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Laplace: capacitor voltage and current, switch a/b

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.

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]

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:

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

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)).

✓ Answer checked by Claude Opus 5.5 (high effort)

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