LC Oscillations & RLC Circuits: JEE Main Guide

Alternating current is one of the most reliably tested chapters in JEE Main Physics — two to three questions appear every session, and the formula set is compact. LC oscillations connect electromagnetic induction to simple harmonic motion, creating a beautiful conceptual bridge. RLC circuits with resonance, impedance, and power factor complete the chapter. This guide covers every testable aspect systematically.

LC Oscillations

An LC circuit (inductor L and capacitor C) exhibits electrical oscillations analogous to a spring-mass mechanical oscillator. Energy alternates between electric field in the capacitor (U_E = Q²/2C) and magnetic field in the inductor (U_B = LI²/2). Total energy is conserved: Q²/2C + LI²/2 = Q₀²/2C = constant. The governing equation: L(d²Q/dt²) + Q/C = 0 — identical in form to SHM. Angular frequency ω = 1/√(LC), so frequency f = 1/(2π√(LC)) and time period T = 2π√(LC).

Charge as a function of time: Q(t) = Q₀cos(ωt). Current: I(t) = −Q₀ω·sin(ωt). JEE Main asks for the time at which energy is equally shared between L and C: U_E = U_B → Q² = Q₀²/2 → Q = Q₀/√2 → at t = T/8 from maximum charge. Energy exchange is at twice the frequency of charge oscillation. For the electromagnetic induction underlying LC circuits, see our Electromagnetic Induction Guide.

Phasors and Impedance in AC Circuits

In AC analysis, voltage and current are represented as rotating phasors. For a resistor: V_R = IR, in phase with I. For an inductor: V_L = IX_L = I·ωL, leads I by 90°. For a capacitor: V_C = IX_C = I/(ωC), lags I by 90°. In a series RLC circuit: V_net = I·Z, where impedance Z = √(R² + (X_L − X_C)²). Phase angle between V and I: tanφ = (X_L − X_C)/R. Take a free mock test on alternating current to practise phasor addition and impedance calculations.

Resonance in Series RLC Circuit

Resonance occurs when X_L = X_C, i.e., ωL = 1/(ωC), giving resonant frequency ω₀ = 1/√(LC). At resonance: Z = R (minimum impedance), current is maximum (I_max = V/R), power factor cosφ = 1, and the circuit is purely resistive. The voltage across L and across C at resonance can exceed the source voltage by the Q-factor: V_L = V_C = Q × V_source, where Q-factor = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C). A high Q-factor means sharp resonance — the circuit selects a narrow frequency band.

Bandwidth: the range of frequencies for which power is at least half the maximum (half-power frequencies): Δω = R/L. Quality factor Q = ω₀/Δω = ω₀L/R. JEE Main uses Q-factor in the context of radio tuning: a high-Q circuit selects one station without interference from adjacent frequencies.

Power in AC Circuits

Average power P = V_rms·I_rms·cosφ, where cosφ is the power factor. V_rms = V₀/√2, I_rms = I₀/√2. Power factor cosφ = R/Z. For purely inductive or capacitive circuits: φ = 90°, cosφ = 0, average power = 0 (no energy dissipated, only reactive power). For purely resistive: φ = 0, cosφ = 1, P = V_rms·I_rms. JEE Main tests: given V₀, I₀, and phase difference, find average power — substitute directly into P = (V₀I₀/2)cosφ.

Transformers

Ideal transformer: V_s/V_p = N_s/N_p = I_p/I_s. Power is conserved: V_p·I_p = V_s·I_s. Step-up transformer: N_s > N_p → V_s > V_p → I_s < I_p. Step-down transformer: N_s < N_p. Real transformers have losses: copper losses (I²R in windings), iron losses (eddy currents and hysteresis in core). JEE Main tests transformer calculations and efficiency: η = (output power)/(input power) × 100%.

Exam Strategy

AC circuits questions are among the most mechanical in JEE Main — the answer almost always follows from one or two formula applications. Practise drawing the phasor diagram first (even a 10-second sketch prevents sign errors), then compute Z, φ, and whatever quantity is asked. For a complete alternating current revision plan, see our Alternating Current Guide. Upgrade for ₹149/month for 200+ AC circuit problems with detailed phasor diagram solutions.