Module II: AC Circuits

1. Generation of Alternating Voltage

How is AC Voltage Generated?

Alternating current (AC) is generated when a coil rotates in a magnetic field, or when a magnetic field rotates around a stationary coil. This process is based on Faraday's Law of Electromagnetic Induction.

When a conductor moves through a magnetic field, an EMF (voltage) is induced in the conductor. As the coil rotates, the direction of the induced EMF reverses every half rotation, creating an alternating voltage.

AC Generation - Rotating Coil in Magnetic Field

N S ω AC Output

Coil rotating in magnetic field produces sinusoidal AC voltage

Instantaneous Value of AC Voltage

v(t) = Vm sin(ωt)

Or with phase angle:

v(t) = Vm sin(ωt + φ)

Where:

2. Basic Definitions

Cycle

One complete set of positive and negative values of an alternating quantity. The waveform repeats after each cycle.

Time Period (T)

The time taken to complete one cycle. Measured in seconds (s).

T = 1/f

Frequency (f)

Number of cycles completed per second. Measured in Hertz (Hz). In India, standard frequency is 50 Hz.

f = 1/T

Angular Frequency (ω)

Rate of change of angle in radians per second.

ω = 2πf = 2π/T rad/s

Amplitude / Peak Value (Vm or Im)

The maximum value reached by the alternating quantity during one cycle.

Peak-to-Peak Value (Vpp)

The difference between maximum positive and maximum negative values.

Vpp = 2 × Vm

AC Waveform with Key Parameters

t v +Vm −Vm Vpp T (One Cycle) Peak (+) Peak (−)

3. Average and RMS Values

Average Value

The average value of an AC quantity over a complete cycle is zero (positive and negative halves cancel out). However, the average value over a half-cycle is meaningful.

The average value represents the DC equivalent that would transfer the same amount of charge over a half-cycle.

Average Value Formula (for sinusoidal wave)

Vavg = (2/π) × Vm = 0.637 × Vm
Iavg = (2/π) × Im = 0.637 × Im

RMS (Root Mean Square) Value

The RMS value is the effective value of an AC quantity. It is defined as the DC equivalent that would produce the same heating effect (power dissipation) in a resistance.

This is the most important value for AC circuits! When we say "220V AC supply", we mean 220V RMS.

RMS Value Formula (for sinusoidal wave)

Vrms = Vm / √2 = 0.707 × Vm
Irms = Im / √2 = 0.707 × Im

Important Relationships

Quantity Formula Value for Sine Wave
Peak Value Vm Vm
RMS Value Vm/√2 0.707 Vm
Average Value 2Vm 0.637 Vm
Form Factor Vrms/Vavg 1.11
Peak Factor (Crest Factor) Vm/Vrms 1.414 (√2)

Example: RMS and Average Values

Problem

An AC voltage has a peak value of 340V. Find: (a) RMS value, (b) Average value, (c) Form factor

Solution
  1. RMS Value: Vrms = Vm/√2 = 340/1.414 = 240.4V
  2. Average Value: Vavg = 0.637 × Vm = 0.637 × 340 = 216.6V
  3. Form Factor = Vrms/Vavg = 240.4/216.6 = 1.11

4. Phasors and Phase Difference

What is a Phasor?

A phasor is a rotating vector that represents a sinusoidal quantity. It rotates at angular velocity ω and its projection on the vertical axis gives the instantaneous value.

Phasors allow us to represent magnitude and phase of AC quantities graphically, making AC circuit analysis much simpler.

Phasor Representation

V θ V sin θ Reference (0°) +90° ω Resulting Waveform

Phase Difference

Phase difference is the angular displacement between two phasors (or two sinusoidal waveforms) of the same frequency.

Phase Difference Between Two Waveforms

V I φ Current lags Voltage by phase angle φ

Example: Phasor Addition

Problem

Two voltages are given as: V₁ = 100 sin(ωt) and V₂ = 50 sin(ωt + 90°). Find the resultant voltage.

Solution
  1. Convert to phasor form (RMS phasors):
    V₁ = 100/√2 ∠0° = 70.7∠0° V
    V₂ = 50/√2 ∠90° = 35.4∠90° V
  2. Convert to rectangular form:
    V₁ = 70.7 + j0
    V₂ = 0 + j35.4
  3. Add: VR = 70.7 + j35.4
  4. Magnitude: |VR| = √(70.7² + 35.4²) = √(4998.5 + 1253.2) = √6251.7 = 79.1V (RMS)
  5. Phase: θ = tan⁻¹(35.4/70.7) = tan⁻¹(0.5) = 26.6°
  6. Peak value = 79.1 × √2 = 111.9V
  7. Result: VR = 111.9 sin(ωt + 26.6°) V

5. Single-Phase AC Circuits (R, L, C, RL, RC, RLC)

5.1 Pure Resistive (R) Circuit

In a pure resistive circuit, voltage and current are in phase (phase difference = 0°).

