Chapter 22: Alternating Current

Introduction

Alternating Current (AC) vs Direct Current (DC)

• AC current magnitude changes continuously and reverses direction periodically.
• DC current flows with constant magnitude and fixed direction.
• AC voltage can be efficiently transformed to different voltages using transformers.

Solved Examples

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

Practice Set

• Q1: Which current reverses direction periodically?
  • A) DC
  • B) AC
  • C) Both
  • D) None
• Q2: Why is AC preferred over DC for power transmission?
  • A) AC is safer
  • B) AC voltage can be easily transformed
  • C) DC is more expensive
  • D) DC cannot flow in circuits
• Q3: In the graph, which line represents DC?
  • A) Wavy line
  • B) Straight line
  • C) Both
  • D) None

Answer Key

• Q1: B - AC reverses direction periodically.
• Q2: B - AC voltage can be transformed easily using transformers.
• Q3: B - Straight line represents DC current.

Memory Aids

• AC = Alternates direction; DC = Direct flow.
• Transformers work only with AC.

Fundamentals of Alternating Current

AC Current and Voltage Representation

• Instantaneous current: \( I = I_m \sin \omega t \) or \( I = I_m \cos \omega t \)
• \( I_m \): Peak current (amplitude)
• \( \omega = 2\pi / T = 2\pi \nu \): Angular frequency (rad/s)
• \( T \): Time period (s), \( \nu \): Frequency (Hz)

Solved Examples

• Ex 1: Calculate angular frequency if frequency \( \nu = 50 \) Hz.
  • \( \omega = 2\pi \times 50 = 314.16 \) rad/s

Practice Set

• Q1: What is the expression for instantaneous current in AC?
  • A) \( I = I_m \sin \omega t \)
  • B) \( I = I_m \cos \omega t \)
  • C) Both A and B
  • D) None
• Q2: If \( T = 0.02 \) s, find frequency \( \nu \).
  • A) 50 Hz
  • B) 20 Hz
  • C) 100 Hz
  • D) 0.02 Hz
• Q3: Calculate \( \omega \) for \( \nu = 60 \) Hz.
  • A) 120 rad/s
  • B) 377 rad/s
  • C) 60 rad/s
  • D) 314 rad/s

Answer Key

• Q1: C - Both sine and cosine represent instantaneous current.
• Q2: A - \( \nu = 1/T = 1/0.02 = 50 \) Hz.
• Q3: B - \( \omega = 2\pi \times 60 = 377 \) rad/s.

Memory Aids

• \( \omega = 2\pi \nu \) always.
• Sine starts at zero; cosine starts at peak.

Average and RMS Values

Definitions and Formulas

• Average value (mean) 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 equals the DC value producing the same heat in a resistor.
• RMS is also called virtual or apparent value.

Solved Examples

• Ex 1: Find RMS current if \( I_m = 10 \) A.
  • \( I_{rms} = 10 / \sqrt{2} = 7.07 \) A
• Ex 2: Calculate average current for half cycle if \( I_m = 10 \) A.
  • \( \langle I \rangle = 2 \times 10 / \pi = 6.37 \) A

Practice Set

• Q1: RMS value of current is:
  • A) \( I_m \)
  • B) \( I_m / 2 \)
  • C) \( I_m / \sqrt{2} \)
  • D) \( 2 I_m / \pi \)
• Q2: Average current over half cycle is:
  • A) \( I_m / 2 \)
  • B) \( 2 I_m / \pi \)
  • C) \( I_m / \sqrt{2} \)
  • D) \( I_m \)
• Q3: If RMS current is 5 A, find peak current \( I_m \).
  • A) 3.54 A
  • B) 7.07 A
  • C) 10 A
  • D) 5 A

Answer Key

• Q1: C - RMS current is \( I_m / \sqrt{2} \).
• Q2: B - Average over half cycle is \( 2 I_m / \pi \).
• Q3: B - \( I_m = I_{rms} \times \sqrt{2} = 5 \times 1.414 = 7.07 \) A.

Memory Aids

• Average half cycle = \( 2 I_m / \pi \).
• RMS = peak / \( \sqrt{2} \).

Phase and Phase Difference

Key Concepts

• Current with phase: \( I = I_m \sin (\omega t \pm \phi) \)
• Initial phase \( \phi \) is constant; instantaneous phase \( \omega t \) varies with time.
• Phase difference between voltage \( \varepsilon = \varepsilon_m \sin (\omega t + \phi_1) \) and current \( I = I_m \sin (\omega t + \phi_2) \) is \( \phi = \phi_2 - \phi_1 \).
• Voltage leads current if \( \varepsilon \) reaches max before \( I \) (\( \varepsilon \) leads, \( I \) lags).
• Voltage lags current if \( \varepsilon \) reaches max after \( I \) (\( \varepsilon \) lags, \( I \) leads).

