Transient response of RL, RC and RLC circuits

Why it matters

Every time a sensor is powered up, a relay switches or an ADC samples, the circuit passes through a transient before it settles. Settling time limits how fast an instrument can measure. The voltage spike when an inductive load is switched off can destroy a transistor. Under-damped ringing shows up as overshoot on a scope. First- and second-order transient analysis predicts all of this from R, L and C.

Key ideas

Continuity (initial) conditions. Energy cannot change instantly with a finite source, so:

• the inductor current is continuous: iL(0+) = iL(0−);
• the capacitor voltage is continuous: vC(0+) = vC(0−).

Everything else (resistor currents, inductor voltage, capacitor current) may jump at the switching instant. Find iL(0−) and vC(0−) from the DC steady state before switching, where L is a short and C is an open.

First-order circuits (one L or one C). Any variable x(t) after a switching at t = 0 follows x(t) = x(∞) + [x(0+) − x(∞)]·e^(−t/τ).

• The time constant is τ = R_th·C or τ = L/R_th, where R_th is the Thevenin resistance seen by the energy-storage element after switching, with independent sources deactivated.
• After 1τ the response has covered 63.2% of the change. After 5τ it has covered 99.3%, which is taken as "settled".
• A larger R gives a larger RC time constant but a smaller L/R time constant.

Natural and forced response. The total response is the natural part (the decaying exponential, set by the circuit) plus the forced part (the steady state, set by the source). Equivalently, it is the zero-input response plus the zero-state response.

Second-order circuits (series RLC with a step input). The characteristic equation is s² + (R/L)·s + 1/(LC) = 0.

• Damping coefficient: α = R/(2L). Undamped natural frequency: ω0 = 1/√(LC). Damping ratio: ζ = α/ω0 = (R/2)·√(C/L).
• Overdamped (ζ > 1): two real roots, a slow non-oscillatory response.
• Critically damped (ζ = 1): a repeated root, the fastest response without overshoot. This happens at R = 2√(L/C).
• Underdamped (ζ < 1): roots −α ± jωd, with damped frequency ωd = √(ω0² − α²). The response rings with an envelope e^(−αt).

For a parallel RLC, α = 1/(2RC), so a larger R gives less damping, the opposite of the series case.

Second-order initial conditions. You need x(0+) and dx/dt(0+). Use dvC/dt(0+) = iC(0+)/C and diL/dt(0+) = vL(0+)/L.

Formulas

• First order: x(t) = x(∞) + [x(0+) − x(∞)]·e^(−t/τ)
• Time constants: τ_RC = R·C (s, with R in Ω and C in F), τ_RL = L/R (s, with L in H).
• Time to go from x(0+) to x1: t = τ·ln[(x(0+) − x(∞))/(x1 − x(∞))]
• Stored energy: w_L = ½·L·i², w_C = ½·C·v² (J).
• Series RLC: α = R/(2L) (s⁻¹), ω0 = 1/√(LC) (rad/s), ζ = (R/2)·√(C/L), ωd = ω0·√(1 − ζ²).
• Parallel RLC: α = 1/(2RC), ζ = (1/(2R))·√(L/C).
• Roots: s1,2 = −α ± √(α² − ω0²).

Worked examples

Example 1 (standard): RC charging from a non-zero initial voltage. Given: a 12 V source, R = 4 kΩ and C = 50 µF. The capacitor holds 2 V at t = 0, when the switch closes. Find vC(t) and the time at which vC = 10 V.

1. Time constant: τ = R·C = 4000 × 50×10⁻⁶ = 0.2 s.
2. Initial and final values: vC(0+) = 2 V, vC(∞) = 12 V.
3. Response: vC(t) = 12 + (2 − 12)·e^(−t/0.2) = 12 − 10·e^(−5t) V.
4. Set vC = 10 V: 10 = 12 − 10·e^(−5t), so e^(−5t) = 0.2.
5. Solve: t = 0.2·ln 5 = 0.322 s.

Answer: vC(t) = 12 − 10·e^(−5t) V; vC = 10 V at t = 0.322 s.

Example 2 (GATE level): inductive kick when a switch opens. Given: a 24 V source with series 4 Ω feeds a 2 H inductor, which is in parallel with a 12 Ω resistor. The circuit is in steady state. At t = 0 a switch disconnects the source and its 4 Ω resistor, leaving the inductor and the 12 Ω resistor as a closed loop. Find iL(t), the voltage across the 12 Ω resistor at t = 0+, and the energy dissipated after t = 0.

1. Before switching, the inductor is a short, so it carries iL(0−) = 24/4 = 6 A and the 12 Ω resistor carries 0.
2. Continuity: iL(0+) = 6 A.
3. After switching, the inductor sees only 12 Ω: τ = L/R = 2/12 = 0.1667 s.
4. Response: iL(t) = 6·e^(−6t) A.
5. The 6 A must now flow through the 12 Ω resistor, against its earlier reference direction. The voltage across it jumps from 0 to a magnitude of 6 × 12 = 72 V, three times the supply voltage.
6. Energy dissipated: w = ½·L·i² = ½ × 2 × 6² = 36 J.

Answer: iL = 6·e^(−6t) A, |v(0+)| = 72 V, 36 J dissipated.

This is why relay coils are fitted with freewheeling diodes.

Example 3 (damping check). Given: a series RLC with R = 2 Ω, L = 0.5 H and C = 0.125 F.

1. α = 2/(2 × 0.5) = 2 s⁻¹.
2. ω0 = 1/√(0.0625) = 4 rad/s.
3. ζ = 2/4 = 0.5, so the circuit is underdamped.
4. ωd = √(16 − 4) = 3.46 rad/s.

Common mistakes

• Assuming a resistor current or an inductor voltage is continuous at t = 0. Only iL and vC are.
• Using the source-side resistance before switching to compute τ. Use R_th seen by L or C after switching.
• Saying a larger R increases the RL time constant. τ = L/R decreases.
• Using the series-RLC damping formula for a parallel RLC.
• Forgetting the second initial condition, dx/dt(0+), in second-order problems.
• Mixing ms and µF: 1 kΩ × 1 µF = 1 ms.

For GATE IN

Expect: iL or vC at a given time after a switch operates (NAT); the time to reach a threshold; energy stored or dissipated; the damping type of an RLC circuit, or the R for critical damping; the initial values of currents and derivatives just after switching. Practise drawing the t = 0− circuit (L short, C open) and the t = ∞ circuit quickly, and finding R_th seen by the storage element.

Quick check

1. What is τ for R = 10 kΩ and C = 10 µF?
2. What fraction of the final change is reached after 3τ?
3. A series RLC has L = 1 H and C = 1 µF. What R gives critical damping?
4. Is the inductor voltage continuous at a switching instant?

Answers: 1. 0.1 s. 2. 1 − e⁻³ = 95.0%. 3. R = 2√(L/C) = 2 kΩ. 4. No; only the inductor current is continuous.