---
title: "Laplace: capacitor voltage and current, switch a/b"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-switch-a-b-v0-r1"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-switch-a-b-v0-r1
site: BCT Engineering
---

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

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

- Asked in: 2073 Shrawan
- 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-b-v0-r1
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

Using Laplace Transform method in the circuit shown in figure below find the voltage and current of capacitor for t > 0.
[Figure: source V0; switch (t = 0) with position a connected directly to the V0 positive terminal and position b connected through R2 to the bottom wire; the switch pole is connected through R1 to the capacitor C, which returns to the bottom wire]

## Short answer

With initial capacitor voltage V0, the discharge loop for t>0 contains R1+R2 in series with C. Using s-domain with initial voltage source gives a decaying exponential: vC(t)=V0 e^−t/τ V, capacitor current i(t)=(V0/(R1+R2))e^−t/τ A (discharge), where τ=(R1+R2)C.

## Answer

**Assumption (figure not fully clear):** the switch has been at a for a long time, so the capacitor is charged to V₀. At t = 0 it moves to b, which disconnects the source and puts R₂ across the series path.

#### Initial condition
v_C(0⁻) = V₀ = v_C(0⁺)

#### s-domain circuit (t > 0)
The loop contains C (with the initial-voltage source V₀/s), R₁ and R₂ in series:

```text
I(s) = (V0/s) / (R1 + R2 + 1/(sC))
     = (V0 / (R1 + R2)) × 1/(s + 1/τ),     τ = (R1 + R2) C
```

#### Capacitor current
**i(t) = (V₀/(R₁ + R₂)) e^(−t/τ), t > 0**, with the sense of a discharge (flowing out of the + plate); in the direction of the original charging current it is −(V₀/(R₁ + R₂)) e^(−t/τ).

#### Capacitor voltage
```text
Vc(s) = V0/s − I(s)/(sC)  =  V0 / (s + 1/τ)
```
**v_C(t) = V₀ e^(−t/τ) V, t > 0**

Check: v_C(0) = V₀ ✓; the initial discharge current is V₀/(R₁ + R₂) ✓; and v_C(∞) = 0, since R₂ is connected to the bottom wire.

For the other reading (capacitor uncharged, switch moved from b to a at t = 0): I(s) = (V₀/R₁)/(s + 1/R₁C), so i = (V₀/R₁)e^(−t/R₁C) and v_C = V₀(1 − e^(−t/R₁C)).

## Check yourself

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

1. What is vC(0+) after the switch moves from a to b?
   A. V0, since the capacitor was fully charged and its voltage cannot jump
   B. 0
   C. V0 R2/(R1+R2)
   D. V0/2
   Answer: A. After a long time at a, the capacitor is charged to V0, and it keeps that voltage at 0+. (Key point: Initial condition)

2. What is the time constant for t > 0?
   A. tau = R1 C
   B. tau = (R1 + R2) C
   C. tau = R2 C
   D. tau = (R1 R2/(R1+R2)) C
   Answer: B. The capacitor discharges through R1 and R2 in series. (Key point: Time constant)

3. What is the capacitor voltage for t > 0?
   A. V0 (1 - e^(-t/tau))
   B. V0 e^(-t/R1C)
   C. V0 e^(-t/tau)
   D. V0 (R1/(R1+R2)) e^(-t/tau)
   Answer: C. Vc(s) = V0/(s + 1/tau), a discharge from V0 to 0 since R2 returns to ground. (Key point: Result)

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