AC Circuits — Impedance, Resonance, Power Factor

AC circuits power the electricity in your home, the radio on your desk, and the Wi-Fi router in your room. Unlike DC circuits where current flows in one direction, in AC (alternating current) the current reverses direction periodically — typically 50 times per second (50 Hz) in India.

The reason AC dominates electricity transmission isn’t historical accident: AC voltage can be stepped up with transformers, reducing current and therefore reducing transmission line losses ( $P = I^2R$ ). DC cannot be transformed this way (without modern power electronics). Understanding AC circuits means understanding everything from why your phone charger gets warm to how radio stations transmit signals.

Key Terms and Definitions

AC voltage/current: Quantities that vary sinusoidally with time. $v(t) = V_0 \sin(\omega t + \phi)$ , where $V_0$ is peak voltage, $\omega = 2\pi f$ is angular frequency, and $\phi$ is phase.

RMS value: Root Mean Square. The equivalent DC value that produces the same average power. For a sinusoidal signal: $V_{rms} = V_0/\sqrt{2}$ , $I_{rms} = I_0/\sqrt{2}$ .

Impedance (Z): The AC equivalent of resistance. It resists current in an AC circuit, but includes effects of inductors and capacitors. Measured in ohms (Ω).

Reactance (X): The opposition to current due to inductors ( $X_L$ ) or capacitors ( $X_C$ ) alone. Unlike resistance, reactance is frequency-dependent.

Phase angle (φ): The phase difference between voltage and current. In a pure resistor, $\phi = 0°$ . In a pure inductor, voltage leads current by 90°. In a pure capacitor, current leads voltage by 90°.

Power factor (cos φ): The fraction of total apparent power that actually does useful work. Ranges from 0 (purely reactive) to 1 (purely resistive).

Resonance: The condition where $X_L = X_C$ , giving minimum impedance and maximum current.

Impedance in RLC Circuits

Z = \sqrt{R^2 + (X_L - X_C)^2}

where:

$X_L = \omega L = 2\pi f L$ (inductive reactance, increases with frequency)
$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$ (capacitive reactance, decreases with frequency)

Current: $I_{rms} = \frac{V_{rms}}{Z}$

Phase angle: $\tan\phi = \frac{X_L - X_C}{R}$

The phasor diagram helps visualise: $R$ is on the horizontal axis, $(X_L - X_C)$ is on the vertical axis, and $Z$ is the hypotenuse.

Individual Components in AC

Resistor in AC

A resistor obeys Ohm’s law instantaneously. Voltage and current are in phase ( $\phi = 0°$ ).

Power dissipated: $P = I_{rms}^2 R = V_{rms}^2/R$ . All supplied power is consumed.

Inductor in AC

An inductor resists changes in current. The voltage across an inductor leads the current by 90° (ELI — Voltage leads current in an Inductor).

$X_L = \omega L$ . At high frequencies, $X_L$ is large — the inductor blocks high-frequency AC. At DC ( $\omega = 0$ ), $X_L = 0$ — the inductor is a short circuit.

Power consumed by a pure inductor: $P = 0$ (energy stored and returned each cycle).

Capacitor in AC

A capacitor resists changes in voltage. The current through a capacitor leads the voltage by 90° (ICE — Current leads voltage in a Capacitor).

$X_C = 1/(\omega C)$ . At high frequencies, $X_C$ is small — the capacitor passes high-frequency AC. At DC, $X_C \to \infty$ — the capacitor blocks DC.

Power consumed by a pure capacitor: $P = 0$ (energy stored and returned each cycle).

Resonance in Series RLC Circuit

At resonance: $X_L = X_C$

\omega_0 = \frac{1}{\sqrt{LC}}, \quad f_0 = \frac{1}{2\pi\sqrt{LC}}

At resonance:

Impedance is minimum: $Z = R$
Current is maximum: $I_{max} = V_{rms}/R$
Voltage across $L$ and $C$ can be very large (resonant voltage amplification)
Phase angle $\phi = 0°$ — circuit behaves like a pure resistor

Quality Factor (Q): Measures the “sharpness” of resonance.

Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} = \frac{1}{R}\sqrt{\frac{L}{C}}

High $Q$ → narrow resonance peak → selective circuit (used in radio tuners).

