Short notes

RL and RC Transient Analysis Short Notes

Switch a 9V battery into an RC circuit (R = 10 kΩ, C = 100 µF) at t = 0 and the capacitor voltage does not jump instantly — it rises as v_C(t) = 9(1 − e^(−t/τ)) where τ = RC = 10 kΩ × 100 µF = 1 s. After one time constant (t = 1s), the voltage is 9 × 0.632 = 5.69V; after five time constants the capacitor is essentially fully charged. That exponential rise, its time constant, and the initial and final values are the complete description of the RC step response.

EEE, ECE, EI

How it works

For an RC circuit with step input V_s applied at t = 0 with zero initial condition: v_C(t) = V_s(1 − e^(−t/RC))u(t). Time constant τ = RC in seconds. For an RL circuit: i_L(t) = (V_s/R)(1 − e^(−Rt/L))u(t), τ = L/R. The complete response = natural response + forced response. Natural response is determined by initial conditions and decays as e^(−t/τ); forced response is the steady-state value. General formula: x(t) = x(∞) + [x(0⁺) − x(∞)]e^(−t/τ), where x(0⁺) is the initial value and x(∞) is the final (steady-state) value.

Key points to remember

Key initial and final conditions: capacitor voltage cannot change instantaneously (v_C(0⁺) = v_C(0⁻)); inductor current cannot change instantaneously (i_L(0⁺) = i_L(0⁻)). At t = 0⁺, a capacitor with zero initial charge acts as a short circuit; a charged capacitor acts as a voltage source. At steady state (t → ∞), a capacitor is open circuit and an inductor is short circuit for DC. For a source-free RC discharge: v_C(t) = V₀e^(−t/RC). Energy stored in capacitor at steady state: W_C = ½CV². Only half the source energy is stored; the other half is dissipated in R regardless of R value.

Exam tip

The examiner always asks you to find the complete response of an RL or RC circuit by identifying x(0⁺) and x(∞) and substituting into the single-formula solution — draw the equivalent circuits at t=0⁺ and t=∞ separately before writing any equation.

More Network Analysis notes