9.07 Transient Response of RL and RC Circuits using Laplace

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9.07 Transient Response of RL & RC Circuits using Laplace Transform

Alright — let’s proceed to Note 9.07: Transient Response of RL & RC Circuits using Laplace Transform! This is one of the highest-yielding topics of Chapter 9 under Instructor 1, routinely carrying 10 to 13 marks on KUET Section B examinations.

While classical time-domain transient analysis requires solving messy, first-order differential equations and using integrating factors to find homogeneous and particular solutions, the unilateral Laplace transform turns these calculus bottlenecks into simple, high-speed algebraic loop equations in the s-domain.

In this note, we will mathematically formulate the step and impulse transient response workflows for first-order RL and RC circuits, master the exact s-domain loop equation derivations, and analyze high-yield solved past year questions involving switch-moving transitions {moving a switch from position 1 to position 2 at $t=0$}.

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1. Transient Response of first-order series RC Circuits

We begin by establishing the complete transient response workflow for a series resistor-capacitor (RC) circuit.

1.1 Time-Domain Differential Equation

Consider a series RC circuit connected to a time-varying input voltage source $v_{in}(t)$. By applying Kirchhoff’s Voltage Law (KVL) around the loop: $v_{in}(t)=v_R(t)+v_C(t)=R\cdot i(t)+v_C(t)$

Since current flowing through the capacitor is $i(t)=C\frac{d{v}_C(t)}{dt}$, we substitute this to find the governing first-order constant-coefficient differential equation: $RC\frac{d{v}_C(t)}{dt}+v_C(t)=v_{in}(t)$

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1.2 s-Domain Loop Equation Derivation

Instead of solving this differential equation in the time domain, let’s take the unilateral Laplace transform of the KVL equation: $V_{in}(s)=R\cdot I(s)+V_C(s)$

Recall from Note 9.06 that the s-domain series Thévenin equivalent of a capacitor with an initial voltage $v_C(0^{-})$ immediately prior to a switch transition is: $V_C(s)=\frac{1}{Cs}I(s)+\frac{v_C(0^{-})}{s}$

Substituting this s-domain modeling equation back into the KVL equation: $V_{in}(s)=R\cdot I(s)+\frac{1}{Cs}I(s)+\frac{v_C(0^{-})}{s}$

This represents the complete s-domain algebraic loop equation. We can visualize this algebraically transformed loop using standard schematic models:

[CIRCUIT: Series RC s-Domain Equivalent Loop with Initial Voltage]
                  I(s)             1/Cs
  o───────────────>───────────────┤├───( + ─ )───o
  │                                       │      │
( + )                                     │   v_C(0⁻)/s
V_in(s)                                  [R]     │
  │                                       │      │
  o───────────────────────────────────────┴──────o

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1.3 Algebraic Solution for Capacitor Voltage $V_C(s)$

We isolate the s-domain loop current $I(s)$: $V_{in}(s)-\frac{v_C(0^{-})}{s}=I(s)\left(R+\frac{1}{Cs}\right)$

$I(s)=\frac{V_{in}(s)-\frac{v_C(0^{-})}{s}}{R+\frac{1}{Cs}}=\frac{s{V}_{in}(s)-v_C(0^{-})}{s\left(R+\frac{1}{Cs}\right)}=\frac{Cs{V}_{in}(s)-C{v}_C(0^{-})}{RCs+1}$

Now, substitute this current expression back into our s-domain capacitor voltage equation $V_C(s)$ to find the general system response: $V_C(s)=\frac{1}{Cs}\left[\frac{Cs{V}_{in}(s)-C{v}_C(0^{-})}{RCs+1}\right]+\frac{v_C(0^{-})}{s}$

$V_C(s)=\frac{V_{in}(s)-\frac{v_C(0^{-})}{s}}{RCs+1}+\frac{v_C(0^{-})}{s}=\frac{V_{in}(s)}{RCs+1}-\frac{v_C(0^{-})}{s(RCs+1)}+\frac{v_C(0^{-})(RCs+1)}{s(RCs+1)}$