Formulas

V = IR
I = V/R
P = I²R = V²/R = VI
R ~ V

V and I in phase

5.2 Pure Inductive (L) Circuit

In a pure inductive circuit, current lags voltage by 90°. The inductor opposes change in current.

Inductive Reactance

XL = ωL = 2πfL (in Ω)
V = I × XL

Power = 0 (Average power consumed is zero - energy is stored and returned)

Remember: "ELI the ICE man" - In inductor (L), voltage (E) leads current (I)

5.3 Pure Capacitive (C) Circuit

In a pure capacitive circuit, current leads voltage by 90°. The capacitor opposes change in voltage.

Capacitive Reactance

XC = 1/(ωC) = 1/(2πfC) (in Ω)
V = I × XC

Power = 0 (Average power consumed is zero)

Remember: "ELI the ICE man" - In capacitor (C), current (I) leads voltage (E)

5.4 Series R-L Circuit

In an R-L series circuit, current lags voltage by angle φ (between 0° and 90°).

R-L Series Circuit Formulas

Z = √(R² + XL²) (Impedance)
I = V/Z
tan φ = XL/R
cos φ = R/Z (Power Factor, lagging)

Impedance Triangle for R-L Circuit

R XL Z φ

Z² = R² + XL²

5.5 Series R-C Circuit

In an R-C series circuit, current leads voltage by angle φ (between 0° and 90°).

R-C Series Circuit Formulas

Z = √(R² + XC²)
I = V/Z
tan φ = XC/R
cos φ = R/Z (Power Factor, leading)

5.6 Series R-L-C Circuit

The R-L-C series circuit combines all three elements. The net reactance determines whether the circuit is inductive or capacitive.

R-L-C Series Circuit Formulas

Z = √(R² + (XL − XC)²)
I = V/Z
tan φ = (XL − XC)/R

Example: R-L-C Series Circuit

Problem

A series RLC circuit has R = 10Ω, L = 0.1H, C = 100μF connected to 230V, 50Hz supply. Find impedance, current, and power factor.

Solution
  1. Calculate XL: XL = 2πfL = 2π × 50 × 0.1 = 31.4Ω
  2. Calculate XC: XC = 1/(2πfC) = 1/(2π × 50 × 100×10⁻⁶) = 31.8Ω
  3. Net reactance: X = XL − XC = 31.4 − 31.8 = −0.4Ω (capacitive)
  4. Impedance: Z = √(R² + X²) = √(100 + 0.16) = √100.16 = 10.01Ω
  5. Current: I = V/Z = 230/10.01 = 22.98A
  6. Power factor: cos φ = R/Z = 10/10.01 = 0.999 (leading)

The circuit is nearly at resonance (XL ≈ XC), so impedance is almost purely resistive!

5.7 Parallel AC Circuits

In parallel AC circuits, voltage is common across all branches. We use admittance (Y) for easier calculations.

Parallel Circuit Formulas

Y = 1/Z (Admittance in Siemens, S)
Y = G + jB

Where:

  • G = Conductance = 1/R = R/Z² (S)
  • B = Susceptance = X/Z² (S)
  • BL = −1/XL (Inductive susceptance, negative)
  • BC = 1/XC (Capacitive susceptance, positive)

6. Power in AC Circuits

Types of Power

Real Power (P)

Also called Active Power or True Power. This is the actual power consumed by the resistance and converted to heat/work.

P = VI cos φ

Unit: Watt (W)

Reactive Power (Q)

Power that oscillates between source and reactive elements (L, C). It does no useful work.

Q = VI sin φ

Unit: VAR (Volt-Ampere Reactive)

Apparent Power (S)

The total power supplied by the source. It's the product of RMS voltage and current.

S = VI

Unit: VA (Volt-Ampere)

Power Triangle

S² = P² + Q²
S = √(P² + Q²)
Power Factor = cos φ = P/S
P (Watts) Q (VAR) S (VA) φ

Power Factor (cos φ)

Example: AC Power Calculation

Problem

A load draws 10A from a 230V supply at 0.8 power factor lagging. Find real power, reactive power, and apparent power.

