ElectromagnetismHigh Yield (2 Qs in JEE)

Alternating Current (AC) Formulas

RMS and peak values, reactance, impedance of LCR circuits, resonance, Q-factor, and power factor.

8 Formulas•606+ Questions in Bank•

Concept Diagram

Series LCR Circuit & Resonance Phasor

Phasor relationships of VR, VL, and VC. At resonance (ωL = 1/ωC), impedance is minimum (Z = R) and current amplitude is maximized.

I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0, \quad I_{\text{avg}} = \frac{2I_0}{\pi} \approx 0.637 I_0 \; (\text{half cycle})

Root-mean-square effective heating current and half-wave rectification average.

Examiner Tip: Domestic AC supply of 220 V means V_rms = 220 V (peak V₀ ≈ 311 V).

X_L = \omega L = 2\pi f L, \quad X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

Frequency-dependent opposition to current.

Examiner Tip: Inductor blocks high frequencies (X_L ∝ f); capacitor blocks DC (X_C -> ∞ at f = 0).

Z = \sqrt{R^2 + (X_L - X_C)^2}, \quad \tan\phi = \frac{X_L - X_C}{R}

Total impedance vector and phase angle between voltage and current.

Examiner Tip: Current leads voltage if X_C > X_L (capacitive); lags if X_L > X_C (inductive).

\omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}, \quad Z_{\text{min}} = R, \quad Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

Condition when inductive and capacitive reactances cancel out.

Examiner Tip: At resonance, current is maximum, impedance is purely resistive, and power factor cos φ = 1.

P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi, \quad \text{Power Factor } \cos\phi = \frac{R}{Z}

Real electrical power dissipated in AC circuit.

Examiner Tip: Wattless current component is I_rms * sin φ.

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}, \quad \eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{V_s I_s}{V_p I_p}

Voltage transformation ratio in mutual induction transformer.

Examiner Tip: Step-up transformer increases voltage but decreases current.

q(t) = Q_0 \cos(\omega_0 t), \quad I(t) = -\omega_0 Q_0 \sin(\omega_0 t), \quad E_{\text{total}} = \frac{q^2}{2C} + \frac{1}{2}LI^2 = \frac{Q_0^2}{2C}

Harmonic energy exchange between electrical field in capacitor and magnetic field in inductor.

Examiner Tip: Analogy with mechanical SHM: mass m ↔ inductance L, spring constant k ↔ 1/C, velocity v ↔ current I.

\Delta\omega = \omega_2 - \omega_1 = \frac{R}{L}, \quad Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

Frequency width between half-power points where current drops to I_max / √2. Smaller bandwidth implies sharper resonance.

Examiner Tip: High Q circuits are required in radio receivers for tuning sharply to a single carrier frequency.

Frequently tested misconceptions in Alternating Current (AC)

•Meters in AC circuits (hot-wire ammeters/voltmeters) read RMS values, NEVER peak or average values.
•Power consumed by ideal inductor or capacitor over a full cycle is ZERO (wattless current).