12.2 Single-Phase AC Circuits & Phasor Analysis

12.2 Single-Phase AC Circuits & Phasor Analysis

Alternating current (AC) circuit theory forms the core of power engineering. In steady-state sinusoidal analysis, time-domain differential equations transform into complex algebraic equations using phasors and complex impedance. This section examines AC waveforms, phasor domain algebra, RLC resonance, active and reactive power, and industrial power factor correction.

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1. Sinusoidal Waveforms & Quantitative Parameters

A sinusoidal voltage wave is defined in the time domain as:

$$v(t) = V_m \sin(\omega t + \phi)$$

where $V_m$ is peak amplitude (V), $\omega = 2 \pi f$ is angular frequency (rad/s), $f = \frac{1}{T}$ is cyclic frequency (Hz), $T$ is period (s), and $\phi$ is phase angle (rad or degrees).

Effective (RMS) & Average Values

• Root-Mean-Square (RMS / Effective) Value: The equivalent DC value that delivers equal average thermal power to a resistor:

$$V_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) dt} = \frac{V_m}{\sqrt{2}} \approx 0.7071 V_m$$

• Average Value (Half-Cycle Rectified):

$$V_{\text{avg}} = \frac{2}{T} \int_{0}^{T/2} V_m \sin(\omega t) dt = \frac{2 V_m}{\pi} \approx 0.6366 V_m$$

• Form Factor & Crest (Peak) Factor:

$$\text{Form Factor} = \frac{V_{\text{rms}}}{V_{\text{avg}}} = \frac{0.7071 V_m}{0.6366 V_m} = \frac{\pi}{2\sqrt{2}} \approx 1.111$$ $$\text{Crest (Peak) Factor} = \frac{V_m}{V_{\text{rms}}} = \frac{V_m}{V_m / \sqrt{2}} = \sqrt{2} \approx 1.414$$

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2. Phasors, Complex Impedance, and Admittance

Using Euler's identity ($e^{j\theta} = \cos\theta + j\sin\theta$), sinusoidal functions map into time-independent complex phasor vectors:

$$\mathbf{V} = V_{\text{rms}} \angle \phi = V_{\text{rms}} \cos\phi + j V_{\text{rms}} \sin\phi$$

Component Impedance in the Phasor Domain

Complex impedance $\mathbf{Z}$ (Ohms, $\Omega$) represents total opposition to sinusoidal current flow:

$$\mathbf{Z} = R + j X = |Z| \angle \theta_z$$

• Resistor ($R$): $\mathbf{Z}_R = R \angle 0^\circ = R + j0$ (Voltage and current in phase).
• Inductor ($L$): $\mathbf{Z}_L = j \omega L = \omega L \angle +90^\circ = 0 + j X_L$ (Voltage leads current by $90^\circ$).
• Capacitor ($C$): $\mathbf{Z}_C = \frac{1}{j \omega C} = -j \frac{1}{\omega C} = \frac{1}{\omega C} \angle -90^\circ = 0 - j X_C$ (Voltage lags current by $90^\circ$).

$$\text{Net Reactance: } X = X_L - X_C = 2\pi f L - \frac{1}{2\pi f C}$$ $$\text{Impedance Magnitude: } |Z| = \sqrt{R^2 + (X_L - X_C)^2}, \quad \theta_z = \arctan\left(\frac{X_L - X_C}{R}\right)$$

Complex Admittance

Admittance $\mathbf{Y}$ (Siemens, $\text{S}$) is the reciprocal of impedance:

$$\mathbf{Y} = \frac{1}{\mathbf{Z}} = G + j B = |Y| \angle -\theta_z$$

where $G = \frac{R}{R^2 + X^2}$ is conductance ($\text{S}$) and $B = \frac{-X}{R^2 + X^2}$ is susceptance ($\text{S}$). For pure parallel branches, inductive susceptance is negative ($B_L = -\frac{1}{\omega L}$) and capacitive susceptance is positive ($B_C = \omega C$).

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3. Series and Parallel RLC Resonance

Resonance occurs when inductive and capacitive reactances cancel, causing the input impedance to become purely resistive.

Series RLC Resonance

In a series RLC circuit, $\mathbf{Z} = R + j(X_L - X_C)$. Resonance occurs when $X_L = X_C$:

$$\omega_0 L = \frac{1}{\omega_0 C} \quad \implies \quad \omega_0 = \frac{1}{\sqrt{L C}} \quad \text{rad/s}$$ $$f_0 = \frac{1}{2 \pi \sqrt{L C}} \quad \text{Hz}$$

Characteristics of Series Resonance:

1. Minimum circuit impedance: $\mathbf{Z}_0 = R + j0$.
2. Maximum line current: $I_0 = \frac{V_{\text{rms}}}{R}$ (in phase with voltage, $\text{PF} = 1.0$).
3. High inductor/capacitor reactive voltages: $V_L = V_C = Q \cdot V_{\text{source}}$ (Voltage magnification).

Quality Factor ($Q$) & Bandwidth (BW):

$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R} \sqrt{\frac{L}{C}}$$ $$\text{Bandwidth (BW)} = f_2 - f_1 = \frac{f_0}{Q} = \frac{R}{2 \pi L} \quad \text{Hz}$$

where $f_1, f_2$ are upper and lower half-power ($-3\ \text{dB}$) cutoff frequencies:

$$f_1, f_2 = f_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} \mp \frac{f_0}{2Q}$$

Parallel RLC Resonance (Anti-Resonance)

For an ideal parallel RLC circuit, admittance is $\mathbf{Y} = \frac{1}{R} + j\left(\omega C - \frac{1}{\omega L}\right)$. At resonance ($\omega_0 = \frac{1}{\sqrt{LC}}$):

• Maximum input impedance: $Z_p = R$.
• Minimum line current drawn from source: $I_{\text{min}} = \frac{V}{R}$.
• High internal branch circulation current: $I_L = I_C = Q \cdot I_{\text{line}}$ (Current magnification).

