First-Order Circuits | Transient Analysis

Tutorial 14 — First-Order RC & RL Circuits

A first-order circuit contains a single energy-storage element (a capacitor or an inductor) and reduces to a first-order differential equation. Every response in this tutorial follows the general formula \(x(t) = x(\infty) + \big[x(0^{+}) - x(\infty)\big]\,e^{-t/\tau}\), where \(\tau = R_{Th}C\) for RC circuits and \(\tau = L/R_{Th}\) for RL circuits. These problems cover time constants, natural (source-free) responses, step responses, energy delivered or absorbed, dependent sources, and sequentially switched networks.

Transient Analysis · Electric Circuit Analysis · 16 solved problems

Demonstrative VideosWalkthrough

Part I — RC Circuits

Problems 1–10 deal with capacitive first-order circuits: time constants, source-free discharge, step (charging) responses, and energy transfer.

Problem 1RC Time Constant

Determine the time constant of the network.

Solution

The time constant is set by the resistance seen by the capacitor. Deactivating the source (short-circuit the 10 V battery), the capacitor sees \(R\) in parallel with \(2R\):

\[ R_{Th} = R \parallel 2R = \frac{R \cdot 2R}{R + 2R} = \frac{2}{3}R \]

Resistance seen by the capacitor for Problem 1

Hence the time constant:

\[ \tau = R_{Th}\,C = \frac{2}{3}RC \ \text{sec} \]

Answer\(\tau = \dfrac{2}{3}RC\ \text{sec}\)

Problem 2RC Discharge — Charge

In the circuit shown, the initial capacitor voltage is 4 V. Switch \(S_1\) is closed at \(t=0\). Find the charge lost by the capacitor from \(t = 25\ \mu s\) to \(t = 100\ \mu s\).

Solution

With the switch closed, the charged capacitor (\(C = 5\ \mu F\)) discharges through \(R = 5\ \Omega\). The discharge current is:

\[ \tau = R\,C = (5)(5\times10^{-6}) = 25\times10^{-6}\ \text{sec}, \qquad i(t) = \frac{V}{R}\,e^{-t/\tau} = \frac{4}{5}\,e^{-t/\tau} \]

The charge lost is the integral of the current over the given interval:

\[ Q = \int_{25\,\mu s}^{100\,\mu s} i(t)\,dt = \frac{4}{5}\,\tau\Big[e^{-25/25} - e^{-100/25}\Big] = (20\times10^{-6})\big(e^{-1} - e^{-4}\big) = 6.99\times10^{-6}\ \text{C} \]

Answer\(Q \approx 6.99\ \mu\text{C}\)

Problem 3RC Switching Transient

\(S_1\) was closed and \(S_2\) was open for a very long time. At \(t=0\), \(S_1\) is opened and \(S_2\) is closed. Determine the voltage across the capacitor at \(t = 5\ \mu s\).

Solution

Initial value — at \(t = 0^{-}\) the capacitor was charged by the 1 V source, so:

\[ V_c(0^{-}) = V_c(0^{+}) = 1\ \text{V} \]

Final value — at \(t = \infty\) (capacitor open), the 3 V source divides across the 1 Ω and 2 Ω:

\[ V_c(\infty) = 3 \times \frac{2}{2 + 1} = 2\ \text{V} \]

Time constant and response — the capacitor (\(C = 10\ \mu F\)) sees \(R_{Th} = 2 \parallel 1 = \tfrac{2}{3}\ \Omega\):

\[ \tau = R_{Th}C = \frac{2}{3}\times 10\times 10^{-6}, \qquad V_c(t) = \big[V_c(0^{+}) - V_c(\infty)\big]e^{-t/\tau} + V_c(\infty) = [1 - 2]e^{-t/\tau} + 2 \]

At \(t = 5\ \mu s\), with \(t/\tau = 0.75\):

\[ V_c(5\,\mu s) = 2 - e^{-0.75} = 1.52\ \text{V} \]

Answer\(V_c(5\,\mu s) \approx 1.52\ \text{V}\)

Problem 4Source-Free RC

The switch has been closed for a long time and opens at \(t=0\). Find \(v(t)\) for \(t \geq 0\).