Solved Examples

• Ex 1: If \( \varepsilon = \varepsilon_m \sin \omega t \), and \( I = I_m \sin (\omega t - \phi) \), current lags voltage by \( \phi \).
• Ex 2: If \( \varepsilon = \varepsilon_m \sin (\omega t + \phi) \), and \( I = I_m \sin \omega t \), voltage leads current by \( \phi \).

Practice Set

• Q1: If current lags voltage, phase difference \( \phi \) is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined
• Q2: Expression for current leading voltage is:
  • A) \( I = I_m \sin (\omega t - \phi) \)
  • B) \( I = I_m \sin (\omega t + \phi) \)
  • C) \( I = I_m \sin \omega t \)
  • D) \( I = I_m \cos \omega t \)
• Q3: Phase difference between voltage and current is \( 30^\circ \). If voltage leads, current expression is:
  • A) \( I = I_m \sin (\omega t - 30^\circ) \)
  • B) \( I = I_m \sin (\omega t + 30^\circ) \)
  • C) \( I = I_m \sin \omega t \)
  • D) \( I = I_m \cos (\omega t + 30^\circ) \)

Answer Key

• Q1: A - Current lags voltage means positive phase difference.
• Q2: B - Current leading voltage: \( I = I_m \sin (\omega t + \phi) \).
• Q3: A - Current lags voltage by 30°, so \( I = I_m \sin (\omega t - 30^\circ) \).

Memory Aids

• Voltage leads current: \( \varepsilon \) max before \( I \).
• Current leads voltage: \( I \) max before \( \varepsilon \).

AC Voltage Applied to Resistor

Key Points

• Voltage: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Current: \( I = I_m \sin \omega t \)
• Voltage and current are in phase (peaks coincide).
• Instantaneous power: \( P_r = I^2 R = I_m^2 R \sin^2 \omega t \)
• Average power over cycle: \( P = \langle P_r \rangle = \frac{1}{2} I_m^2 R = I_{rms}^2 R = \frac{\varepsilon_{rms}^2}{R} \)

Solved Examples

• Ex 1: Calculate average power if \( I_m = 4 \) A and \( R = 5 \) \( \Omega \).
  • \( P = \frac{1}{2} \times 4^2 \times 5 = 40 \) W
• Ex 2: Find RMS voltage if \( \varepsilon_m = 220 \) V.
  • \( \varepsilon_{rms} = 220 / \sqrt{2} = 155.56 \) V

Practice Set

• Q1: In a resistor, current and voltage are:
  • A) Out of phase
  • B) In phase
  • C) 90° out of phase
  • D) 180° out of phase
• Q2: Average power dissipated in resistor is:
  • A) \( I_m^2 R \)
  • B) \( \frac{1}{2} I_m^2 R \)
  • C) \( I_{rms} R \)
  • D) \( \varepsilon_m^2 / R \)
• Q3: If \( I_m = 6 \) A and \( R = 10 \) \( \Omega \), find average power.
  • A) 180 W
  • B) 360 W
  • C) 720 W
  • D) 60 W

Answer Key

• Q1: B - Current and voltage are in phase in resistor.
• Q2: B - Average power is \( \frac{1}{2} I_m^2 R \).
• Q3: A - \( P = \frac{1}{2} \times 6^2 \times 10 = 180 \) W.

Memory Aids

• In resistor, voltage and current peaks align.
• Average power = half of peak power.

Representation of AC Current and Voltage by Rotating Vectors–Phasors

Concepts

• Phasor: A vector rotating at angular speed \( \omega \) representing sinusoidal quantities.
• Voltage \( \varepsilon \) and current \( I \) in AC circuits are represented by phasors with magnitudes \( \varepsilon_m \) and \( I_m \).
• Vertical component of phasor gives instantaneous value: \( \varepsilon_m \sin \omega t \), \( I_m \sin \omega t \).
• Phasor diagram shows phase relationship; angle between voltage and current phasors is phase difference \( \phi \).
• For resistor-only circuit, voltage and current phasors are in phase (\( \phi = 0 \)).

Solved Examples

• Ex 1: In a resistor AC circuit, if \( \varepsilon_m = 10\,V \) and \( I_m = 2\,A \), 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: Find instantaneous current at \( \omega t = 60^\circ \) if \( I_m = 3\,A \).
  • \( I = I_m \sin \omega t = 3 \times \sin 60^\circ = 3 \times 0.866 = 2.598\,A \).

