Physics · JEE Advanced only

RC, LR, LC and LCR Circuits with DC and AC Sources (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 8-10 hours. Spend 2-3 hours on DC transients (RC and LR circuits with exponential behavior), 2 hours on LC oscillations, 3-4 hours on AC steady state analysis including impedance, phasors and resonance, and 1-2 hours practicing mixed problems..

Last updated: 26 October 2026

Direct answer

RC, LR, LC and LCR circuits describe how resistors, inductors and capacitors behave when connected to DC or AC sources.

RC, LR, LC and LCR circuits describe how resistors, inductors and capacitors behave when connected to DC or AC sources. With DC, these circuits show transient behavior where currents and voltages change exponentially with time constants. With AC, they exhibit steady-state behavior involving impedance, phase differences and resonance phenomena.

Official syllabus

What the JEE Advanced syllabus says

“RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: LCR series circuit, resonance, power in AC circuits.

JEE Main focuses mainly on steady-state AC analysis and resonance in LCR circuits. JEE Advanced expects you to handle transient analysis with differential equations, LC oscillations, and the complete time-domain behavior of these circuits from switch-on to steady state.

Key ideas

What you must understand

Time constant τ = RC for RC circuits and τ = L/R for LR circuits determines how fast the circuit reaches steady state
In DC circuits, capacitors block current at steady state (open circuit) while inductors allow full current (short circuit)
LC circuits produce undamped oscillations with frequency ω = 1/√(LC), and energy oscillates between electric (capacitor) and magnetic (inductor) forms
For AC sources, impedance Z combines R, X_L = ωL and X_C = 1/ωC, with current lagging or leading voltage depending on circuit type
Resonance in series LCR occurs when X_L = X_C, giving minimum impedance Z = R and maximum current
Quality factor Q = ω₀L/R = 1/(ω₀CR) measures sharpness of resonance and also equals the voltage magnification across L or C at resonance

Formulas

Formulas and key facts

RC charging: q(t) = CE(1 − e^(−t/RC)), i(t) = (E/R)e^(−t/RC)
RC discharging: q(t) = Q₀e^(−t/RC), i(t) = (Q₀/RC)e^(−t/RC)
LR growth: i(t) = (E/R)(1 − e^(−Rt/L)), decay: i(t) = I₀e^(−Rt/L)
LC oscillation: q(t) = Q₀cos(ωt), ω = 1/√(LC), total energy = Q₀²/2C = constant
Series LCR impedance: Z = √(R² + (ωL − 1/ωC)²)
Phase angle: tan φ = (X_L − X_C)/R, current lags voltage if X_L > X_C
Resonant frequency: ω₀ = 1/√(LC), at resonance Z = R and current is maximum
Power in AC: P = I²R = (V²R)/Z² = VI cos φ, where cos φ is power factor

Avoid

Common mistakes

Forgetting that at t = 0, inductor opposes sudden current change (acts as open circuit) while capacitor opposes sudden voltage change (acts as short circuit if uncharged)
Confusing time constant formulas: τ = RC for capacitor circuits but τ = L/R for inductor circuits
In resonance problems, assuming voltage across L and C are zero. They are actually equal and opposite, each can be Q times the source voltage
Using wrong sign in phase angle: current LAGS voltage in inductive circuits (φ positive) and LEADS in capacitive circuits (φ negative)

Worked example

A JEE Advanced-level example

Question. An uncharged capacitor of 2 μF is connected in series with a resistor of 1 MΩ and a 10 V DC source at t = 0. Find: (a) the time constant, (b) the charge on capacitor at t = 2 s, and (c) the energy stored in the capacitor at steady state.

Show solution

Time constant τ = RC = (1 × 10⁶ Ω)(2 × 10⁻⁶ F) = 2 s
Charge at time t: q(t) = CE(1 − e^(−t/τ)) = (2 × 10⁻⁶)(10)(1 − e^(−2/2))
q(2 s) = 20 × 10⁻⁶ × (1 − e⁻¹) = 20 × 10⁻⁶ × (1 − 0.368) = 20 × 10⁻⁶ × 0.632 = 12.64 μC
At steady state, capacitor is fully charged: Q_max = CE = (2 × 10⁻⁶)(10) = 20 μC
Energy stored at steady state: U = Q²/(2C) = (20 × 10⁻⁶)²/(2 × 2 × 10⁻⁶) = (400 × 10⁻¹²)/(4 × 10⁻⁶) = 100 × 10⁻⁶ J = 100 μJ

Answer: (a) τ = 2 s, (b) q = 12.64 μC, (c) U = 100 μJ

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is RC, LR, LC and LCR Circuits with DC and AC Sources in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is LCR series circuit, resonance, power in AC circuits.

How long does RC, LR, LC and LCR Circuits with DC and AC Sources take to prepare for JEE Advanced?

About 8-10 hours. Spend 2-3 hours on DC transients (RC and LR circuits with exponential behavior), 2 hours on LC oscillations, 3-4 hours on AC steady state analysis including impedance, phasors and resonance, and 1-2 hours practicing mixed problems., if you already know the related JEE Main topic.