Solution

Initial value — for \(t < 0\) (switch closed, capacitor open), the source voltage divides across the 10 kΩ and 2 kΩ:

\[ v_o = v(0) = \frac{2}{10 + 2}(24) = 4\ \text{V} \]

For \(t > 0\) the source is disconnected; the capacitor discharges through the 2 kΩ resistor:

\[ \tau = RC = (2\times10^{3})(40\times10^{-6}) = \frac{2}{25}\ \text{s}, \qquad v(t) = v_o\,e^{-t/\tau} = 4\,e^{-12.5t}\ \text{V} \]

Answer\(v(t) = 4e^{-12.5t}\ \text{V}\)

Problem 5Source-Free RC — Decay Time

For the circuit below, find \(v_o(t)\) for \(t > 0\). Determine the time necessary for the capacitor voltage to decay to one-third of its value at \(t = 0\).

Solution

Initial value — for \(t < 0\) (switch closed, capacitor open), the 60 V source divides across the 9 kΩ and 3 kΩ:

\[ v(0^{-}) = \frac{3}{3 + 9}(60) = 15\ \text{V} \]

For \(t > 0\) the source is removed, giving a source-free RC circuit with \(R = 3\ \text{k}\Omega\) and \(C = 20\ \mu F\):

\[ \tau = RC = (3\times10^{3})(20\times10^{-6}) = 0.06\ \text{s}, \qquad v_o(t) = 15\,e^{-16.667t}\ \text{V} \]

Let \(t_0\) be the time at which \(v_o\) falls to one-third of its initial value. Because this ratio depends only on \(\tau\), the initial amplitude cancels:

\[ \frac{1}{3}v_o(0) = v_o(0)\,e^{-16.667\,t_0} \;\Rightarrow\; e^{16.667\,t_0} = 3 \;\Rightarrow\; t_0 = \frac{\ln 3}{16.667} = 65.92\ \text{ms} \]

Answer\(v_o(t) = 15e^{-16.667t}\ \text{V},\quad t_0 = 65.92\ \text{ms}\)

Problem 6RC with AC Sources

Determine the current \(i(t)\) through the capacitor.

Solution

Applying source transformation, the two sources combine into a single voltage source \(2\sqrt{2}\sin(5t - 45^{\circ})\) in series with the 2 Ω resistor and 0.1 F capacitor. The series impedance at \(\omega = 5\) rad/s is:

\[ Z = \sqrt{R^{2} + \frac{1}{(C\omega)^{2}}} = \sqrt{4 + \frac{1}{(0.1\times5)^{2}}} = 2\sqrt{2}\ \Omega\ \angle\,{-45^{\circ}} \]

The capacitor current is the source divided by this impedance:

\[ i = \frac{2\sqrt{2}\sin(5t - 45^{\circ})}{2\sqrt{2}\,\angle(-45^{\circ})} = \sin 5t\ \text{A} \]

Answer\(i(t) = \sin 5t\ \text{A}\)

Problem 7Source-Free RC

The capacitor shown is initially charged to 10 V. The switch closes at \(t = 0\). Find \(v_c(t)\) at \(t = 10\) ms.

Solution

Closing the switch connects \(R = 10\ \text{k}\Omega\) across the charged capacitor (\(C = 1\ \mu F\)), producing a source-free discharge:

\[ v_c(t) = V_o\,e^{-t/\tau} = 10\,e^{-100t}, \qquad \tau = RC = \frac{1}{100}\ \text{s} \]

At \(t = 10\) ms:

\[ v_c(10\,\text{ms}) = 10\,e^{-(100\times10\times10^{-3})} = 10\,e^{-1} = 3.678\ \text{V} \]

Answer\(v_c(10\,\text{ms}) = 3.678\ \text{V}\)

Problem 8RC Step Response

The capacitor initially has a charge of 10 coulomb. Find the current in the circuit 1 second after the switch \(S\) is closed.

Solution

Initial and final values — the initial capacitor voltage comes from its stored charge (\(C = 0.5\ \text{F}\)), and the final value is the source voltage:

\[ V_c(0^{-}) = V_c(0^{+}) = \frac{Q}{C} = \frac{10}{0.5} = 20\ \text{V}, \qquad V_c(\infty) = 100\ \text{V} \]

With \(\tau = RC = (2)(0.5) = 1\) s, the capacitor voltage and current are:

\[ \begin{aligned} V_c(t) &= V_c(\infty) + \big[V_c(0^{+}) - V_c(\infty)\big]e^{-t/RC} = 100 + (20 - 100)e^{-t} = 100 - 80e^{-t}\\ i_c(t) &= C\,\frac{dV_c(t)}{dt} = 0.5\,(80\,e^{-t}) = 40\,e^{-t} \end{aligned} \]

At \(t = 1\) s:

\[ i_c(1) = 40\,e^{-1} = 14.71\ \text{A} \]

Answer\(i_c(1\,\text{s}) = 14.71\ \text{A}\)

Problem 9RC Energy — Bridge Network

The initial charge on the 1 F capacitor is zero. Find the energy (in joules) transformed from the DC source until steady state is reached.