$V_C(s)=\frac{V_{in}(s)}{RCs+1}+\frac{RC{v}_C(0^{-})}{RCs+1}$

To align with standard pole-zero rational formats, divide the numerator and denominator by the time constant $RC$ {where $\tau =RC$ represents the circuit charging time constant}: $V_C(s)=\left(\frac{\frac{1}{RC}}{s+\frac{1}{RC}}\right)V_{in}(s)+\frac{v_C(0^{-})}{s+\frac{1}{RC}}$

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1.4 Step Response Evaluation

Let’s evaluate the circuit’s response to a DC step input voltage of amplitude $V_0$ applied at $t=0$: $v_{in}(t)=V_{0}u(t)\leftrightarrow V_{in}(s)=\frac{V_0}{s}$

Substituting $V_{in}(s)$ into our general $V_C(s)$ expression: $V_C(s)=\frac{\frac{V_0}{RC}}{s\left(s+\frac{1}{RC}\right)}+\frac{v_C(0^{-})}{s+\frac{1}{RC}}$

Applying Partial Fraction Expansion to the first term (ZSR): $\frac{\frac{1}{RC}}{s\left(s+\frac{1}{RC}\right)}=\frac{A}{s}+\frac{B}{s+\frac{1}{RC}}$

• For residue $A$: $A={\frac{\frac{1}{RC}}{s+\frac{1}{RC}}|}_{s=0}=1$
• For residue $B$: $B={\frac{\frac{1}{RC}}{s}|}_{s=-\frac{1}{RC}}=-1$

Thus, the s-domain capacitor voltage equation is: $V_C(s)=V_0\left[\frac{1}{s}-\frac{1}{s+\frac{1}{RC}}\right]+\frac{v_C(0^{-})}{s+\frac{1}{RC}}$

Taking the inverse unilateral Laplace transform back to the time domain: $v_C(t)=\underset{\text{Zero-State Response (ZSR)}}{\underbrace{V_0\left(1-e^{-t/RC}\right)u(t)}}+\underset{\text{Zero-Input Response (ZIR)}}{\underbrace{v_C(0^{-})e^{-t/RC}u(t)}}$

NOTE

Physical Interpretation of ZSR and ZIR:

• Zero-State Response (ZSR): Represents the circuit’s response to the external step source, assuming no initial energy was stored on the capacitor plates ($v_C(0^{-})=0$ V). The capacitor voltage charges exponentially toward $V_0$ with time constant $\tau =RC$.
• Zero-Input Response (ZIR): Represents the circuit’s natural discharge response due strictly to the initial voltage $v_C(0^{-})$ stored on the capacitor plates, assuming the external input source is short-circuited.

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2. Transient Response of first-order series RL Circuits

We now apply the same Laplace workflow to derive the transient response of a series resistor-inductor (RL) circuit.

2.1 Time-Domain Differential Equation

Applying KVL around a series loop containing a resistor $R$, inductor $L$, and input voltage source $v_{in}(t)$: $v_{in}(t)=v_R(t)+v_L(t)=R\cdot i(t)+L\frac{di(t)}{dt}$

This is the governing first-order linear differential equation of the series RL network.

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2.2 s-Domain Loop Equation Derivation

Taking the unilateral Laplace transform of the KVL loop equation: $V_{in}(s)=R\cdot I(s)+V_L(s)$

Recall from Note 9.06 that the s-domain Thévenin series model of an inductor carrying an initial current $i(0^{-})$ immediately prior to a transient switch action is: $V_L(s)=Ls\cdot I(s)-L\cdot i(0^{-})$

Substitute this s-domain model into the KVL loop equation: $V_{in}(s)=R\cdot I(s)+Ls\cdot I(s)-L\cdot i(0^{-})$

This represents the complete s-domain algebraic loop equation.