Solution
  1. Apparent Power: S = VI = 230 × 10 = 2300 VA
  2. Real Power: P = VI cos φ = 2300 × 0.8 = 1840 W
  3. Find sin φ: sin φ = √(1 − cos²φ) = √(1 − 0.64) = 0.6
  4. Reactive Power: Q = VI sin φ = 2300 × 0.6 = 1380 VAR

Verify: S = √(P² + Q²) = √(1840² + 1380²) = √(3385600 + 1904400) = √5290000 = 2300 VA ✓

7. Admittance (Y)

Admittance is the reciprocal of impedance. It represents how easily a circuit allows current to flow.

Admittance Formulas

Y = 1/Z = G + jB (in Siemens, S)
|Y| = 1/|Z| = √(G² + B²)

Components:

  • Conductance (G) = Real part of Y = R/(R² + X²)
  • Susceptance (B) = Imaginary part of Y = −X/(R² + X²)

8. Resonance (Theory Only)

8.1 Series Resonance

What is Series Resonance?

Series resonance occurs when XL = XC in a series RLC circuit. At this condition:

Resonant Frequency

fr = 1/(2π√LC)

At resonance: ωrL = 1/(ωrC)

Characteristics of Series Resonance

8.2 Parallel Resonance

What is Parallel Resonance?

Parallel resonance occurs in a parallel LC circuit when the susceptances are equal (BL = BC). At this condition:

Resonant Frequency (Ideal Case)

fr = 1/(2π√LC)

Characteristics of Parallel Resonance

Property Series Resonance Parallel Resonance
Impedance at resonance Minimum (= R) Maximum
Current at resonance Maximum Minimum
Power factor Unity (1) Unity (1)
Resonant frequency fr = 1/(2π√LC) fr = 1/(2π√LC)
Application Tuning circuits, filters Oscillators, tank circuits

9. Three-Phase Systems

9.1 Generation of Three-Phase Voltages

Three-phase voltage is generated by three identical coils placed 120° apart, rotating in a uniform magnetic field. This produces three sinusoidal voltages of equal magnitude but displaced by 120° in phase.

Three-Phase Voltages

VR = Vm sin(ωt)
VY = Vm sin(ωt − 120°)
VB = Vm sin(ωt − 240°) = Vm sin(ωt + 120°)

Three-Phase Waveforms

R Y B 120° 240°

Three phases displaced by 120° each

Advantages of Three-Phase System

9.2 Star (Y) Connection

In star connection, one end of each phase winding is connected to a common point called the neutral point (N). The other ends are the line terminals (R, Y, B).

Star (Y) Connection

N R Y B Vph

Star Connection Relationships

VL = √3 × Vph
IL = Iph

Where:

9.3 Delta (Δ) Connection

In delta connection, the three phase windings are connected end-to-end to form a closed triangle. The line terminals are taken from the junction points.

Delta (Δ) Connection

R Y B Vph = VL

Delta Connection Relationships

VL = Vph
IL = √3 × Iph

9.4 Three-Phase Power

Three-Phase Power Formulas

P = √3 × VL × IL × cos φ (Watts)
Q = √3 × VL × IL × sin φ (VAR)
S = √3 × VL × IL (VA)

These formulas apply to both Star and Delta connections!

Parameter Star (Y) Connection Delta (Δ) Connection
Line Voltage (VL) √3 × Vph Vph
Line Current (IL) Iph √3 × Iph
Neutral Available Not available
Phase Voltage VL/√3 (lower) VL (higher)
Used for Transmission, distribution Motors, transformers

Example: Three-Phase Power

Problem

A three-phase, star-connected load draws a line current of 20A from a 415V, 50Hz supply at 0.8 power factor lagging. Calculate: (a) Phase voltage, (b) Total power consumed.

Solution
  1. For star connection: Vph = VL/√3 = 415/√3 = 239.6V
  2. Line current = Phase current = 20A (for star)
  3. Total Power: P = √3 × VL × IL × cos φ
    P = √3 × 415 × 20 × 0.8
    P = 1.732 × 415 × 20 × 0.8
    P = 11,501 W = 11.5 kW

Summary: AC Circuits Key Formulas

Quantity Formula Unit
RMS Value Vrms = Vm/√2 = 0.707 Vm V or A
Average Value Vavg = 0.637 Vm V or A
Inductive Reactance XL = 2πfL = ωL Ω
Capacitive Reactance XC = 1/(2πfC) = 1/(ωC) Ω
Impedance (Series RLC) Z = √(R² + (XL − XC)²) Ω
Resonant Frequency fr = 1/(2π√LC) Hz
Real Power P = VI cos φ W
Reactive Power Q = VI sin φ VAR
Apparent Power S = VI VA
3-Phase Power P = √3 VL IL cos φ W