Practice Set

• Q1: What does a phasor represent in AC circuits?
  • A) Constant voltage
  • B) Rotating vector representing sinusoidal quantity
  • C) Direct current
  • D) Resistance
• Q2: In a purely resistive AC circuit, the phase difference between voltage and current is:
  • A) 90°
  • B) 180°
  • C) 0°
  • D) 45°
• Q3: If the peak voltage is 20 V, what is the instantaneous voltage at \( \omega t = 45^\circ \)?
  • A) 10 V
  • B) 14.14 V
  • C) 20 V
  • D) 0 V

Answer Key

• Q1: B — Phasor is a rotating vector representing sinusoidal voltage or current.
• Q2: C — Voltage and current are in phase in a resistive circuit, so phase difference is 0°.
• Q3: B — Instantaneous voltage \( = 20 \times \sin 45^\circ = 20 \times 0.707 = 14.14\,V \).

Memory Aids

• Phasor = Rotating vector \( \rightarrow \) represents AC sinusoid.
• Resistor circuit: voltage and current phasors align (0° phase difference).
• Instantaneous value = peak × \( \sin \omega t \).

AC Voltage Applied to an Inductor

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°)

Current Amplitude

• Amplitude of current: \( I_m = \frac{\varepsilon_m}{\omega L} \)
• \( \omega = 2 \pi \nu \) (angular frequency in rad/s)

Inductive Reactance

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

Solved Examples

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

Practice Set

• Q1: What is the phase difference between voltage and current in a pure inductive circuit?
  • 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) \( \frac{L}{\omega} \)
  • D) \( \omega / L \)
• Q3: For an inductor of 0.2 H connected to 60 Hz AC supply, calculate \( X_L \).
  • A) 75.4 \( \Omega \)
  • B) 37.7 \( \Omega \)
  • C) 12.56 \( \Omega \)
  • D) 18.85 \( \Omega \)

Answer Key

• Q1: C) \( \pi/2 \) — Current lags voltage by 90° in pure inductance.
• Q2: B) \( \omega L \) — Inductive reactance is product of angular frequency and inductance.
• Q3: B) 75.4 \( \Omega \) — \( X_L = 2 \pi \times 60 \times 0.2 = 75.4 \Omega \).

Memory Aids

• Current lags voltage by 90° in inductors.
• Inductive reactance \( X_L = 2 \pi \nu L \).
• Higher frequency or inductance increases \( X_L \), reducing current amplitude.

Power in an Inductive Circuit

Instantaneous Power

• Instantaneous power: \( p_l = I \varepsilon = - \frac{I_m \varepsilon_m}{2} \sin (2 \omega t) \)
• Power oscillates sinusoidally, changing sign over time

Average Power

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

Solved Examples

• Ex 1: Calculate average power for \( I_m = 2\,A \), \( \varepsilon_m = 10\,V \).
  • Use \( P_l = 0 \) for pure inductive load.

Practice Set

• Q1: What is the average power consumed by a pure inductor in AC circuit?
  • A) Zero
  • B) Maximum
  • C) Half of maximum
  • D) Depends on frequency
• 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 does average power in an inductor equal zero?
  • A) No current flows
  • B) Voltage and current are in phase
  • C) Energy is stored and returned each cycle
  • D) Resistance is zero

Answer Key

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

Memory Aids

• Average power in pure inductors = 0.
• Instantaneous power oscillates at twice the supply frequency.
• Inductors store energy in magnetic field, return it to circuit.

Phasor and Waveform Representation

Phasor Diagram

• Voltage and current represented as rotating vectors (phasors)
• Current phasor lags voltage phasor by \( \pi/2 \)
• Phasor length represents amplitude

Waveform Graphs

• Voltage: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Current: \( I = I_m \sin (\omega t - \pi/2) \)
• Graphs show sinusoidal variation with phase difference

Solved Examples

• Ex 1: Identify phase difference from phasor diagram where current vector is perpendicular and behind voltage vector.
  • Phase difference = \( \pi/2 \) lag of current.

Practice Set

• Q1: In phasor diagram, current vector lags voltage vector by:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: The waveform of current in an inductor is:
  • A) In phase with voltage
  • B) Leading voltage by 90°
  • C) Lagging voltage by 90°
  • D) Opposite phase to voltage
• Q3: Which graph correctly shows voltage and current in an inductive AC circuit?
  • A) Both in phase
  • B) Current leads voltage
  • C) Current lags voltage by 90°
  • D) Voltage lags current by 90°

Answer Key

• Q1: C) 90° — Current lags voltage by 90° in inductors.
• Q2: C) Lagging voltage by 90° — Characteristic of inductive load.
• Q3: C) Current lags voltage by 90° — Correct waveform relation.

Memory Aids

• Phasor current vector is 90° behind voltage vector in inductors.
• Waveform lag = \( \pi/2 \) radians or 90°.
• Voltage leads current in inductive circuits.

