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Laplace: switch a to b, capacitor current and voltage

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.

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.]

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.

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:

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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