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4. AC Power Triangle & Power Factor Correction

When sinusoidal voltage $\mathbf{V} = V \angle \alpha$ supplies current $\mathbf{I} = I \angle \beta$, the phase difference is $\theta = \alpha - \beta$.

Complex Power & Power Components

Complex Power $\mathbf{S}$ (Volt-Amperes, $\text{VA}$) is calculated using the conjugate of the current phasor:

$$\mathbf{S} = \mathbf{V} \mathbf{I}^* = V I \angle (\alpha - \beta) = P + j Q$$

Lagging vs. Leading PF: Power factor is lagging when current lags voltage (inductive load, $Q > 0$). Power factor is leading when current leads voltage (capacitive load, $Q < 0$).

Power Factor Correction (Capacitor Sizing)

Low lagging power factor increases distribution current $I = \frac{P}{V \cos\theta}$, creating high line $I^2 R$ losses and excessive voltage drop. Connecting shunt capacitors across inductive loads supplies reactive power locally.

$$Q_C = P \left( \tan\theta_1 - \tan\theta_2 \right) \quad \text{VAR}$$ $$\text{Required Shunt Capacitance: } C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{Q_C}{2 \pi f V_{\text{rms}}^2} \quad \text{Farads}$$

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Solved Board Exam Examples

Example 1: Series RLC Circuit & Power Triangle

Problem: A series circuit containing $R = 12\ \Omega$, $L = 0.15\ \text{H}$, and $C = 100\ \mu\text{F}$ is connected to a $230\ \text{V}$, $60\ \text{Hz}$ single-phase AC supply. Calculate: (a) total impedance $\mathbf{Z}$, (b) RMS current $I$, (c) active power $P$, (d) reactive power $Q$, and (e) power factor $\text{PF}$.

Solution:

1. Calculate inductive and capacitive reactances at $f = 60\ \text{Hz}$ ($\omega = 2\pi(60) = 376.99\ \text{rad/s}$): $$X_L = \omega L = 376.99 \times 0.15 = 56.55\ \Omega$$ $$X_C = \frac{1}{\omega C} = \frac{1}{376.99 \times 100 \times 10^{-6}} = \frac{1}{0.0377} = 26.53\ \Omega$$
2. Calculate net reactance $X$ and complex impedance $\mathbf{Z}$: $$X = X_L - X_C = 56.55 - 26.53 = 30.02\ \Omega$$ $$\mathbf{Z} = 12 + j 30.02 = \sqrt{12^2 + 30.02^2} \angle \arctan\left(\frac{30.02}{12}\right) = 32.33\ \Omega \angle 68.2^\circ$$
3. Compute circuit RMS current $I$: $$I = \frac{V}{|Z|} = \frac{230}{32.33} = 7.114\ \text{A}$$
4. Compute Power Factor and Power Components: $$\text{PF} = \cos(68.2^\circ) = 0.3714 \quad (\text{lagging})$$ $$P = V I \cos\theta = 230 \times 7.114 \times 0.3714 = 607.7\ \text{W}$$ $$Q = V I \sin\theta = 230 \times 7.114 \times \sin(68.2^\circ) = 1519.4\ \text{VAR (inductive)}$$ $$S = V I = 230 \times 7.114 = 1636.2\ \text{VA}$$

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Example 2: Industrial Power Factor Correction

Problem: An industrial plant consumes $100\ \text{kW}$ at a lagging power factor of $0.65$ from a $460\ \text{V}$, $60\ \text{Hz}$ supply. Determine: (a) the kVAR rating of a parallel capacitor bank needed to correct the overall power factor to $0.95$ lagging, and (b) the required capacitance $C$ in microfarads.

Solution:

1. Find initial phase angle $\theta_1$ and initial reactive power $Q_1$: $$\theta_1 = \arccos(0.65) = 49.46^\circ \implies \tan\theta_1 = 1.1691$$ $$Q_1 = P \tan\theta_1 = 100\ \text{kW} \times 1.1691 = 116.91\ \text{kVAR}$$
2. Find target phase angle $\theta_2$ and target reactive power $Q_2$ for $\text{PF} = 0.95$: $$\theta_2 = \arccos(0.95) = 18.19^\circ \implies \tan\theta_2 = 0.3287$$ $$Q_2 = P \tan\theta_2 = 100\ \text{kW} \times 0.3287 = 32.87\ \text{kVAR}$$
3. Calculate required capacitive kVAR rating $Q_C$: $$Q_C = Q_1 - Q_2 = 116.91 - 32.87 = 84.04\ \text{kVAR} = 84,040\ \text{VAR}$$
4. Calculate required shunt capacitance $C$: $$C = \frac{Q_C}{2 \pi f V^2} = \frac{84,040}{2 \pi \times 60 \times (460)^2} = \frac{84,040}{376.991 \times 211,600} = \frac{84,040}{79,771,326} = 1.0535 \times 10^{-3}\ \text{F} = 1053.5\ \mu\text{F}$$