Power in AC Circuits

P_{avg} = V_{rms} I_{rms} \cos\phi

where $\cos\phi$ is the power factor.

Pure resistor: $\phi = 0°$ , $\cos\phi = 1$ , $P = V_{rms} I_{rms}$ (maximum power)
Pure inductor/capacitor: $\phi = 90°$ , $\cos\phi = 0$ , $P = 0$ (no power consumed)
Series RLC at resonance: $\phi = 0°$ , $\cos\phi = 1$

Apparent power: $S = V_{rms} I_{rms}$ (in VA — volt-amperes)

Real power: $P = S\cos\phi$ (in Watts)

Reactive power: $Q = S\sin\phi$ (in VAR — volt-ampere reactive)

Solved Examples

Easy (CBSE Class 12): Finding reactance

Q: A 50 Hz AC supply is connected to a 0.2 H inductor. Find $X_L$ .

$X_L = 2\pi f L = 2\pi \times 50 \times 0.2 = 20\pi \approx 62.8 \text{ }\Omega$ .

Medium (JEE Main): Series RLC calculation

Q: A series RLC circuit has $R = 10\text{ }\Omega$ , $L = 0.1\text{ H}$ , $C = 100\text{ }\mu\text{F}$ . Find impedance at 50 Hz.

$X_L = 2\pi \times 50 \times 0.1 = 10\pi \approx 31.4\text{ }\Omega$

$X_C = \frac{1}{2\pi \times 50 \times 100 \times 10^{-6}} = \frac{1}{0.0314} \approx 31.8\text{ }\Omega$

$Z = \sqrt{10^2 + (31.4 - 31.8)^2} = \sqrt{100 + 0.16} \approx \sqrt{100.16} \approx 10\text{ }\Omega$

This is nearly resonance! $Z \approx R$ confirms it.

Hard (JEE Advanced): Q-factor and voltage amplification

Q: In a resonant series RLC circuit with $L = 100$ mH, $C = 25$ μF, $R = 5$ Ω. Find the resonant frequency and Q-factor.

$f_0 = \frac{1}{2\pi\sqrt{0.1 \times 25 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{2.5 \times 10^{-6}}} = \frac{1}{2\pi \times 1.58 \times 10^{-3}} \approx 100.7\text{ Hz}$

$Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{5}\sqrt{\frac{0.1}{25 \times 10^{-6}}} = \frac{1}{5}\sqrt{4000} = \frac{63.25}{5} \approx 12.6$

The voltage across $L$ or $C$ at resonance is $Q$ times the source voltage.

Exam-Specific Tips

JEE Main: 3–4 marks on AC circuits every year. Most common: finding impedance, resonant frequency, power factor, and the behaviour of components at different frequencies. Phasor diagrams are frequently tested — learn to draw and interpret them.

CBSE Class 12 Board: Chapter 7 (Alternating Current). Key topics: RMS values, impedance of RLC, resonance, transformer principle, and power factor. A 5-mark question on resonance or transformer is standard every year.

NEET: AC circuits appear but are less heavily tested than mechanics or optics. Focus on RMS values, resonance condition, and the basic distinction between resistive/inductive/capacitive behaviour.

Common Mistakes to Avoid

Mistake 1: Using peak values instead of RMS in power calculations. $P = V_{rms} I_{rms} \cos\phi$ , not $V_0 I_0 \cos\phi$ . Power involves RMS values.

Mistake 2: Adding reactances like resistances. $Z \neq R + X_L + X_C$ . The correct formula is $Z = \sqrt{R^2 + (X_L - X_C)^2}$ — this is the Pythagorean sum, not arithmetic.

Mistake 3: Forgetting that at resonance, $Z = R$ (not zero). At resonance, $X_L = X_C$ , so they cancel, leaving only $R$ . $Z$ is minimum but not zero (unless $R = 0$ , which is ideal).

Mistake 4: Confusing “ELI the ICE man” mnemonic. ELI: In an inductor (L), voltage (E) leads current (I). ICE: In a capacitor (C), current (I) leads voltage (E). The mnemonic gives the sequence E, L, I and I, C, E.

Mistake 5: Forgetting to account for power factor in real power calculations. A circuit with 230 V, 10 A, and power factor 0.7 consumes $230 \times 10 \times 0.7 = 1610$ W, not 2300 W.

