Question

A capacitor of 10.0 ?F and a resistor of 90.0 Ω are quickly connected in series to a battery of 8.00 V. What is the charge ? on the capacitor 0.00200 s after the connection is made?

Answer 1

Charge on the capacitor, Q = Qo * [1 - exp(-t/)]
Where Qo is the maximum charge the capacitor will have, t is the time taken to charge and is the time constant
Qo = C * V
= (10 * 10^-6) * 8
= 80 * 10^-6 C
The time constant of the circuit, = R * C
Where R is the resistance and C is the capacitance.
= 90 * 10 * 10^-6
= 0.9 * 10^-3 s

Substituting values,
Q = (80 * 10^-6) * [1 - exp[-(2 * 10^-3) / (0.9 * 10^-3)]]
= (80 * 10^-6) * [1 - 0.108]
= 7.13 * 10^-5 C