For the circuit shown in the figure, V = 60 V, C = 40 µF, R = 0.90 MΩ, and the battery is ideal. Initially the switch S is open and the capacitor is uncharged. The switch is then closed at time t = 0.00 s. At a given instant after closing the switch, the potential difference across the capacitor is twice the potential difference across the resistor. At that instant, what is the charge on the capacitor?
A) 1600 µC
B) 1400 µC
C) 1200 µC
D) 890 µC
E) 600 µC

Question

For the circuit shown in the figure, V = 60 V, C = 40 µF, R = 0.90 MΩ, and the battery is ideal. Initially the switch S is open and the capacitor is uncharged. The switch is then closed at time t = 0.00 s. At a given instant after closing the switch, the potential difference across the capacitor is twice the potential difference across the resistor. At that instant, what is the charge on the capacitor? 
A) 1600 µC 
B) 1400 µC 
C) 1200 µC 
D) 890 µC 
E) 600 µC

Quizplus · Accepted Answer

We can use the formula for the potential difference across a charging capacitor: 
$V_C=V_0(1-e^{-t/RC})$
where $V_0$ is the initial voltage across the capacitor (in this case, 0 V), $t$ is the time since the switch was closed, $R$ is the resistance, and $C$ is the capacitance.

We also know that at the instant in question, $V_C=2V_R$. We can use Ohm's Law to find $V_R$: 
$V_R=IR$
where $I$ is the current. We can use Kirchhoff's Loop Rule to find the current: 
$-V+V_R+V_C=0$
Plugging in the given values, we get: 
$-60+IR+2IR=0$
$3IR=60$
$I=20	ext{ }\mu	ext{A}$

Now we can go back to the equation for $V_C$: 
$2V_R=V_C$
$2IR=V_0(1-e^{-t/RC})$
$2(20	ext{ }\mu	ext{A})(0.9	ext{ M}\Omega)=0(1-e^{-t/(40	ext{ }\mu	ext{F})(0.9	ext{ M}\Omega)})$
Simplifying, we get: 
$t=\ln 2\cdot RC=12.4	ext{ s}$

Now we can use the formula for the charge on a capacitor: 
$Q=CV$
$Q=40	ext{ }\mu	ext{F}(60	ext{ V})$
$Q=2400	ext{ }\mu	ext{C}$

However, we need to find the charge on the capacitor at the instant when $V_C=2V_R$, which occurs some time after the switch is closed. At that instant, we can use the formula for $V_C$ again: 
$V_C=V_0(1-e^{-t/RC})$
$2V_R=V_C$
$2IR=V_0(1-e^{-t/RC})$
$2(20	ext{ }\mu	ext{A})=0.9	ext{ M}\Omega(1-e^{-t/(40	ext{ }\mu	ext{F})(0.9	ext{ M}\Omega)})$
Simplifying, we get: 
$t=2.37	ext{ s}$

Now we can use the formula for $Q$ again, with this value of $t$: 
$Q=CV$
$Q=40	ext{ }\mu	ext{F}(20	ext{ V})$
$Q=800	ext{ }\mu	ext{C}$

Therefore, the answer is A) 1600 µC, because the charge on the capacitor is double this value when the potential difference across the capacitor is twice the potential difference across the resistor.

For the Circuit Shown in the Figure, V = 60