[CIRCUIT: Series RL s-Domain Equivalent Loop with Initial Current]
                  I(s)             Ls
  o───────────────>───────────────[UUUU]───( ─ + )───o
  │                                       │      │
( + )                                     │   L·i(0⁻)
V_in(s)                                  [R]     │
  │                                       │      │
  o───────────────────────────────────────┴──────o

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2.3 Algebraic Solution for Loop Current $I(s)$

We group the s-domain current terms and isolate $I(s)$: $V_{in}(s)+L\cdot i(0^{-})=I(s)[Ls+R]$

$I(s)=\frac{V_{in}(s)}{Ls+R}+\frac{L\cdot i(0^{-})}{Ls+R}$

Divide the numerator and denominator of both terms by the inductance $L$ to format poles clearly {where $\tau =L/R$ represents the RL circuit time constant}: $I(s)=\left(\frac{\frac{1}{L}}{s+\frac{R}{L}}\right)V_{in}(s)+\frac{i(0^{-})}{s+\frac{R}{L}}$

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2.4 Step Response Evaluation

If the series RL circuit is subjected to a constant DC step input voltage of amplitude $V_0$ applied at $t=0$: $v_{in}(t)=V_{0}u(t)\leftrightarrow V_{in}(s)=\frac{V_0}{s}$

Substitute $V_{in}(s)$ into our general $I(s)$ equation: $I(s)=\frac{\frac{V_0}{L}}{s\left(s+\frac{R}{L}\right)}+\frac{i(0^{-})}{s+\frac{R}{L}}$

Applying Partial Fraction Expansion to the first term (ZSR): $\frac{\frac{1}{L}}{s\left(s+\frac{R}{L}\right)}=\frac{A}{s}+\frac{B}{s+\frac{R}{L}}$

• For residue $A$: $A={\frac{\frac{1}{L}}{s+\frac{R}{L}}|}_{s=0}=\frac{1}{R}$
• For residue $B$: $B={\frac{\frac{1}{L}}{s}|}_{s=-\frac{R}{L}}=-\frac{1}{R}$

Substituting the evaluated residues back into $I(s)$: $I(s)=\frac{V_0}{R}\left[\frac{1}{s}-\frac{1}{s+\frac{R}{L}}\right]+\frac{i(0^{-})}{s+\frac{R}{L}}$

Taking the inverse unilateral Laplace transform back to the time domain: $i(t)=\underset{\text{Zero-State Response (ZSR)}}{\underbrace{\frac{V_0}{R}\left(1-e^{-\frac{R}{L}t}\right)u(t)}}+\underset{\text{Zero-Input Response (ZIR)}}{\underbrace{i(0^{-})e^{-\frac{R}{L}t}u(t)}}$

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2.5 Impulse Response Evaluation

Now let’s find the system’s impulse response {the current response when the input is a unit impulse delta function}: $v_{in}(t)=\delta (t)\leftrightarrow V_{in}(s)=1$

Substitute $V_{in}(s)=1$ and assume zero initial current ($i(0^{-})=0$): $I(s)=\frac{\frac{1}{L}}{s+\frac{R}{L}}$

Taking the inverse unilateral Laplace transform: $i(t)={\mathcal{L}}^{-1}\left\{\frac{\frac{1}{L}}{s+\frac{R}{L}}\right\}=\frac{1}{L}{e}^{-\frac{R}{L}t}u(t)\text{ A}$

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3. High-Yield Switch-Moving Transition Solved Problems

On KUET examinations, the most common first-order transient questions involve switch-moving transitions {where a switch has been at position 1 for a long time to establish steady-state, and is moved to position 2 at $t=0$}. Let’s solve two highly tested exam patterns step-by-step.

3.1 Problem 1: The 5-Mark series RL switch transition [KUET PYQ 2021, 2018, 2017]

Question: In the circuit shown below, determine the current $i(t)$ for $t\ge 0$ when the switch is moved from position 1 to position 2 at $t=0$. Initially, the switch S has been at position 1 for a long time to reach steady-state.

Given Circuit Parameters:

• DC Voltage Source: $V=10\text{ V}$
• Resistors: $R_1=5\Omega$, $R_2=5\Omega$
• Inductor: $L=2\text{ H}$

Step 1: Analyze the circuit at $t<0$ (Switch at Position 1)

For a long time prior to $t=0$, the switch S is at position 1. Under DC steady-state, the inductor acts as a perfect short circuit {its frequency-domain impedance $Ls$ becomes $j(0)L=0$}.

Looking at the loop connected to position 1:

• The 10V DC source is connected in series with resistor $R_1=5\Omega$ and the inductor $L=2\text{ H}$.
• Resistor $R_2=5\Omega$ is open-circuited (disconnected from the loop).

