---
title: "Laplace: switch a to b, capacitor current and voltage"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-switch-a-to-b-two"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-switch-a-to-b-two
site: BCT Engineering
---

# Laplace: switch a to b, capacitor current and voltage

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

- Asked in: 2074 Chaitra, 8 marks
- Video: [Circuit Analysis using Laplace Transform](https://www.youtube.com/watch?v=YQlcHUDaNS8) — Electrical Engineering Authority
- 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-switch-a-to-b-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

Keeping the switch at position 'a' for a long time, if the switch is moved to position 'b' at t = 0 in the circuit shown in figure below, find expressions for current through and voltage across capacitor using Laplace Transform method.
[Figure: 10 V source in series with a 1 Ω resistor connected to position a of the switch. The switch pole (t = 0) is connected to a 1 F capacitor to the bottom wire. Position b is connected through a 1 Ω resistor to a second 1 F capacitor (right side) to the bottom wire.]

## Short answer

Current is i(t)=10e^(−2t) A. Capacitor voltages: v_C1(t)=5+5e^(−2t) V, v_C2(t)=5−5e^(−2t) V. Final charge sharing gives 5 V each, time constant C_eq R = 0.5 s.

## Answer

**Before t = 0 (switch at a, steady state):** the 10 V source charges the first capacitor C₁ through 1 Ω; in steady state no current flows, so v_C1(0⁻) = 10 V. The second capacitor C₂ is uncharged (v_C2(0) = 0).

**For t > 0 (switch at b):** C₁ (10 V) is connected to C₂ through the 1 Ω resistor. The source is disconnected.

#### s-domain equation
Series loop: C₁ (with initial source 10/s), 1 Ω and C₂ (uncharged), total capacitive impedance 1/s + 1/s = 2/s.

```text
I(s) = (10/s) / (1 + 1/s + 1/s) = 10 / (s + 2)
```

#### Capacitor current and voltage
Current is the discharge current of C₁ (leaving the + plate):

**i(t) = 10 e^(−2t) A, t > 0**

Voltage across C₁ = 10/s − I/s:

```text
V1(s) = 10/s − 10/[s (s + 2)] = 10/s − 5/s + 5/(s + 2) = 5/s + 5/(s + 2)
```

**v_C1(t) = 5 + 5 e^(−2t) V**, and for the second capacitor V2(s) = I/s = 5/s − 5/(s+2):

**v_C2(t) = 5 − 5 e^(−2t) V**

Check: τ = R C_eq = 1 × (1×1)/(1+1) = 0.5 s (so e^(−2t)). Both voltages end at 5 V (charge sharing: 10 C = 2C × V → V = 5 V), and v_C1 − v_C2 = 10e^(−2t) = R i ✓. i(0) = 10/1 = 10 A.

## Check yourself

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

1. What are the capacitor voltages just before the switch moves?
   A. C1 = 10 V and C2 = 0 V
   B. C1 = 0 and C2 = 10 V
   C. Both 10 V
   D. Both 5 V
   Answer: A. At steady state with the switch at a, no current flows and C1 is charged to the source 10 V; C2 is uncharged. (Key point: Initial conditions)

2. What is the loop current for t > 0?
   A. 10 e^(-t) A
   B. 10 e^(-2t) A
   C. 5 e^(-2t) A
   D. 10 (1 - e^(-2t)) A
   Answer: B. I(s) = (10/s)/(1 + 2/s) = 10/(s+2), where the C's in series give 2/s. (Key point: Current)

3. What do both capacitors approach and why?
   A. 10 V each, because C1 is a source
   B. 0 V, since the source is removed
   C. 5 V each, by charge sharing (10C = 2C x V)
   D. 10 V and 0 V unchanged
   Answer: C. Charge is conserved; equal capacitors share it equally, giving 5 V each. vC1 = 5 + 5e^(-2t), vC2 = 5 - 5e^(-2t). (Key point: Final voltages)

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