3.3 Transient Response (RL, RC, RLC)

Continuity rules set the initial conditions

Two physical constraints drive every transient problem:

• Capacitor voltage cannot jump (a jump needs infinite current): v_C(0+) = v_C(0-).
• Inductor current cannot jump (a jump needs infinite voltage): i_L(0+) = i_L(0-).

The standard procedure: analyze the circuit just before switching (t = 0-) at its old DC steady state to find the continuous quantity, carry that value across the switch to t = 0+, then analyze the new DC steady state (t = infinity) to find the final value. At DC steady state a capacitor is an open circuit (no current) and an inductor is a short circuit (no voltage). These two facts let you find both the t = 0- starting point and the t = infinity endpoint by inspection, leaving only the time constant to compute.

First-order time constants and the universal formula

A circuit with one independent energy-storage element is first order. Its time constant tau (seconds) is:

• RC circuit: tau = R*C, where R is the Thevenin resistance seen by the capacitor (kill independent sources, look back from the capacitor's terminals).
• RL circuit: tau = L/R, where R is the Thevenin resistance seen by the inductor.

The universal first-order formula describes any voltage or current in the circuit:

x(t) = x(infinity) + [x(0+) - x(infinity)] * e^(-t/tau).

Engineers treat the transient as effectively complete after about five time constants. A common trap is using the bare source resistance instead of the full Thevenin resistance seen by the storage element. Another trap is mismatched units: capacitance is usually given in microfarads (1 microF = 1e-6 F) and the product R*C in ohm-farads yields seconds, so a 10 kohm by 100 microF combination gives 10,000 * 100e-6 = 1 s, not 1,000,000 s.

The larger the time constant, the slower the response: bigger R or C (or bigger L) stretches the exponential. This is why an RC low-pass filter with a large tau smooths fast signals, and why a power-supply filter capacitor holds voltage between rectified peaks.

Worked RC example and natural vs. forced response

Worked example: A 100 microfarad capacitor, initially charged to 50 V, discharges through a 10 kohm resistor at t = 0. Here tau = RC = 10,000 * 100e-6 = 1 s, x(0+) = 50 V, and x(infinity) = 0 V. So v_C(t) = 0 + (50 - 0)e^(-t/1) = 50e^(-t) V. After one time constant (t = 1 s), v_C = 500.368 = 18.4 V (about 37% of the start). After 5 s the capacitor is essentially discharged.

The natural response is how the circuit decays on stored energy with no source (the discharge above: v(t) = V_0e^(-t/RC)). The step (forced) response is the reaction to a DC source applied at t = 0 (charging: v(t) = V_s(1 - e^(-t/RC))). The complete response is the sum of natural (transient) and forced (steady-state) parts — the universal formula already combines them, which is why it works whether the element is charging or discharging.

For an inductor the dual relations hold: charging current i(t) = I_final*(1 - e^(-t/tau)) and decaying current i(t) = I_0e^(-t/tau), with tau = L/R. Worked RL example: an inductor with 2 A initially, in a loop with R = 4 ohm and L = 8 H and no source, decays as i(t) = 2e^(-t/2) A (tau = 8/4 = 2 s); after 2 s the current is 2*0.368 = 0.74 A. Always identify which quantity is continuous (v_C or i_L), find its t = 0+ and t = infinity values, compute tau from the Thevenin resistance at the storage element, then assemble x(t) with the universal formula.

Second-order RLC response

A circuit with both L and C is second order. Its behavior depends on the damping ratio zeta, comparing the neper (damping) frequency alpha to the undamped natural frequency omega_0:

• Series RLC: alpha = R/(2L), omega_0 = 1/sqrt(LC).
• Parallel RLC: alpha = 1/(2RC), omega_0 = 1/sqrt(LC).
• zeta = alpha / omega_0.

Underdamped responses ring (oscillate while decaying); critically damped responses settle fastest without oscillation, which is often the design target. The damped (ringing) frequency omega_d = sqrt(omega_0^2 - alpha^2) is always less than omega_0, and as alpha shrinks toward zero the response approaches a pure undamped oscillation at omega_0. Note that in the series case a larger R increases alpha and pushes toward overdamping, whereas in the parallel case a larger R decreases alpha and pushes toward underdamping — the resistance plays opposite roles, a frequent FE trap.

Worked example: a series RLC with R = 200 ohm, L = 50 mH, C = 0.2 microF. Here alpha = R/(2L) = 200/(20.05) = 2000 rad/s and omega_0 = 1/sqrt(LC) = 1/sqrt(0.050.2e-6) = 1/sqrt(1e-8) = 10,000 rad/s. Since alpha < omega_0 (2000 < 10,000), the circuit is underdamped and rings at omega_d = sqrt(10,000^2 - 2000^2) = sqrt(1e8 - 4e6) = 9,798 rad/s, just below omega_0.