Question

In the circuit shown in figure the capacitor is charged with a cell of 5 V. If the switch is closed at $t=0$, then at $t=12s$, charge on the capacitor is

• A. $\left(0.37\right)10\mu C$
• B. $(0.37{)}^{2}10\mu C$
• C. $\left(0.63\right)10\mu C$
• D. $(0.63{)}^{2}10\mu C$

Answer 1

The correct option is B $(0.37{)}^{2}10\mu C$

When S is closed, the capacitor will start to discharge.

We know that the charge on the capacitor at time t is during discharging given by

$q\left(t\right)=q_0{e}^{-t/CR}$

here, $q_0=CV=2\times {10}^{-6}\times 5=10\mu C$,

time constant $\tau =CR=2\times {10}^{-6}\times 3\times {10}^6=6s$

so, at $t=12s,q\left(12\right)=10e^{-12/6}=10\left(1/e^2\right)=\left(1/e{)}^2\right(10)=(0.37{)}^2\left(10\right)\mu C$