Solution

Reducing the resistor bridge by a delta–star transformation, the capacitor sees an equivalent resistance of 5 Ω in series with the 10 V source, so \(\tau = RC = (5)(1) = 5\) s. The charging current is:

\[ i(t) = \frac{V}{R}\,e^{-t/\tau} = \frac{10}{5}\,e^{-t/5} = 2\,e^{-0.2t}\ \text{A} \]

Reduced circuit after delta-star for Problem 9

The energy supplied by the source is the integral of \(v\,i\) over all time. (Equivalently, \(E = \int V i\,dt = V\,Q_\infty = CV^{2}\), independent of \(R\).)

\[ E = \int_{0}^{\infty} 10 \times 2\,e^{-0.2t}\,dt = 100\ \text{J} \]

Answer\(E = 100\ \text{J}\)

Problem 10RC Energy

Find the energy absorbed by the 4 Ω resistor in the interval \((0, \infty)\). Given \(V_c(0) = 6\) V.

Solution

The capacitor current follows the general first-order form \(i_C(t) = i_C(\infty) + \big[i_C(0^{+}) - i_C(\infty)\big]e^{-t/\tau}\), with the given initial voltage \(V_C(0^{-}) = V_C(0^{+}) = 6\) V.

Boundary values and time constant — at \(0^{+}\) the resistor carries \((10-6)/4\); at steady state the capacitor blocks DC; and \(\tau = RC = (4)(2)\):

\[ i_C(0^{+}) = \frac{10 - 6}{4} = 1\ \text{A}, \qquad i_C(\infty) = 0, \qquad \tau = RC = 8\ \text{s} \]

Transient and steady-state setups for Problem 10

Substituting gives the current, and the energy is the integral of \(i_C^{2}R\):

\[ i_C(t) = e^{-t/8}\ \text{A}, \qquad E = \int_{0}^{\infty} i_C^{2}(t)\,R\,dt = \int_{0}^{\infty} 4\,e^{-t/4}\,dt = 4(-4)\big[e^{-t/4}\big]_{0}^{\infty} = 16\ \text{J} \]

Answer\(E = 16\ \text{J}\)

Part II — RL Circuits

Problems 11–14 deal with inductive first-order circuits, where the time constant is \(\tau = L/R_{Th}\) and the inductor current cannot change instantaneously.

Problem 11RL with Dependent Source

Assuming that \(i(0) = 10\) A, calculate \(i(t)\) and \(i_x(t)\).

Solution

Method 1 — because of the dependent source, apply a 1 V test source \(v_o\) at the inductor terminals and solve the mesh equations:

\[ \begin{aligned} 2(i_1 - i_2) + 1 &= 0\\ 6 i_2 - 2 i_1 - 3 i_1 &= 0\\ i_1 = -3\ \text{A}, &\quad i_o = -i_1 = 3\ \text{A} \end{aligned} \]

Hence the Thevenin resistance, time constant and natural response:

\[ R_{eq} = R_{Th} = \frac{v_o}{i_o} = \frac{1}{3}\ \Omega, \qquad \tau = \frac{L}{R_{eq}} = \frac{1/2}{1/3} = \frac{3}{2}\ \text{s}, \qquad i(t) = i(0)\,e^{-t/\tau} = 10\,e^{-(2/3)t}\ \text{A},\ t>0 \]

Method 2 — alternatively, write the circuit's differential equation directly:

\[ \begin{aligned} \frac{1}{2}\frac{di_1}{dt} + 2(i_1 - i_2) &= 0 \;\Rightarrow\; \frac{di_1}{dt} + 4 i_1 - 4 i_2 = 0\\ 6 i_2 - 2 i_1 - 3 i_1 = 0 &\;\Rightarrow\; i_2 = \frac{5}{6}i_1\\ \frac{di_1}{dt} + \frac{2}{3}i_1 = 0 &\;\Rightarrow\; i(t) = i(0)\,e^{-(2/3)t} = 10\,e^{-(2/3)t}\ \text{A},\ t>0 \end{aligned} \]

Differential-equation setup for Problem 11

Finally, the inductor voltage gives \(i_x\) through the 2 Ω resistor:

\[ v = L\frac{di}{dt} = 0.5(10)\!\left(-\frac{2}{3}\right)e^{-(2/3)t} = -\frac{10}{3}e^{-(2/3)t}\ \text{V}, \qquad i_x(t) = \frac{v}{2} = -1.6667\,e^{-(2/3)t}\ \text{A},\ t>0 \]

Answer\(i(t) = 10e^{-(2/3)t}\ \text{A},\quad i_x(t) = -1.6667e^{-(2/3)t}\ \text{A}\)

Problem 12Source-Free RL

The switch has been closed for a long time. At \(t = 0\) the switch is opened. Calculate \(i(t)\) for \(t > 0\).