AC Voltage Applied to a Capacitor

Voltage and Current Relationship

• Voltage across capacitor: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Current through capacitor: \( I = I_m \sin (\omega t + \frac{\pi}{2}) \)
• Current leads voltage by \( \frac{\pi}{2} \) radians (90°)
• Peak current: \( I_m = \varepsilon_m \omega C \)

Capacitive Reactance

• Capacitive reactance: \( X_c = \frac{1}{\omega C} = \frac{1}{2 \pi \nu C} \)
• Acts like resistance opposing AC current

Power in Capacitor

• Instantaneous power: \( p_c = \frac{I_m \varepsilon_m}{2} \sin 2 \omega t \)
• Average power over a cycle: \( P_c = 0 \) (no net energy loss)

Solved Examples

Ex 1

• Given \( \varepsilon_m = 10\,V \), \( C = 5 \mu F \), \( \nu = 50\,Hz \)
• Calculate \( I_m \): \( I_m = \varepsilon_m \omega C = 10 \times 2\pi \times 50 \times 5 \times 10^{-6} = 0.0157\,A \)

Practice Set

Q1

In an AC circuit with a capacitor, the current leads the voltage by:

• A) 0
• B) \( \frac{\pi}{4} \)
• C) \( \frac{\pi}{2} \)
• D) \( \pi \)

Q2

Capacitive reactance \( X_c \) is given by:

• A) \( \omega C \)
• B) \( \frac{1}{\omega C} \)
• C) \( \omega L \)
• D) \( R \)

Q3

A capacitor of 10 \( \mu F \) is connected to 100 V AC of frequency 50 Hz. The capacitive reactance is:

• A) 318 \( \Omega \)
• B) 3180 \( \Omega \)
• C) 31.8 \( \Omega \)
• D) 3.18 \( \Omega \)

Answer Key

• Q1: C) \( \frac{\pi}{2} \) — Current leads voltage by 90° in capacitor.
• Q2: B) \( \frac{1}{\omega C} \) — Definition of capacitive reactance.
• Q3: A) 318 \( \Omega \) — \( X_c = \frac{1}{2\pi \times 50 \times 10 \times 10^{-6}} = 318 \Omega \).

Memory Aids

• Current leads voltage by 90° in capacitor.
• Capacitive reactance \( X_c = \frac{1}{2\pi \nu C} \).
• Average power in capacitor is zero.

Phasor Diagram and Waveform

Phasor Representation

• Phasors represent AC quantities as rotating vectors.
• Length = amplitude; angle = phase.
• Current phasor leads voltage phasor by \( 90^\circ \) in capacitor.

Waveform Graphs

• Voltage and current are sinusoidal waves.
• Current wave peaks before voltage wave by \( \frac{\pi}{2} \).
• Phase difference visible as time shift between peaks.

Solved Examples

Ex 1

• Given voltage \( \varepsilon = 10 \sin \omega t \), current leads by \( 90^\circ \)
• Current phasor at \( 90^\circ \) ahead of voltage phasor.

Practice Set

Q1

In a phasor diagram for a capacitor, the current vector is:

• A) In phase with voltage
• B) Lagging voltage by 90°
• C) Leading voltage by 90°
• D) Opposite to voltage

Q2

The phase difference between current and voltage in a capacitor is:

• A) 0°
• B) 45°
• C) 90°
• D) 180°

Q3

In waveform graphs, if current leads voltage by \( \frac{\pi}{2} \), the current wave peaks:

• A) At the same time as voltage
• B) After voltage
• C) Before voltage
• D) Never peaks

Answer Key

• Q1: C) Leading voltage by 90° — Current leads voltage in capacitor.
• Q2: C) 90° — Phase difference in capacitor.
• Q3: C) Before voltage — Current wave peaks earlier.

Memory Aids

• Phasor length = amplitude; angle = phase.
• Current phasor leads voltage by 90° in capacitor.
• Wave peak order shows phase lead/lag.

Combination of RL, RC and LC in AC Circuit

Basic Circuit Types

• R-L: Resistor and Inductor in series
• R-C: Resistor and Capacitor in series
• L-C: Inductor and Capacitor in series

Impedance and Phase

• R-L: \( Z = \sqrt{R^2 + (\omega L)^2} \), voltage leads current by \( \phi = \tan^{-1} \frac{\omega L}{R} \)
• R-C: \( Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \), current leads voltage by \( \phi = \tan^{-1} \frac{1}{\omega CR} \)
• L-C: \( Z = |\omega L - \frac{1}{\omega C}| \), voltage and current phase depends on which reactance dominates

Frequency Behavior

• At low frequency, inductive reactance \( \omega L \) is small, capacitive reactance \( \frac{1}{\omega C} \) is large
• At high frequency, inductive reactance is large, capacitive reactance is small
• Circuits behave like dominant element at frequency extremes

