Classical: Vc and ic with 10 A source

Assumptions: the 10 A current source is in parallel with a 20 Ω resistor. At t = 0 the switch connects this to a second 20 Ω resistor in parallel with C = 10 μF. The capacitor is initially uncharged.

Differential equation

After the switch closes, the 10 A source feeds the parallel combination of the two 20 Ω resistors and the capacitor. KCL at the top node:

10 = v/20 + v/20 + C dv/dt
10⁻⁵ dv/dt + 0.1 v = 10
dv/dt + 10⁴ v = 10⁶

Solution

• Particular (final) value: v_p = 10⁶/10⁴ = 100 V (= 10 A × (20 ∥ 20) = 10 × 10).
• Complementary: s = −10⁴, so v_c = K e^(−10⁴t); time constant τ = 10 Ω × 10 μF = 100 μs.
• v_C(0⁺) = 0, so K = −100.

V_c(t) = 100 (1 − e^(−10⁴ t)) V,   t ≥ 0

Capacitor current

i_c(t) = C dv_C/dt = 10×10⁻⁶ × 100 × 10⁴ e^(−10⁴t) = 10 e^(−10⁴ t) A

Check: at t = 0⁺ the capacitor is a short, so the whole 10 A flows through it ✓. At t → ∞ i_c → 0 and the 10 A divides equally between the two 20 Ω resistors (5 A each, 100 V across them ✓).

Answer: V_c(t) = 100(1 − e^(−10⁴t)) V and i_c(t) = 10e^(−10⁴t) A.