Solution

Initial value — for \(t < 0\) the switch is closed, the inductor acts as a short, and the 16 Ω resistor is shorted out:

\[ R_{eq} = \frac{4\times12}{4+12} = 3\ \Omega, \qquad i_1 = \frac{40}{2+3} = 8\ \text{A}, \qquad i(0^{-}) = \frac{12}{12+4}\,i_1 = 6\ \text{A} \]

For \(t > 0\) the source is disconnected; the inductor current starts at \(i(0^{+}) = i(0^{-}) = 6\) A and decays through the remaining resistance:

\[ R_{eq} = (12 + 4)\parallel 16 = 8\ \Omega, \qquad \tau = \frac{L}{R_{eq}} = \frac{2}{8} = \frac{1}{4}\ \text{s}, \qquad i(t) = i(0)\,e^{-t/\tau} = 6\,e^{-4t}\ \text{A} \]

Answer\(i(t) = 6e^{-4t}\ \text{A},\ t>0\)

Problem 13RL Complete Response

Find \(i_o\), \(v_o\) and \(i\) for all time, assuming the switch was open for a long time.

Solution

For \(t < 0\) the switch is open, the inductor is a short, and the 6 Ω resistor is shorted:

\[ i_o = 0, \qquad i(t) = \frac{10}{2 + 3} = 2\ \text{A}, \qquad v_o(t) = 3\,i(t) = 6\ \text{V}, \qquad i(0) = 2\ \text{A} \]

For \(t > 0\) the source-free network has \(R_{Th} = 3 \parallel 6 = 2\ \Omega\), giving:

\[ \tau = \frac{L}{R_{Th}} = 1\ \text{s}, \qquad i(t) = i(0)\,e^{-t/\tau} = 2\,e^{-t}\ \text{A}, \qquad v_o(t) = -v_L = -L\frac{di}{dt} = 4\,e^{-t}\ \text{V} \]

The 6 Ω current follows from the inductor voltage, and assembling all intervals:

\[ i_o(t) = \frac{v_L}{6} = -\frac{2}{3}e^{-t}\ \text{A},\ t>0 \]

\[ i_o(t) = \begin{cases} 0\ \text{A}, & t < 0\\ -\tfrac{2}{3}e^{-t}\ \text{A}, & t > 0 \end{cases} \quad v_o(t) = \begin{cases} 6\ \text{V}, & t < 0\\ 4e^{-t}\ \text{V}, & t > 0 \end{cases} \quad i(t) = \begin{cases} 2\ \text{A}, & t < 0\\ 2e^{-t}\ \text{A}, & t \geq 0 \end{cases} \]

Answer\(i(t) = 2e^{-t}\,\text{A},\ v_o = 4e^{-t}\,\text{V},\ i_o = -\tfrac{2}{3}e^{-t}\,\text{A}\ (t>0)\)

Problem 14RL with Two Switches

At \(t = 0\), switch-1 is closed, and switch-2 is closed 4 s later. Find \(i(t)\) for \(t > 0\), and calculate \(i\) at \(t = 2\) s and \(t = 5\) s.

Solution

Analyse the three intervals \(t \leq 0\), \(0 \leq t \leq 4\), and \(t \geq 4\) separately. For \(t < 0\) both switches are open, so:

\[ i(0^{-}) = i(0) = i(0^{+}) = 0 \]

For \(0 \leq t \leq 4\), \(S_1\) is closed (with \(S_2\) still open), so the 4 Ω and 6 Ω are in series. Assuming \(S_1\) stays closed:

\[ i(\infty) = \frac{40}{4+6} = 4\ \text{A}, \qquad R_{Th} = 4 + 6 = 10\ \Omega, \qquad \tau = \frac{L}{R_{Th}} = \frac{5}{10} = \frac{1}{2}\ \text{s} \]