Solved Examples

Ex 1

• R = 10 \( \Omega \), L = 0.1 H, \( \nu = 50 Hz \)
• Calculate impedance: \( Z = \sqrt{10^2 + (2\pi \times 50 \times 0.1)^2} = \sqrt{100 + 98.7} = 14.1 \ \Omega \)

Practice Set

Q1

In an R-L circuit, the phase angle \( \phi \) between voltage and current is:

• A) 0°
• B) Between 0° and 90° (voltage leads current)
• C) 90° (current leads voltage)
• D) 180°

Q2

In an R-C circuit, the impedance is:

• A) \( \sqrt{R^2 + (\omega L)^2} \)
• B) \( \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \)
• C) \( R + \omega L \)
• D) \( R + \frac{1}{\omega C} \)

Q3

In an L-C circuit, at resonance frequency \( \nu_0 \), the impedance is:

• A) Maximum
• B) Minimum (ideally zero)
• C) Equal to R
• D) Infinite

Answer Key

• Q1: B) Between 0° and 90° — Voltage leads current in R-L circuit.
• Q2: B) \( \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \) — Formula for R-C impedance.
• Q3: B) Minimum (ideally zero) — At resonance, inductive and capacitive reactances cancel.

Memory Aids

• R-L: Voltage leads current; impedance \( Z = \sqrt{R^2 + (\omega L)^2} \).
• R-C: Current leads voltage; impedance \( Z = \sqrt{R^2 + (1/\omega C)^2} \).
• L-C: Resonance when \( \omega L = 1/\omega C \), impedance minimum.

AC Voltage Applied to a Series LCR Circuit

Fundamentals

• Voltage applied: \( \varepsilon = \varepsilon_m \sin \omega t \)
• Voltage drops: \( V_R \) (resistor), \( V_L \) (inductor), \( V_C \) (capacitor)
• Voltages across capacitor and inductor are opposite in direction
• Total voltage relation: \( \varepsilon_m^2 = I_m^2 \left[R^2 + (X_C - X_L)^2 \right] \)
• 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 between voltage and current: \( \tan \phi = \frac{X_C - X_L}{R} \)
• 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 \( Z \) for \( R=10\Omega, X_C=30\Omega, X_L=20\Omega \).
  • \( Z = \sqrt{10^2 + (30-20)^2} = \sqrt{100 + 100} = \sqrt{200} = 14.14\Omega \)
• Ex 2: Find current amplitude if \( \varepsilon_m = 100V \) and \( Z=14.14\Omega \).
  • \( I_m = \frac{100}{14.14} = 7.07A \)

Practice Set

• Q1: In a series LCR circuit, if \( X_C = 40\Omega, X_L = 30\Omega, R = 10\Omega \), what is the impedance \( Z \)?
  • A) 10 \( \Omega \)
  • B) 20 \( \Omega \)
  • C) 14.14 \( \Omega \)
  • D) 50 \( \Omega \)
• Q2: For the above circuit, if \( \varepsilon_m = 100V \), what is the current amplitude \( I_m \)?
  • A) 5 A
  • B) 7.07 A
  • C) 10 A
  • D) 2.5 A
• Q3: If \( \tan \phi = 1 \) and \( R = 10\Omega \), what is \( X_C - X_L \)?
  • A) 5 \( \Omega \)
  • B) 10 \( \Omega \)
  • C) 15 \( \Omega \)
  • D) 20 \( \Omega \)

Answer Key

• Q1: C; \( Z = \sqrt{10^2 + (40-30)^2} = \sqrt{100 + 100} = 14.14\Omega \)
• Q2: B; \( I_m = \frac{100}{14.14} = 7.07A \)
• Q3: B; \( \tan \phi = \frac{X_C - X_L}{R} = 1 \Rightarrow X_C - X_L = R = 10\Omega \)

Memory Aids

• Impedance formula: \( Z = \sqrt{R^2 + (X_C - X_L)^2} \)
• Current amplitude: \( I_m = \varepsilon_m / Z \)
• Phase angle: \( \tan \phi = (X_C - X_L)/R \)
• Current leads if capacitive, lags if inductive

Resonance

Key Points

• Resonance occurs when inductive reactance equals capacitive reactance: \( X_L = X_C \)
• At resonance, impedance is minimum: \( Z = R \)
• Current amplitude at resonance: \( I_m = \frac{\varepsilon_m}{R} \)
• Resonant angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
• Resonant frequency: \( \nu_r = \frac{1}{2\pi \sqrt{LC}} \)
• Resonance does not occur in RL or RC circuits
• Higher resistance lowers peak current at resonance

Solved Examples

• Ex 1: Find resonant frequency for \( L=1mH, C=1nF \).
  • \( \omega_0 = \frac{1}{\sqrt{1\times10^{-3} \times 1\times10^{-9}}} = 10^6 \text{ rad/s} \)
  • \( \nu_r = \frac{1}{2\pi \times 10^{-6}} \approx 159 kHz \)
• Ex 2: Calculate current amplitude at resonance if \( \varepsilon_m=100V, R=100\Omega \).
  • \( I_m = \frac{100}{100} = 1A \)

