---
title: "Laplace: series RLC with 6e^(-2t) source"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-series-rlc-exponential-source-6e"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-series-rlc-exponential-source-6e
site: BCT Engineering
---

# Laplace: series RLC with 6e^(-2t) source

Electrical Circuit Theory · Chapter 4: Transient analysis in RLC circuit by Laplace Transform

- Asked in: 2069 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-series-rlc-exponential-source-6e
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

In a series RLC, as shown in figure below find the value of current for t > 0, also find the voltage across capacitor for t > 0, using Laplace transform method.
[Figure: source v(t) = 6e^(−2t) with a switch (t = 0), a 5 Ω resistor, a 1 H inductor and a ¼ F capacitor with an initial voltage of 2 V (+ on the top plate) in a single loop with current i]

## Short answer

Write s-domain loop equation with source 6/(s+2) and initial capacitor voltage 2 V opposing. Solve I(s), partial fractions give three exponential modes: i(t)=−8/3 e^−t+6e^−2t−10/3 e^−4t A. Integrate to obtain vC(t)=32/3 e^−t−12e^−2t+10/3 e^−4t V.

## Answer

**Assumptions:** R = 5 Ω, L = 1 H, C = ¼ F, no initial inductor current, and the capacitor is connected so that i charges its + plate, i.e. its initial voltage v_C(0) = 2 V opposes the loop current direction in the usual way (drop convention).

#### s-domain equation
Source V(s) = 6/(s + 2); impedance Z(s) = R + sL + 1/(sC) = s + 5 + 4/s. The charged capacitor is a series source 2/s opposing the current.

```text
I(s) = [ 6/(s + 2) − 2/s ] / (s + 5 + 4/s)
     = [ 4(s − 1) / (s (s + 2)) ] × s / (s² + 5s + 4)
     = 4 (s − 1) / [ (s + 1)(s + 2)(s + 4) ]
```

#### Partial fractions for the current
```text
K1 = 4(−2) / ((1)(3))   = −8/3     (s = −1)
K2 = 4(−3) / ((−1)(2))  = 6        (s = −2)
K3 = 4(−5) / ((−3)(−2)) = −10/3    (s = −4)
```
**i(t) = −(8/3) e^(−t) + 6 e^(−2t) − (10/3) e^(−4t) A, t > 0**

(i(0) = −2.667 + 6 − 3.333 = 0 ✓.)

#### Capacitor voltage
v_C = v_C(0) + (1/C)∫i dt, so in the s-domain Vc(s) = 2/s + 4 I(s)/s:

```text
Vc(s) = 2 (s² + 7s + 22) / [ (s + 1)(s + 2)(s + 4) ]
      = (32/3)/(s + 1) − 12/(s + 2) + (10/3)/(s + 4)
```
**v_C(t) = (32/3) e^(−t) − 12 e^(−2t) + (10/3) e^(−4t) V, t > 0**

Check: v_C(0) = 10.667 − 12 + 3.333 = 2 V ✓. Substituting both expressions in the loop equation i′ + 5i + v_C = 6e^(−2t) (L = 1) gives equality at every t; for example at t = 0.5 s both sides equal 2.207. The formulas also agree with a numerical solution of the loop equations (error below 10⁻¹²).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the s-domain source and total impedance?
   A. V(s) = 6/(s+2) and Z = s + 5 + 4/s
   B. V(s) = 6/s and Z = s + 5 + 4s
   C. V(s) = 6/(s+2) and Z = 5 + 4s + 1/s
   D. V(s) = 6/s and Z = s + 5 + 1/(4s)... 4/s
   Answer: A. 6e^(-2t) transforms to 6/(s+2); 1/(sC) = 4/s with C = 1/4, L = 1. (Key point: Source and impedance)

2. How is the 2 V initial capacitor voltage treated?
   A. It is added to the source as +2/s
   B. A series source 2/s that opposes the current, so I = [6/(s+2) - 2/s]/Z
   C. It is ignored
   D. It is added as 2 to the impedance
   Answer: B. The charged capacitor appears as V0/s in series, here taken opposing the loop current. (Key point: Capacitor initial voltage)

3. How is vC(s) found from I(s)?
   A. Vc(s) = 4 I(s)
   B. Vc(s) = sL I(s)
   C. Vc(s) = 2/s + 4 I(s)/s (initial voltage plus integral of current)
   D. Vc(s) = 2 I(s)
   Answer: C. vC = vC(0) + (1/C) integral of i, so Vc = 2/s + I/(sC) with 1/C = 4. (Key point: Capacitor voltage)

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