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Laplace: i(t) with 12 V, 4 Ω, 1 F, 2 Ω

Short answer

Initial capacitor voltage is 4 V. After switch closes, s-domain node equation gives V(s). Current through 4 Ω is V/4: i(t) = 0.4286 + 0.5714 e^(−0.875t) A, or 3/7 + (4/7)e^(−7t/8) A, t ≥ 0.

Using Laplace transform method, find the current i(t) for t > 0 in the circuit shown in the figure below. [Figure: 12 V source with an 8 Ω resistor in the top wire; a 4 Ω resistor as a shunt branch carrying i(t) (downward); a 1 F capacitor as a shunt branch; and a switch (t = 0) in series with a 2 Ω resistor as the last shunt branch]

Assumption: the switch is open for a long time before t = 0, so the capacitor is charged by the 12 V source through 8 Ω and the 4 Ω resistor. The branch with 4 Ω carries i(t) downward.

Initial condition (t = 0⁻)

No current flows in the capacitor, so 8 Ω and 4 Ω form a divider:

v_C(0⁻) = 12 × 4/(8 + 4) = 4 V = v_C(0⁺)

For t > 0 (switch closed, extra 2 Ω in parallel)

Let V be the common node voltage. KCL (C = 1 F):

(12 − V)/8 = V/4 + V/2 + C dV/dt
dV/dt + 0.875 V = 1.5

s-domain (including the initial condition V(0) = 4):

(s + 0.875) V(s) = 1.5/s + 4
V(s) = (4 s + 1.5) / [ s (s + 0.875) ]
     = 1.7143/s + 2.2857/(s + 0.875)

V(t) = 1.7143 + 2.2857 e^(−0.875t) V

Current i(t) in the 4 Ω resistor

i = V/4:

i(t) = 0.4286 + 0.5714 e^(−0.875t) A = 3/7 + (4/7) e^(−7t/8) A, t ≥ 0

Check: i(0) = 1 A = 4 V/4 Ω ✓; i(∞): with C open the final node voltage is 12 × (4∥2)/(8 + 4∥2) = 12 × 1.333/9.333 = 1.714 V, so i = 1.714/4 = 0.4286 A ✓. τ = R_eq C = (8 ∥ 4 ∥ 2) × 1 = 1.143 s ✓ (1/0.875).

✓ Answer checked by Claude Opus 5.5 (high effort)

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