Question

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible internal resistance. find the time constant of the circuit. what is the charge of the capacitor 1.95 time constants after the circuit is closed? what is the charge after a long time?

Answer 1

In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$, the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$, we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$. We can see this also from the previous formule, by using $t=\infty$:
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$