A 4.0-μF capacitor is charged to 6.0-V. It is then connected in series with a 3.0-MΩ resistor and connected to a 12-V battery. How long after being connected to the battery will the voltage across the capacitor be 9.0 V?
A)5.5 s
B)7.2 s
C)8.3 s
D)12 s
E)17 s

Question

Quizplus · Accepted Answer

The time constant τ for an RC circuit is given by τ = R*C. Here, R = 3.0 MΩ and C = 4.0 μF, so τ = 3.0 * 10^6 * 4.0 * 10^-6 = 12 s. The voltage across the capacitor as a function of time t is given by V(t) = V_final + (V_initial - V_final) * e^(-t/τ). Solving for t when V(t) = 9.0 V, V_initial = 6.0 V, V_final = 12.0 V, and τ = 12 s, we find t ≈ 8.3 s.

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