A series RC circuit has a time constant of 0.900 s. The battery has an emf of 24.0 V, and the maximum current in the circuit is 0.500 mA.
-What is the charge stored in the capacitor 2.00 s after the switch is closed?
A)401 μC
B)18.8 μC
C)108 μC
D)450 μC

Question

Quizplus · Accepted Answer

We can use the formula for charge in a capacitor as a function of time:

Q(t) = Q_max * (1 - e^(-t/τ))

where Q_max is the maximum charge that the capacitor can hold (which is equal to the maximum current multiplied by the time constant), and τ is the time constant (which is equal to the product of the resistance and capacitance in the circuit).

In this problem, we are given τ = 0.900 s, I_max = 0.500 mA, and E = 24.0 V. Therefore, we can calculate:

Q_max = I_max * τ = 0.500 mA * 0.900 s = 0.450 μC

Now we can use the formula above to find the charge in the capacitor after 2.00 s:

Q(2.00 s) = 0.450 μC * (1 - e^(-2.00 s/0.900 s)) = 401 μC

Therefore, the answer is (A) 401 μC.

A Series RC Circuit Has a Time Constant of 0