Practice Set

• Q1: At resonance, what is the impedance of a series LCR circuit?
  • A) \( R \)
  • B) \( X_L \)
  • C) \( X_C \)
  • D) \( 0 \)
• Q2: For \( L=2mH, C=0.5nF \), find resonant frequency \( \nu_r \).
  • A) 159 kHz
  • B) 159 MHz
  • C) 225 kHz
  • D) 225 MHz
• Q3: If \( R=50\Omega, \varepsilon_m=200V \), what is current amplitude at resonance?
  • A) 2 A
  • B) 4 A
  • C) 0.25 A
  • D) 1 A

Answer Key

• Q1: A; At resonance, \( Z=R \) as \( X_L = X_C \)
• Q2: B; \( \nu_r = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{2\times10^{-3} \times 0.5\times10^{-9}}} \approx 159 \text{ MHz} \)
• Q3: A; \( I_m = \frac{\varepsilon_m}{R} = \frac{200}{50} = 4A \)

Memory Aids

• Resonance: \( X_L = X_C \)
• Minimum impedance: \( Z = R \)
• Resonant frequency: \( \nu_r = \frac{1}{2\pi \sqrt{LC}} \)
• Current max at resonance: \( I_m = \varepsilon_m / R \)

Quality Factor

Essentials

• Quality factor (Q) measures sharpness of resonance
• Defined as 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 and sharper resonance peak

Solved Examples

• Ex 1: Calculate Q for \( L=1mH, C=1nF, R=100\Omega, \omega_0=10^6 \text{ rad/s} \).
  • \( Q = \frac{10^6 \times 1\times10^{-3}}{100} = 10 \)
• Ex 2: Find Q using \( Q = \frac{1}{R} \sqrt{\frac{L}{C}} \) with same values.
  • \( Q = \frac{1}{100} \sqrt{\frac{1\times10^{-3}}{1\times10^{-9}}} = \frac{1}{100} \times 1000 = 10 \)

Practice Set

• Q1: For \( L=2mH, C=0.5nF, R=50\Omega, \omega_0=10^6 \text{ rad/s} \), find Q.
  • A) 40
  • B) 80
  • C) 20
  • D) 10
• Q2: Which of the following increases Q?
  • A) Increase R
  • B) Increase L
  • C) Increase C
  • D) Decrease L
• Q3: If Q = 50 and \( R=10\Omega \), what is \( \omega_0 L \)?
  • A) 500
  • B) 5
  • C) 50
  • D) 100

Answer Key

• Q1: B; \( Q = \frac{\omega_0 L}{R} = \frac{10^6 \times 2\times10^{-3}}{50} = 40 \)
• Q2: B; Increasing inductance \( L \) increases Q
• Q3: A; \( Q = \frac{\omega_0 L}{R} \Rightarrow \omega_0 L = Q \times R = 50 \times 10 = 500 \)

Memory Aids

• Q-factor: \( Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R} \sqrt{\frac{L}{C}} \)
• Higher Q = sharper resonance
• Q inversely proportional to resistance

Power in AC Circuit: The Power Factor

Voltage and Current in AC Circuit

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

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, \( \phi = \tan^{-1} \left( \frac{X_C - X_L}{R} \right) \).
• At resonance in LCR circuit: \( X_C = X_L = 0 \), \( \phi = 0 \), \( \cos \phi = 1 \), 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 AC circuit with \( \varepsilon_m = 100\,V \), \( I_m = 5\,A \), and phase angle \( \phi = 30^\circ \).

• Use \( P = \frac{\varepsilon_m I_m}{2} \cos \phi \)
• \( P = \frac{100 \times 5}{2} \times \cos 30^\circ = 250 \times 0.866 = 216.5\,W \)

Ex 2

Find power factor of a choke coil with \( R = 10\,\Omega \), \( L = 0.1\,H \), and frequency \( f = 50\,Hz \).

• Angular frequency: \( \omega = 2\pi f = 314 \; rad/s \)
• \( \cos \phi = \frac{R}{\sqrt{R^2 + (\omega L)^2}} = \frac{10}{\sqrt{10^2 + (314 \times 0.1)^2}} = \frac{10}{\sqrt{100 + 985.96}} = \frac{10}{32.3} = 0.31 \)

Practice Set

Q1

In a purely resistive AC circuit, the power factor is:

• A) 0
• B) 0.5
• C) 1
• D) Depends on frequency

Q2

For an LCR circuit at resonance, the phase angle \( \phi \) is:

• A) 90°
• B) 0°
• C) 45°
• D) 180°

Q3

A choke coil has \( R = 5\,\Omega \), \( L = 0.2\,H \), and operates at \( 50\,Hz \). What is its power factor?

