Electric circuit responses

When a circuit containing inductance or capacitance is switched, voltages and currents do not jump instantly to their final values: energy stored in L and C must change gradually, producing a transient that dies out to leave the steady state. Understanding these responses is essential for switching surges, relay timing, filters, power supplies and electronic timing circuits. This topic covers DC transients and steady-state behaviour of R-L, R-C and R-L-C circuits.

Key concepts

• Steady state vs transient: steady state is the condition after transients have decayed (theoretically infinite time; practically about 5 time constants). Total response = transient (natural) response + steady-state (forced) response.
• Rules at switching:
  • Current through an inductor cannot change instantaneously, $v_L = L\,di/dt$. At t = 0⁺, an uncharged inductor acts as an open circuit; in DC steady state it acts as a short circuit.
  • Voltage across a capacitor cannot change instantaneously, $i_C = C\,dv/dt$. At t = 0⁺, an uncharged capacitor acts as a short circuit; in DC steady state it is an open circuit.
• R-L circuit (DC switching on): $i(t) = \frac{V}{R}(1 - e^{-t/\tau})$, time constant $\tau = L/R$. At t = τ current reaches 63.2 % of its final value. Decay (source removed): $i = I_0e^{-t/\tau}$ — current falls to 36.8 % at t = τ. Inductor voltage $v_L = Ve^{-t/\tau}$.
• R-C circuit (charging): $v_C = V(1 - e^{-t/\tau})$, $i = \frac{V}{R}e^{-t/\tau}$, $\tau = RC$. Discharge: $v_C = V_0e^{-t/\tau}$.
• Energy stored: inductor $\frac12LI^2$ (magnetic field), capacitor $\frac12CV^2$ (electric field). In charging a capacitor from a DC source through R, half the energy supplied is lost in R regardless of R.
• R-L-C series circuit (second-order): KVL gives $L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} = 0$ (source-free). Characteristic roots $s = -\alpha \pm\sqrt{\alpha^2 - \omega_0^2}$, with $\alpha = R/2L$ and $\omega_0 = 1/\sqrt{LC}$.
  • Overdamped (α > ω₀, $R > 2\sqrt{L/C}$): two real roots, slow non-oscillatory response.
  • Critically damped (α = ω₀, $R = 2\sqrt{L/C}$): fastest response without overshoot.
  • Underdamped (α < ω₀): damped oscillations at $\omega_d = \sqrt{\omega_0^2 - \alpha^2}$, envelope $e^{-\alpha t}$.
  • Damping ratio $\zeta = \frac{R}{2}\sqrt{\frac{C}{L}}$ (series RLC); ζ > 1 over, = 1 critical, < 1 under, 0 = undamped (sustained oscillation, R = 0).
• Parallel RLC: $\alpha = 1/(2RC)$; same classification with ω₀ = 1/√(LC).
• Steady-state AC response: with sinusoidal sources, after transients die, analyse with phasors and impedances: $Z_R = R$, $Z_L = j\omega L$, $Z_C = 1/(j\omega C) = -j/(\omega C)$. In an R-L circuit current lags voltage by $\phi = \tan^{-1}(\omega L/R)$; in R-C it leads by $\tan^{-1}(1/\omega CR)$.
• AC switching transient: when a sinusoidal voltage is switched onto an R-L circuit, the current contains a decaying DC offset whose size depends on the switching instant; maximum offset (nearly double peak current) occurs when switched at voltage zero for a highly inductive circuit — explaining the DC offset in fault currents.
• Laplace method: transform the circuit (L → sL with initial-current source, C → 1/sC with initial-voltage source), solve algebraically and invert.

High-yield points

• $\tau_{RL} = L/R$, $\tau_{RC} = RC$; at t = τ a rising response reaches 63.2 %, a decaying one falls to 36.8 %.
• Inductor current and capacitor voltage cannot change instantaneously.
• DC steady state: inductor = short circuit, capacitor = open circuit.
• Series RLC: $\alpha = R/2L$, $\omega_0 = 1/\sqrt{LC}$; critical damping at $R = 2\sqrt{L/C}$.
• Stored energy: $\frac12LI^2$ and $\frac12CV^2$; charging C through R wastes half the supplied energy.

Formulas

$\tau_{RL} = \frac{L}{R},\qquad \tau_{RC} = RC\ (\text{seconds})$

$x(t) = x(\infty) + [x(0^+) - x(\infty)]e^{-t/\tau}\quad\text{(general first-order response)}$

$\alpha = \frac{R}{2L},\quad \omega_0 = \frac{1}{\sqrt{LC}},\quad R_{critical} = 2\sqrt{\frac{L}{C}}$

Worked example

A 100 V DC source is switched onto R = 10 Ω in series with L = 0.5 H. Find τ, final current, current at t = 0.05 s and the energy finally stored.

1. τ = L/R = 0.5/10 = 0.05 s.
2. Final current = 100/10 = 10 A.
3. At t = τ: $i = 10(1 - e^{-1}) = 10\times0.632 = 6.32$ A.
4. Energy = ½ × 0.5 × 10² = 25 J.

Exam tips

• After one time constant: rise to 63.2 %, decay to 36.8 %; after 5τ the response is about 99.3 % complete.
• Inductor: open at t = 0⁺ (if no initial current), short in DC steady state; capacitor: the reverse.
• Increasing R makes τ_RL smaller but τ_RC larger — a common trap.
• Critical damping in series RLC: $R = 2\sqrt{L/C}$.
• Inductor current and capacitor voltage are the continuous (state) variables at switching.

Exam-style questions

From Calmon's question bank, which has 24 questions on this topic.

1. For a series RLC circuit with $L = 0.1$ H and $C = 10\,\mu F$, the resistance that gives a critically damped response is:

• A. $100\,\Omega$
• B. $400\,\Omega$
• C. $20\,\Omega$
• D. $200\,\Omega$

Answer: D. Critical damping: $R = 2\sqrt{L/C} = 2\sqrt{0.1/10^{-5}} = 2 \times 100 = 200\,\Omega$.

2. A 50 V DC source is switched onto a series circuit of $R = 10\,\Omega$ and $L = 2$ H. The time constant and steady-state current are:

• A. 0.2 s and 25 A
• B. 0.2 s and 5 A
• C. 5 s and 5 A
• D. 20 s and 0.2 A

Answer: B. $\tau = L/R = 2/10 = 0.2$ s; $I_{ss} = V/R = 50/10 = 5$ A.

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