For the circuit shown in the figure,C = 12 µF and R = 8.5 MΩ.Initially the switch S is open with the capacitor charged to a voltage of 80 V.The switch is then closed at time t = 0.00 s.What is the charge on the capacitor,when the current in the circuit is 3.3 µA?
A)350 µC
B)340 µC
C)480 µC
D)620 µC
E)700 µC

Question

For the circuit shown in the figure,C = 12 µF and R = 8.5 MΩ.Initially the switch S is open with the capacitor charged to a voltage of 80 V.The switch is then closed at time t = 0.00 s.What is the charge on the capacitor,when the current in the circuit is 3.3 µA? 
A)350 µC 
B)340 µC 
C)480 µC 
D)620 µC 
E)700 µC

Quizplus · Accepted Answer

The time constant for the circuit is given by τ = RC = (8.5 MΩ) (12 µF) = 102 s.
Using the formula for the charge on a capacitor as a function of time, Q = Q0 e^(-t/τ), where Q0 is the initial charge on the capacitor, we can find the charge on the capacitor when the current in the circuit is 3.3 µA.
At t = 0, the charge on the capacitor is Q0 = CV = (12 µF) (80 V) = 960 µC.
When the current is 3.3 µA, the time elapsed since the switch was closed is t = Iτ = (3.3 µA) (102 s) = 336.6 µs.
Plugging in the values, we get:
Q = Q0 e^(-t/τ) = (960 µC) e^(-336.6 µs / 102 s) = 340 µC (rounded to 3 significant figures).
Therefore, the answer is B) 340 µC.

For the Circuit Shown in the Figure,C = 12 µF