Apply Ohm’s Law to calculate the steady-state inductor current: $i_L(0^{-})=\frac{V}{R_1}=\frac{10\text{ V}}{5\Omega }=2\text{ A}$

By the conservation of flux linkage, the current flowing through the inductor coils cannot change instantaneously: $i_L(0^{+})=i_L(0^{-})=2\text{ A}$

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Step 2: Draw the s-Domain Equivalent Circuit for $t\ge 0$ (Switch at Position 2)

At $t=0$, the switch S moves instantaneously to position 2.

• The 10V DC source and resistor $R_1=5\Omega$ are completely disconnected from the circuit.
• The inductor $L=2\text{ H}$ is now connected in series with the second resistor $R_2=5\Omega$ in a closed, sourceless loop.

We transform this transient circuit into the s-domain:

• The resistor $R_2$ remains a constant impedance: $Z_{R2}(s)=5\Omega$.
• The inductor is modeled using the Thévenin series equivalent: an impedance $Z_L(s)=Ls=2s$ in series with an independent voltage source of value $L\cdot i_L(0^{-})=2\text{ H}\times 2\text{ A}=4\text{ V}$.
• The polarity of this series voltage source points in the direction of the initial current flow (opposing KVL drop).

[CIRCUIT: s-Domain Loop for Position 2 Series RL Circuit]
          I(s)      2s
  o────────>───────[UUUU]───( ─ + )───o
  │                                    │
 [5Ω]                                4 V
  │                                    │
  o────────────────────────────────────o

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Step 3: Formulate and Solve the s-Domain Loop Equation

Apply KVL around the s-domain loop in the direction of $I(s)$: $R_{2}I(s)+LsI(s)-L\cdot i_L(0^{-})=0$

Substitute our parameters ($R_2=5$, $L=2$, $L\cdot i_L(0^{-})=4$): $5I(s)+2sI(s)-4=0$

Group the current terms and isolate $I(s)$: $I(s)[2s+5]=4⟹I(s)=\frac{4}{2s+5}$

Divide the numerator and denominator by 2 to isolate the pole coefficient: $I(s)=\frac{2}{s+2.5}$

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Step 4: Perform the Inverse Laplace Transform

Taking the inverse unilateral Laplace transform back to the time domain: $i(t)={\mathcal{L}}^{-1}\left\{\frac{2}{s+2.5}\right\}=2e^{-2.5t}u(t)\text{ A}$

TIP

Quick Classical Verification: Let’s verify this using the classical time-domain equation: $i(t)=i(0)e^{-t/\tau }=i_L(0^{-})e^{-\frac{R_2}{L}t}=2e^{-\frac{5}{2}t}=2e^{-2.5t}\text{ A}$ The s-domain algebraic solution perfectly and elegantly matches the classical derivation!

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3.2 Problem 2: Causal series RC switch transition with dual resistors

Question: A series RC circuit consists of a 10V DC source, a switch S, resistors $R_1=10\Omega$ and $R_2=20\Omega$, and a capacitor $C=0.1\text{ F}$. Initially, the switch S has been at position 1 for a long time. At $t=0$, the switch S moves to position 2. Determine the transient capacitor voltage $v_C(t)$ and the loop current $i(t)$ for $t\ge 0$.

Step 1: Analyze the circuit at $t<0$ (Switch at Position 1)

With the switch at position 1 for a long time, the capacitor is fully charged. Under DC steady-state, the capacitor acts as an open circuit {its s-domain impedance $1/Cs$ approaches infinity as $s\to 0$}: $v_C(0^{-})=V_{dc}=10\text{ V}$

By the conservation of charge, the voltage across the capacitor plates cannot change instantaneously: $v_C(0^{+})=v_C(0^{-})=10\text{ V}$

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Step 2: Draw the s-Domain Equivalent Circuit for $t\ge 0$ (Switch at Position 2)

At $t=0$, the switch S connects the capacitor $C=0.1\text{ F}$ in series with the resistor $R_2=20\Omega$ in a closed loop. We model this discharging loop in the s-domain:

• Resistor impedance: $Z_{R2}(s)=20\Omega$.
• Capacitor impedance: $Z_C(s)=\frac{1}{Cs}=\frac{1}{0.1s}=\frac{10}{s}$.
• Capacitor initial voltage series step source: $\frac{v_C(0^{-})}{s}=\frac{10}{s}\text{ V}$.

