses include copper, hysteresis, eddy current, and flux losses.

Solved Examples

• Ex 1: N_p=200, N_s=100, V_p=240 V. Find V_s.
  • V_s = V_p × ɍrac{N_s}{N_p} = 240 × ɍrac{100}{200} = 120 V.

Practice Set

• Q1: For step-down transformer, ɍrac{N_s}{N_p} is:
  • A) > 1
  • B) = 1
  • C) < 1
  • D) = 0
• Q2: If V_p=220 V, N_p=110, N_s=55, find V_s.
  • A) 110 V
  • B) 220 V
  • C) 55 V
  • D) 275 V
• Q3: Efficiency η of transformer is given by?
  • A) ɍrac{V_p I_p}{V_s I_s} × 100%
  • B) ɍrac{V_s I_s}{V_p I_p} × 100%
  • C) ɍrac{I_s}{I_p} × 100%
  • D) ɍrac{V_s}{V_p} × 100%

Answer Key

• Q1: C - Step-down transformer has ɍrac{N_s}{N_p} < 1.
• Q2: A - V_s = 220 × ɍrac{55}{110} = 110 V.
• Q3: B - Efficiency η = ɍrac{V_s I_s}{V_p I_p} × 100%.

Memory Aids

• Step-down: Secondary turns < Primary turns.
• Voltage decreases, current increases.
• Efficiency formula: η = ɍrac{V_s I_s}{V_p I_p} × 100%.
• Losses: Copper, hysteresis, eddy current, flux leakage.

Concepts

• Converts high AC voltage to low AC voltage.
• Secondary turns (N_s < N_p), so (V_s < V_p).
• Voltage ratio: ( V_s/V_p = N_s/N_p ).
• Ideal transformer power: ( V_pI_p = V_sI_s ).
• Efficiency: ( η = P_out/P_in × 100% = V_sI_s/V_pI_p × 100% ).
• Transformer obeys conservation of energy: increases voltage by decreasing current and vice versa.
• Energy losses include copper, hysteresis, eddy current, and flux losses.

Solved Examples

• Ex 1: (N_p=200), (N_s=100), (V_p=240 V). Find (V_s).
  • (V_s = V_p × N_s/N_p = 240 × 100/200 = 120 V).

Practice Set

• Q1: For step-down transformer, ( N_s/N_p ) is:
  • A) > 1
  • B) = 1
  • C) < 1
  • D) = 0
• Q2: If (V_p=220 V), (N_p=110), (N_s=55), find (V_s).
  • A) 110 V
  • B) 220 V
  • C) 55 V
  • D) 275 V
• Q3: Efficiency (η) of transformer is given by?
  • A) (V_pI_p/V_sI_s × 100%)
  • B) (V_sI_s/V_pI_p × 100%)
  • C) (I_s/I_p × 100%)
  • D) (V_s/V_p × 100%)

Answer Key

• Q1: C - Step-down transformer has ( N_s/N_p < 1 ).
• Q2: A - (V_s = 220 × 55/110 = 110 V).
• Q3: B - Efficiency (η = V_sI_s/V_pI_p × 100%).

Memory Aids

• Step-down: Secondary turns < Primary turns.
• Voltage decreases, current increases.
• Efficiency formula: (η = V_sI_s/V_pI_p × 100%).
• Losses: Copper, hysteresis, eddy current, flux leakage.

Concepts

• Converts high AC voltage to low AC voltage.
• Secondary turns \(N_s < N_p\), so \(V_s < V_p\).
• Voltage ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \).
• Ideal transformer power: \( V_p I_p = V_s I_s \).
• Efficiency: \( \eta = \frac{P_{out}}{P_{in}} \times 100\% = \frac{V_s I_s}{V_p I_p} \times 100\% \).
• Transformer obeys conservation of energy: increases voltage by decreasing current and vice versa.
• Energy losses include copper, hysteresis, eddy current, and flux losses.

Solved Examples

• Ex 1: \(N_p=200\), \(N_s=100\), \(V_p=240\,V\). Find \(V_s\).
  • \(V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{200} = 120\,V\).

Practice Set

• Q1: For step-down transformer, \( \frac{N_s}{N_p} \) is:
  • A) > 1
  • B) = 1
  • C) < 1
  • D) = 0
• Q2: If \(V_p=220\,V\), \(N_p=110\), \(N_s=55\), find \(V_s\).
  • A) 110 V
  • B) 220 V
  • C) 55 V
  • D) 275 V
• Q3: Efficiency \(\eta\) of transformer is given by?
  • A) \(\frac{V_p I_p}{V_s I_s} \times 100\%\)
  • B) \(\frac{V_s I_s}{V_p I_p} \times 100\%\)
  • C) \(\frac{I_s}{I_p} \times 100\%\)
  • D) \(\frac{V_s}{V_p} \times 100\%\)

Answer Key

• Q1: C - Step-down transformer has \( \frac{N_s}{N_p} < 1 \).
• Q2: A - \(V_s = 220 \times \frac{55}{110} = 110\,V\).
• Q3: B - Efficiency \(\eta = \frac{V_s I_s}{V_p I_p} \times 100\%\).

Memory Aids

• Step-down: Secondary turns < Primary turns.
• Voltage decreases, current increases.
• Efficiency formula: \(\eta = \frac{V_s I_s}{V_p I_p} \times 100\%\).
• Losses: Copper, hysteresis, eddy current, flux leakage.