For a Thevenin equivalent with V_th = 12 V and R_th = 2 Ω feeding a capacitor of 100 μF that is initially uncharged, what is the current through the circuit at the instant the capacitor is connected (t=0+)?

For a Thevenin equivalent with V_th = 12 V and R_th = 2 Ω feeding a capacitor of 100 μF that is initially uncharged, what is the current through the circuit at the instant the capacitor is connected (t=0+)?

• A. 6 A
• B. 12 A
• C. 0 A
• D. 3 A

Answer: (A)

When a capacitor is connected to a Thevenin source, the initial current is governed by the Thevenin resistance because the capacitor voltage cannot change instantly. Since the capacitor is uncharged, its voltage at the very first moment is 0 V, so it looks like a short circuit. The current at t = 0+ is simply V_th divided by R_th. Here that gives 12 V / 2 Ω = 6 A. The time constant is τ = R_th × C = 2 Ω × 100 μF = 0.0002 s, so the current will then decay from 6 A as the capacitor charges, following i(t) = 6 e^{-t/τ}.

When a capacitor is connected to a Thevenin source, the initial current is governed by the Thevenin resistance because the capacitor voltage cannot change instantly. Since the capacitor is uncharged, its voltage at the very first moment is 0 V, so it looks like a short circuit. The current at t = 0+ is simply V_th divided by R_th.

Here that gives 12 V / 2 Ω = 6 A.

The time constant is τ = R_th × C = 2 Ω × 100 μF = 0.0002 s, so the current will then decay from 6 A as the capacitor charges, following i(t) = 6 e^{-t/τ}.