[CIRCUIT: s-Domain Discharging RC Loop with Initial Voltage Step]
          I(s)     10/s
  o────────>───────┤├───( + ─ )───o
  │                               │
 [20Ω]                          10/s
  │                               │
  o───────────────────────────────o

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Step 3: Formulate and Solve the s-Domain Loop Equation

Apply KVL around the discharging loop in the direction of $I(s)$: $R_{2}I(s)+\frac{1}{Cs}I(s)+\frac{v_C(0^{-})}{s}=0$

Substitute the parameters ($R_2=20$, $C=0.1$, $v_C(0^{-})=10$): $20I(s)+\frac{10}{s}I(s)+\frac{10}{s}=0$

Isolate the current term $I(s)$: $I(s)\left[20+\frac{10}{s}\right]=-\frac{10}{s}⟹I(s)\left[\frac{20s+10}{s}\right]=-\frac{10}{s}$

Multiply both sides by $s$ to eliminate the denominator: $I(s)[20s+10]=-10⟹I(s)=\frac{-10}{20s+10}=\frac{-0.5}{s+0.5}$

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Step 4: Perform the Inverse Laplace Transform for Current and Voltage

A. Current Response $i(t)$

Taking the inverse unilateral Laplace transform of $I(s)$: $i(t)={\mathcal{L}}^{-1}\left\{\frac{-0.5}{s+0.5}\right\}=-0.5e^{-0.5t}u(t)\text{ A}$ {The negative sign physically indicates that the current is flowing out of the capacitor as it discharges, in the opposite direction of the charging current}.

B. Capacitor Voltage Response $v_C(t)$

Using our s-domain capacitor voltage modeling equation: $V_C(s)=\frac{1}{Cs}I(s)+\frac{v_C(0^{-})}{s}$

Substitute our expressions for $I(s)$ and $v_C(0^{-})$: $V_C(s)=\frac{10}{s}\left(\frac{-0.5}{s+0.5}\right)+\frac{10}{s}=\frac{-5}{s(s+0.5)}+\frac{10}{s}$

Apply Partial Fraction Expansion to the first term: $\frac{-5}{s(s+0.5)}=\frac{A}{s}+\frac{B}{s+0.5}$

• $A={\frac{-5}{s+0.5}|}_{s=0}=-10$
• $B={\frac{-5}{s}|}_{s=-0.5}=10$

Reassembling the expanded terms: $V_C(s)=\left[\frac{-10}{s}+\frac{10}{s+0.5}\right]+\frac{10}{s}=\frac{10}{s+0.5}$

Taking the inverse unilateral Laplace transform: $v_C(t)={\mathcal{L}}^{-1}\left\{\frac{10}{s+0.5}\right\}=10e^{-0.5t}u(t)\text{ V}$

The capacitor voltage decays exponentially from its initial charged state of $10\text{ V}$ down to $0\text{ V}$ with a discharging time constant $\tau =R_{2}C=20\Omega \times 0.1\text{ F}=2\text{ s}$ {which corresponds to a decay exponent coefficient of $\lambda =1/\tau =0.5{\text{ s}}^{-1}$}!

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4. ECE 2108 Laboratory Connection: MATLAB transient Simulation

In ECE 2108 Laboratory Experiment 3, you utilize MATLAB to computationally solve first-order differential equations and simulate circuit transients. The following script utilizes the symbolic engine dsolve and the control system step commands to simulate and plot the step and impulse responses of first-order RL/RC networks.

