Alternating current

AC: RMS values, reactance and impedance

An alternating voltage varies sinusoidally: v = V₀ sin ωt, with ω = 2πf. In India, mains supply has f = 50 Hz.

RMS values: the root-mean-square value is the steady (DC) value that would give the same heating. I_rms = I₀/√2 ≈ 0.707 I₀ and V_rms = V₀/√2. Meters and the 230 V rating of mains are RMS values. Over a full cycle, the average of a sinusoidal current is zero.

Behaviour of circuit elements:

• Resistor: V and I in phase; opposition = R.
• Inductor: current lags voltage by 90° (π/2). Inductive reactance X_L = ωL = 2πfL; it rises with frequency. An inductor passes DC freely.
• Capacitor: current leads voltage by 90°. Capacitive reactance X_C = 1/(ωC) = 1/(2πfC); it falls with frequency. A capacitor blocks DC (X_C → ∞ when f = 0).

Series LCR circuit: the impedance is Z = √[R² + (X_L – X_C)²] and I₀ = V₀/Z. The phase angle φ between V and I is given by tan φ = (X_L – X_C)/R. If X_L > X_C the circuit is inductive (I lags); if X_C > X_L it is capacitive (I leads). Phasor diagrams represent these as rotating vectors.

Worked examples:

• Peak 325 V: V_rms = 325/√2 ≈ 230 V.
• f = 50 Hz, L = 0.1 H: X_L = 2π × 50 × 0.1 ≈ 31.4 Ω.
• R = 30 Ω, X_L = 80 Ω, X_C = 40 Ω: Z = √(900 + 1600) = 50 Ω.

Key points

• I_rms = I₀/√2; V_rms = V₀/√2.
• X_L = ωL (∝ f); X_C = 1/ωC (∝ 1/f).
• Z = √[R² + (X_L – X_C)²]; tan φ = (X_L – X_C)/R.
• Inductor: I lags V by 90°. Capacitor: I leads V by 90°.

Exam tips

• Use the memory aid “CIVIL”: in C, I leads V; V leads I in L.
• Check whether the question gives peak or RMS values.

Common mistakes

• Adding R, X_L and X_C directly instead of using the square-root formula.
• Thinking a capacitor passes DC.
• Using f instead of ω = 2πf in reactance formulas.