Hence over this interval:

\[ i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} = 4 + (0-4)e^{-2t} = 4\big(1 - e^{-2t}\big)\ \text{A}, \quad 0 \leq t \leq 4 \]

For \(t \geq 4\), \(S_2\) closes and connects the 10 V source through the 2 Ω. Since inductor current is continuous, the new initial value is:

\[ i(4) = i(4^{-}) = 4\big(1 - e^{-8}\big) \simeq 4\ \text{A} \]

To find the new steady state, let \(v\) be the node voltage at P and apply KCL:

\[ \frac{40 - v}{4} + \frac{10 - v}{2} = \frac{v}{6} \;\Rightarrow\; v = \frac{180}{11}\ \text{V}, \qquad i(\infty) = \frac{v}{6} = \frac{30}{11} = 2.727\ \text{A} \]

The Thevenin resistance at the inductor terminals and the new time constant:

\[ R_{Th} = 4 \parallel 2 + 6 = \frac{4\times2}{6} + 6 = \frac{22}{3}\ \Omega, \qquad \tau = \frac{L}{R_{Th}} = \frac{5}{22/3} = \frac{15}{22}\ \text{s} \]

Shifting the exponential by the 4 s delay:

\[ i(t) = i(\infty) + [i(4) - i(\infty)]e^{-(t-4)/\tau} = 2.727 + 1.273\,e^{-1.4667(t-4)}\ \text{A}, \quad t \geq 4 \]

Putting all intervals together:

\[ i(t) = \begin{cases} 0, & t \leq 0\\ 4\big(1 - e^{-2t}\big), & 0 \leq t \leq 4\\ 2.727 + 1.273\,e^{-1.4667(t-4)}, & t \geq 4 \end{cases} \]

Evaluating at the requested times:

\[ i(2) = 4\big(1 - e^{-4}\big) = 3.93\ \text{A}, \qquad i(5) = 2.727 + 1.273\,e^{-1.4667} = 3.02\ \text{A} \]

Answer\(i(2) = 3.93\ \text{A},\quad i(5) = 3.02\ \text{A}\)

Additional Practice Problems

Two self-contained worked examples with schematics, illustrating the step (energizing) response that complements the source-free and switching cases above: an RC circuit charging from zero and an RL circuit building up current from zero.

Problem 15RC Step Response (Charging)

A 12 V source charges an initially uncharged 50 µF capacitor through a 4 kΩ resistor. The switch closes at \(t = 0\). Find \(v_c(t)\) for \(t \geq 0\) and the time to reach 90% of the final voltage.

Solution

The capacitor starts uncharged and charges toward the source voltage, so the boundary values and time constant are:

\[ v_c(0) = 0, \qquad v_c(\infty) = 12\ \text{V}, \qquad \tau = RC = (4\times10^{3})(50\times10^{-6}) = 0.2\ \text{s} \]

Applying the general first-order formula:

\[ v_c(t) = v_c(\infty) + [v_c(0) - v_c(\infty)]e^{-t/\tau} = 12\big(1 - e^{-5t}\big)\ \text{V} \]

The time to reach \(0.9\times12 = 10.8\) V:

\[ 0.9 = 1 - e^{-5t_{0.9}} \;\Rightarrow\; t_{0.9} = -\frac{\ln 0.1}{5} = 0.461\ \text{s} \]

Answer\(v_c(t) = 12\big(1 - e^{-5t}\big)\ \text{V},\quad t_{0.9} = 0.461\ \text{s}\)

Problem 16RL Step Response (Energizing)

A 24 V source energizes a 2 H inductor through an 8 Ω resistor. The inductor current is zero before the switch closes at \(t = 0\). Find \(i(t)\) for \(t \geq 0\) and the voltage across the inductor.

Solution

The inductor current starts at zero and builds toward its DC steady-state value \(V/R\); the time constant is \(L/R\):

\[ i(0) = 0, \qquad i(\infty) = \frac{24}{8} = 3\ \text{A}, \qquad \tau = \frac{L}{R} = \frac{2}{8} = 0.25\ \text{s} \]

Applying the general first-order formula:

\[ i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} = 3\big(1 - e^{-4t}\big)\ \text{A} \]

The inductor voltage follows from \(v_L = L\,di/dt\):

\[ v_L(t) = L\frac{di}{dt} = 2\,(3)(4)\,e^{-4t} = 24\,e^{-4t}\ \text{V} \]

Answer\(i(t) = 3\big(1 - e^{-4t}\big)\ \text{A},\quad v_L(t) = 24e^{-4t}\ \text{V}\)