 Method 1: Symbolic Differential Equation Solver (dsolve)
% Solve: L * di/dt + R * i = V0 with i(0) = 0
syms i(t)
eqn = L * diff(i, t) + R * i == V0;
cond = i(0) == 0;
i_sol(t) = dsolve(eqn, cond);
 
fprintf('Symbolic Current Equation i(t):\n');
disp(i_sol(t));
 
%% Method 2: State-Space & Transfer Function step Response
% Transfer Function H(s) = I(s)/V(s) = (1/L) / (s + R/L)
num = [1/L];
den = [1, R/L];
sys = tf(num, den);
 
% Generate time vector
t_sim = 0:0.01:4;
 
% Compute step response (multiplied by V0 to match DC step)
[i_step, t_out] = step(V0 * sys, t_sim);
 
% Plot the transient response
figure;
plot(t_out, i_step, 'LineWidth', 2.5, 'Color', [0, 0.4470, 0.7410]);
grid on;
title('Series RL Circuit Step Current Response [ECE 2108]');
xlabel('Time t (seconds)');
ylabel('Current i(t) (Amperes)');
legend('MATLAB Step Response');

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5. Common Mistakes That Cost Marks

WARNING

Marks-Deduction Pitfalls to Avoid in Exam Transients:

1. Shifting-Time Confusion during Switch Moves: In switch-moving transitions, always define $t=0$ as the moment the switch moves to position 2. Do not carry over time-shifted notations (like $u(t-t_0)$) unless explicitly instructed — start the Laplace analysis fresh from $t=0^{-}$ and use standard unilateral bounds.
2. Omit Inductor Voltage Source Polarities: When converting an inductor with initial current $i_L(0^{-})$ to the s-domain Thévenin series model, the series voltage source $-L{i}_L(0^{-})$ acts as a voltage boost pointing in the direction of the initial current. If you draw this source with the positive terminal opposing current flow, your KVL equation will yield an incorrect sign, resulting in a 3 to 4 mark penalty.
3. Mixing Radian and Cyclic Frequencies: When evaluating the time constant $\tau =RC$ or $\tau =L/R$, do not confuse decay coefficients $\lambda =1/\tau$ with frequency limits. Keep units clearly marked as Neper/sec or rad/s.

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6. PYQ Bank — Verbatim Questions & Answer Plans

Q1: The 9-Mark switch-moving series RL transient [PYQ 2023 Q7c]

Question: In the circuit of Fig. 7(c), find the current $i(t)$ when the switch is at position 2. The switch s is moved from position 1 to position 2 at time $t=0$. Initially the switch has been at position 1 for a long time. (09 Marks)

• Answer Plan:
  1. Analyze position 1 for $t<0$ in DC steady-state: short-circuit the inductor to find $i_L(0^{-})=\frac{V_{dc}}{R_1}$.
  2. For $t\ge 0$, draw the s-domain series equivalent loop with $R_2$, inductor impedance $Ls$, and the series voltage source $L{i}_L(0^{-})$ pointing in the direction of $i_L(0^{-})$.
  3. Formulate the loop equation: $R_{2}I(s)+LsI(s)-L{i}_L(0^{-})=0$.
  4. Solve for $I(s)=\frac{L{i}_L(0^{-})}{Ls+R_2}$.
  5. Perform inverse Laplace transform to obtain the decaying exponential current $i(t)=i_L(0^{-})e^{-\frac{R_2}{L}t}u(t)$.

Q2: The 5-Mark series RL 10V circuit [PYQ 2021 Q5c]

Question: In the circuit drawn below, find the current $i(t)$ when the switch is at position 2. The switch S moved from position 1 to position 2 at time $t=0$. Initially the switch has been at position 1 for a long time. {Circuit: 10V source, $R_1=5\Omega$, $R_2=5\Omega$, $L=2\text{ H}$}. (05 Marks)

• Answer Plan:
  1. Follow the detailed solution derivation in Section 3.1 of this note.
  2. State the initial condition $i_L(0^{-})=10\text{ V}/5\Omega =2\text{ A}$.
  3. Formulate $I(s)=\frac{4}{2s+5}=\frac{2}{s+2.5}$.
  4. Write the final time-domain expression: $i(t)=2e^{-2.5t}u(t)\text{ A}$.

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7. Self-Check Before Moving On

• Can you derive the s-domain loop equation for a first-order RC circuit including initial conditions? (Section 1.2)
• Do you understand why the series voltage source representing initial inductor current points in the direction of the initial flow? (Section 2.2)
• Can you solve a 10-mark KUET exam switch transition RL problem step-by-step using unilateral Laplace transforms? (Section 3.1)
• Do you know how to write a MATLAB script utilizing symbolic engines to solve first-order differential transients? (Section 4)

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