AC circuits: phasors, power and power factor

Why it matters

Motors, solenoids, transformers and switch-mode supplies all run on or draw from sinusoidal AC. Phasors turn the differential equations of R, L and C circuits into simple complex-number algebra, and the power triangle tells you how much of the current drawn actually does work. A poor power factor means larger cables, larger switchgear and tariff penalties in an industrial plant.

Key ideas

Sinusoids and RMS. A sinusoidal voltage v(t) = V_m·sin(ωt + θ) has amplitude V_m, angular frequency ω = 2πf and phase θ. Its RMS (effective) value is the DC value that would dissipate the same heat in a resistor; for a sinusoid V_rms = V_m/√2. Mains "230 V" is an RMS value (peak about 325 V). Meters read RMS, and power formulas use RMS.

Phasors. In a linear circuit driven at one frequency, every voltage and current is a sinusoid at that same frequency, so only magnitude and phase need tracking. A phasor is the complex number V = V_rms∠θ = V_rms(cos θ + j sin θ). Differentiation becomes multiplication by jω, so circuit laws (KCL, KVL, Thevenin, nodal, mesh) apply unchanged with impedances in place of resistances. Phasors only apply in sinusoidal steady state, not to transients or non-sinusoidal waves (use superposition of harmonics for those).

Impedance. Z = V/I in ohms, Z = R + jX.

• Resistor: Z = R, voltage and current in phase.
• Inductor: Z = jωL, current lags voltage by 90°. X_L rises with frequency (open circuit at very high f, short at DC).
• Capacitor: Z = 1/(jωC) = −j/(ωC), current leads voltage by 90°. X_C falls with frequency (open at DC).
• Admittance Y = 1/Z = G + jB in siemens is easier for parallel branches. Memory aid "ELI the ICE man": E leads I in L; I leads E in C.

Series resonance. In a series RLC circuit X_L = X_C at f_0 = 1/(2π√(LC)); impedance is minimum (= R), current is maximum and in phase with the supply. The quality factor Q = ω_0·L/R measures the sharpness, and bandwidth = f_0/Q. In a parallel resonant circuit the impedance is maximum at resonance.

Power in AC circuits. With v and i out of phase by φ (angle of Z, current lagging for inductive loads):

• Real (active) power P, in W, is the average power converted to heat or work.
• Reactive power Q, in VAR, is energy that surges back and forth between source and the fields of L and C each cycle with zero average. Inductive loads absorb Q (positive); capacitors supply it.
• Apparent power S, in VA, is V_rms·I_rms; it sets cable, transformer and generator ratings.
• They form a right-angled power triangle: S² = P² + Q². In complex form S = V·I* = P + jQ (I* is the conjugate).

Power factor. pf = P/S. For sinusoidal waves pf = cos φ, described as lagging (inductive, current lags) or leading (capacitive). Most industrial loads (induction motors) are lagging. For the same P, a lower pf means more current, more I²R loss in lines and more voltage drop. Power factor correction connects shunt capacitors that supply part of the load's reactive power locally. For non-sinusoidal currents (rectifier loads), pf also includes a distortion factor, so pf ≠ cos φ.

Formulas

V_rms = V_m / √2 and I_rms = I_m / √2 (sinusoids only)

ω = 2π·f

• ω: rad/s; f: Hz.

X_L = ω·L; X_C = 1 / (ω·C)

• X_L, X_C: Ω; L: H; C: F.

Z = R + j(X_L − X_C); |Z| = √(R² + (X_L − X_C)²); φ = tan⁻¹((X_L − X_C)/R)

• Series RLC; φ positive means inductive (current lags).

f_0 = 1 / (2π·√(L·C)); Q = ω_0·L / R; BW = f_0 / Q

• Series resonance; BW in Hz.