• A) 0.16
• B) 0.5
• C) 0.8
• D) 1

Answer Key

• Q1: C) 1 — In resistive circuits, voltage and current are in phase, so power factor is 1.
• Q2: B) 0° — At resonance, inductive and capacitive reactances cancel, phase angle is zero.
• Q3: A) 0.16 — \( \cos \phi = \frac{R}{\sqrt{R^2 + (\omega L)^2}} = \frac{5}{\sqrt{25 + (2\pi \times 50 \times 0.2)^2}} = 0.16 \)

Memory Aids

• Power factor \( = \cos \phi \), angle between voltage and current.
• Resistive circuit: power factor = 1 (max power).
• Pure inductive/capacitive: power factor = 0 (no real power).
• Choke coil reduces current with minimal power loss.
• At resonance, LCR circuit behaves like resistive circuit.

LC Oscillations

Basic Concept

• LC circuit: capacitor (C) connected to inductor (L).
• Initially charged capacitor with charge \( Q_0 \).
• When switch closes, capacitor discharges, current builds magnetic field in inductor.
• Energy oscillates between electric field in capacitor and magnetic field in inductor.
• Magnetic energy: \( U_B = \frac{1}{2} L I^2 \), Electric energy: \( U_E = \frac{Q^2}{2C} \).
• Total energy constant: \( U = U_E + U_B = \text{constant} \).


Mathematical Description

• Energy conservation: \( \frac{dU}{dt} = 0 \)
• Governing equation: \( \frac{Q}{C} + L \frac{d^2 Q}{dt^2} = 0 \)
• Solution: \( Q(t) = Q_0 \cos (\omega_0 t + \phi) \), where \( \omega_0 = \frac{1}{\sqrt{LC}} \)
• Current: \( I(t) = -\frac{dQ}{dt} = I_0 \sin (\omega_0 t + \phi) \), \( I_0 = \omega_0 Q_0 \)
• Initial conditions: \( Q(0) = Q_0, I(0) = 0 \) imply \( \phi = 0 \)

Energy Oscillations

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


Mechanical Analogy

• LC oscillations analogous to spring-mass system oscillations.
• Energy oscillates between potential energy in spring and kinetic energy of mass.
• Both systems exhibit simple harmonic motion with energy conservation.


Solved Examples

Ex 1

Find the angular frequency \( \omega_0 \) of an LC circuit with \( L = 0.5\,H \) and \( C = 20\,\mu F \).

• \( \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 20 \times 10^{-6}}} = 316.2\,rad/s \)

Ex 2

Calculate maximum current \( I_0 \) if maximum charge \( Q_0 = 10^{-4} C \) and \( \omega_0 = 5000\,rad/s \).

• \( I_0 = \omega_0 Q_0 = 5000 \times 10^{-4} = 0.5\,A \)

Practice Set

Q1

In an LC circuit, the frequency of oscillation is given by:

• A) \( \frac{1}{2\pi \sqrt{LC}} \)
• B) \( \frac{1}{\sqrt{LC}} \)
• C) \( 2\pi \sqrt{LC} \)
• D) \( \sqrt{LC} \)

Q2

The total energy in an ideal LC circuit is:

• A) Constant and oscillates between electric and magnetic forms
• B) Constant and stored only in capacitor
• C) Constant and stored only in inductor
• D) Decreases with time due to resistance

Q3

For an LC circuit with \( L = 1\,mH \) and \( C = 1\,\mu F \), what is the angular frequency \( \omega_0 \)?

• A) 1000 rad/s
• B) 31622 rad/s
• C) 100000 rad/s
• D) 100 rad/s

Answer Key

• Q1: A) \( \frac{1}{2\pi \sqrt{LC}} \) — Frequency is inverse of period \( T = 2\pi \sqrt{LC} \).
• Q2: A) Constant and oscillates between electric and magnetic forms — Energy transfers between capacitor and inductor.
• Q3: B) 31622 rad/s — \( \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10^{-3} \times 1 \times 10^{-6}}} = 31622 \)

Memory Aids

• LC oscillation frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \)
• Energy oscillates between capacitor (electric) and inductor (magnetic).
• Current lags charge by 90° (sine vs cosine).
• LC circuit analogous to spring-mass SHM system.

Transformers

Overview

• Device converting low AC voltage to high AC voltage and vice versa.
• Operates on mutual induction principle.
• Ideal transformer relation: \( \frac{V_s}{V_p} = \frac{I_p}{I_s} = \frac{N_s}{N_p} = K \) (transformation ratio).

Principle

• Changing current in primary coil changes magnetic flux.
• Induces emf in secondary coil due to mutual induction.

