For the circuit shown in the figure, the switch S is initially open and the capacitor voltage is 80 V. The switch is then closed at time t = 0. How long after closing the switch will the current in the resistor be 7.0 µA?
A) 87 s
B) 95 s
C) 78 s
D) 69 s
E) 61 s

Question

For the circuit shown in the figure, the switch S is initially open and the capacitor voltage is 80 V. The switch is then closed at time t = 0. How long after closing the switch will the current in the resistor be 7.0 µA? 
A) 87 s 
B) 95 s 
C) 78 s 
D) 69 s 
E) 61 s

Quizplus · Accepted Answer

Since the capacitor is initially charged to 80 V, the initial current through the resistor is 0 A. When the switch is closed, the capacitor will start to discharge through the resistor. The current through the resistor is given by:
$$I_R = \frac{V_C}{R}\cdot e^{-\frac{t}{RC}}$$
where $V_C$ is the initial voltage on the capacitor, $R$ is the resistance, $C$ is the capacitance, and $t$ is the time since the switch was closed.
We are given that $I_R = 7.0 \mu 	ext{A}$ and $V_C = 80 	ext{ V}$, and $R = 50 	ext{ k}\Omega$. We can solve for the time $t$:
$$7.0 \mu 	ext{A} = \frac{80 	ext{ V}}{50 	ext{ k}\Omega}\cdot e^{-\frac{t}{(50 	ext{ k}\Omega)(10 	ext{ nF})}}$$
Solving for $t$ yields $t \approx 87 	ext{ s}$. Therefore, the answer is choice A.

For the Circuit Shown in the Figure, the Switch S