Practice Questions

Q1: Find the capacitive reactance of a 50 μF capacitor at 50 Hz.

$X_C = 1/(2\pi \times 50 \times 50 \times 10^{-6}) = 1/(0.01571) \approx 63.7\text{ }\Omega$ .

Q2: A series LCR circuit has $L = 0.5$ H, $C = 5$ μF. Find the resonant frequency.

$f_0 = \frac{1}{2\pi\sqrt{0.5 \times 5 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{2.5 \times 10^{-6}}} \approx \frac{1}{2\pi \times 1.58 \times 10^{-3}} \approx 100.7\text{ Hz}$ .

Q3: What is the power consumed by a pure inductor connected to AC supply?

Zero. A pure inductor has $\cos\phi = \cos 90° = 0$ , so $P = V_{rms}I_{rms} \times 0 = 0$ . Energy is stored in the magnetic field and returned to the circuit each cycle.

Q4: The rms voltage of an AC source is 220 V. What is the peak voltage?

$V_0 = V_{rms} \times \sqrt{2} = 220\sqrt{2} \approx 311\text{ V}$ .

FAQs

Q: Why is 50 Hz used for electricity in India? It’s a legacy standard adopted from British electrical systems. 50 Hz is a good compromise — high enough that AC motors work efficiently, low enough that transformer design is manageable. The US uses 60 Hz for historical reasons (Edison vs. Westinghouse competition).

Q: Why does an inductor block high-frequency AC but pass DC? $X_L = \omega L$ increases with frequency. At DC ( $\omega = 0$ ), $X_L = 0$ — no opposition. As frequency increases, the inductor increasingly resists the current. This makes inductors useful as “low-pass filters” (blocking high-frequency noise).

Q: What’s the practical importance of power factor? Industries are charged for both real and apparent power in some billing systems. A low power factor means you’re drawing more current than necessary for the same real power, wasting energy in transmission line losses. Industries use capacitor banks to improve power factor.

Additional Concepts

Power factor correction

An inductive load (like a motor) has a lagging power factor. To correct it, a capacitor is connected in parallel. The capacitor supplies leading reactive power that partially cancels the lagging reactive power of the inductor, bringing $\cos\phi$ closer to 1.

Why this matters: a factory with power factor 0.5 draws twice the current it would at power factor 1.0 for the same real power. Double the current means four times the $I^2R$ losses in transmission lines. Electricity boards penalise low power factor.

Skin effect

At high frequencies, AC current tends to flow on the surface of a conductor rather than through its bulk. This is the skin effect. It increases the effective resistance because less cross-section is being used.

This is why high-frequency cables use stranded wire (more surface area) or hollow conductors. At 50 Hz, the skin depth in copper is about 9.3 mm — large enough that solid wires work fine for household wiring.

Choke coil vs resistor for controlling current

A choke coil (pure inductor with minimal resistance) limits AC current without dissipating power (since $P = 0$ for ideal inductor). A resistor also limits current but wastes energy as heat. This is why tube light circuits use a choke, not a resistor.

CBSE boards ask: “Why is a choke coil preferred over a resistor for controlling AC current?” Answer: A choke limits current via inductive reactance ( $X_L = \omega L$ ) with negligible power loss. A resistor would waste energy as heat. The choke is more energy-efficient.

Additional Practice Questions

Q5. A factory has a power factor of 0.6. What capacitor value would improve it to 0.95 if the load is 100 kW at 50 Hz, 400 V?

Current at 0.6 pf: $I_1 = P/(V\cos\phi_1) = 100000/(400 \times 0.6) = 417$ A. Reactive power at 0.6: $Q_1 = P\tan\phi_1 = 100000 \times 1.333 = 133.3$ kVAR. At 0.95: $Q_2 = P\tan\phi_2 = 100000 \times 0.329 = 32.9$ kVAR. Capacitor must supply: $Q_C = Q_1 - Q_2 = 100.4$ kVAR. $C = Q_C/(\omega V^2) = 100400/(2\pi \times 50 \times 160000) = 2 \times 10^{-3}$ F = 2 mF.

Q6. At what frequency does an RLC circuit with $L = 1$ H and $C = 1\,\mu$ F resonate?

$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10^{-6}}} = \frac{1}{2\pi \times 10^{-3}} = \frac{1000}{2\pi} \approx 159$ Hz.