Solved Examples

• Ex 1: If \(N_p=100\), \(N_s=200\), and \(V_p=120\,V\), 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: Transformer works on the principle of:
  • A) Electromagnetic induction
  • B) Mutual induction
  • C) Self induction
  • D) Electrostatic induction
• Q2: For an ideal transformer, if \(N_s > N_p\), then it is a:
  • A) Step-down transformer
  • B) Step-up transformer
  • C) Autotransformer
  • D) None of these
• Q3: A transformer has 500 turns in primary and 100 turns in secondary. If primary voltage is 240 V, find secondary voltage.
  • A) 48 V
  • B) 1200 V
  • C) 240 V
  • D) 50 V

Answer Key

• Q1: B) Mutual induction - Transformer operates by emf induced in secondary coil due to changing flux from primary coil.
• Q2: B) Step-up transformer - \(N_s > N_p\) means voltage increases.
• Q3: A) 48 V - \(V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{500} = 48 V\).

Memory Aids

• Transformation ratio \(K = \frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{I_p}{I_s}\).
• Step-up: \(N_s > N_p\), voltage increases.
• Step-down: \(N_s < N_p\), voltage decreases.

Step-up Transformer

Key Points

• Converts low AC voltage to high AC voltage.
• Primary coil voltage: \(V_p = -N_p \frac{d\phi}{dt}\).
• Secondary coil emf: \(\varepsilon_s = -N_s \frac{d\phi}{dt}\).
• Voltage ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} > 1 \).
• Power conservation: \(I_s V_s = I_p V_p\).
• Does not work with DC (no changing flux).


Solved Examples

• Ex 1: Primary voltage = 100 V, \(N_p=50\), \(N_s=150\). Find secondary voltage.
  • \(V_s = V_p \times \frac{N_s}{N_p} = 100 \times \frac{150}{50} = 300 V\).

Practice Set

• Q1: In a step-up transformer, the ratio \(\frac{N_s}{N_p}\) is:
  • A) Less than 1
  • B) Equal to 1
  • C) Greater than 1
  • D) Zero
• Q2: If primary voltage is 120 V and secondary voltage is 360 V, what is the turns ratio?
  • A) 1:3
  • B) 3:1
  • C) 1:1
  • D) 2:1
• Q3: Why transformers do not work on DC?
  • A) No current flows
  • B) No changing magnetic flux
  • C) Resistance is too high
  • D) Core saturates

Answer Key

• Q1: C) Greater than 1 - Step-up transformer increases voltage.
• Q2: A) 1:3 - \(\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{360}{120} = 3\).
• Q3: B) No changing magnetic flux - DC produces constant flux, no emf induced.

Memory Aids

• Step-up: \(N_s > N_p\), voltage increases.
• Voltage formula: \(V = -N \frac{d\phi}{dt}\).
• DC → no flux change → no emf.

Step-down Transformer

Key Points

• Converts high AC voltage to low AC voltage.
• Turns ratio: \(N_s < N_p\), so \(V_s < V_p\).
• Voltage ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} < 1 \).
• Ideal transformer power: \(V_p I_p = V_s I_s\).
• Efficiency: \(\eta = \frac{P_{out}}{P_{in}} \times 100\% = \frac{V_s I_s}{V_p I_p} \times 100\%\).
• Energy losses include copper, hysteresis, eddy current, and flux losses.


Solved Examples

• Ex 1: Primary voltage = 240 V, \(N_p=500\), \(N_s=100\). Find secondary voltage.
  • \(V_s = 240 \times \frac{100}{500} = 48 V\).
• Ex 2: If output power is 480 W and input power is 500 W, find efficiency.
  • \(\eta = \frac{480}{500} \times 100 = 96\%\).

Practice Set

• Q1: In a step-down transformer, the turns ratio \(\frac{N_s}{N_p}\) is:
  • A) Greater than 1
  • B) Equal to 1
  • C) Less than 1
  • D) Zero
• Q2: Efficiency of an ideal transformer is:
  • A) 0%
  • B) 50%
  • C) 100%
  • D) 75%
• Q3: Which loss is minimized by using laminated iron core?
  • A) Copper loss
  • B) Hysteresis loss
  • C) Eddy current loss
  • D) Flux loss

Answer Key

• Q1: C) Less than 1 - Step-down transformer reduces voltage.
• Q2: C) 100% - Ideal transformer has no energy loss.
• Q3: C) Eddy current loss - Laminated core reduces eddy currents.

Memory Aids

• Step-down: \(N_s < N_p\), voltage decreases.
• Efficiency formula: \(\eta = \frac{V_s I_s}{V_p I_p} \times 100\%\).
• Losses: Copper (resistance), Hysteresis (magnetisation), Eddy currents (core), Flux leakage.