P = V·I·cos φ = I²·R (W) Q = V·I·sin φ = I²·X (VAR) S = V·I = √(P² + Q²) (VA)

• V, I: RMS values (V, A).

pf = cos φ = P / S = R / |Z|

Q_C = P·(tan φ1 − tan φ2); C = Q_C / (ω·V²)

• Capacitor kVAR needed to raise pf from cos φ1 to cos φ2; C in F; V: RMS voltage across the capacitor (V).

Worked examples

Example 1 (standard: series RL circuit). A coil with R = 30 Ω and L = 0.1273 H is connected to 230 V, 50 Hz. Find the impedance, current, power factor, P, Q and S.

1. X_L = 2π·f·L = 2π × 50 × 0.1273 = 40.0 Ω.
2. |Z| = √(30² + 40²) = 50 Ω, φ = tan⁻¹(40/30) = 53.13°.
3. I = V/|Z| = 230/50 = 4.6 A, lagging the voltage by 53.13°.
4. pf = cos 53.13° = R/|Z| = 30/50 = 0.6 lagging.
5. P = I²R = 4.6² × 30 = 634.8 W; Q = I²X_L = 4.6² × 40 = 846.4 VAR; S = V·I = 230 × 4.6 = 1058 VA.
6. Check: √(634.8² + 846.4²) = 1058 VA.

I = 4.6 A, pf = 0.6 lagging, P = 634.8 W, Q = 846.4 VAR, S = 1058 VA.

Example 2 (GATE level: power factor correction). A single-phase 230 V, 50 Hz workshop load draws 10 kW at 0.7 lagging. Find the capacitance needed in parallel to raise the power factor to 0.95 lagging, and the supply current before and after.

1. tan φ1 = tan(cos⁻¹ 0.7) = 1.0202; tan φ2 = tan(cos⁻¹ 0.95) = 0.3287.
2. Q1 = P·tan φ1 = 10 000 × 1.0202 = 10 202 VAR; Q2 = 10 000 × 0.3287 = 3287 VAR.
3. Q_C = Q1 − Q2 = 6915 VAR (the capacitor supplies this).
4. C = Q_C/(ω·V²) = 6915/(314.16 × 230²) = 6915/(314.16 × 52 900) = 4.161 × 10⁻⁴ F.
5. Current before: I1 = P/(V·cos φ1) = 10 000/(230 × 0.7) = 62.1 A.
6. Current after: I2 = 10 000/(230 × 0.95) = 45.8 A. Real power is unchanged.

C ≈ 416 µF; the supply current falls from 62.1 A to 45.8 A.

Common mistakes

• Mixing peak and RMS values: P = V·I·cos φ uses RMS; using peaks gives twice the power.
• Forgetting ω = 2πf and using f directly in X_L or X_C.
• Adding magnitudes of voltages in a series RL or RC circuit; phasors add as complex numbers (V_R and V_L are 90° apart).
• Calling S = P + Q arithmetically; the relation is S² = P² + Q².
• Using cos φ = P/S but taking φ as the angle between two currents instead of between the voltage and current of the same element.
• Thinking a capacitor changes real power. Ideal capacitors only change Q and the line current.
• Over-correcting: too much capacitance makes the pf leading and the current rises again.

For GATE ME

Expect impedance and current of series and parallel RLC circuits, phase angle and power factor, real/reactive/apparent power, resonant frequency and bandwidth, and capacitor sizing for power factor correction. Practise complex-number arithmetic on the calculator (polar ↔ rectangular), and always state whether a power factor is lagging or leading.

Quick check

1. A sinusoid has a peak of 100 V. What is its RMS value?
2. Does current lead or lag the voltage in a capacitor, and by how much?
3. A load has P = 3 kW and Q = 4 kVAR. Find S and pf.
4. What is the impedance of a series RLC circuit at resonance?
5. Does adding a power factor correction capacitor change the real power drawn?

Answers: 1. 70.7 V. 2. Leads by 90°. 3. S = 5 kVA, pf = 0.6 lagging. 4. Purely resistive, equal to R. 5. No, only the reactive power and line current change.