Question

An uncharged capacitor of capacitance $C$ is connected in the circuit diagram as shown and switch $S$ is closed at $t=0$. If the current in branch $B C$ as a function of time is given by:
$I=I_{0}$ (ampere) $e^{-\frac{t}{\tau( in \mu s)}}$, then find the numerical value of $I_{0} \tau .$

Answer 1

For time constant,
$R_{ eq }=\frac{4}{3}$
$\tau=\frac{4}{3} \times 4=\frac{16}{3} \mu s$
Charge on capacitor in steady state
$=4 \Delta V_{B C}=4 \times 4=16 \,\mu C$
Charge as function of time $=16\left(1-e^{-\frac{t}{16 / 3}}\right)$

$I_{B C}=\frac{d Q}{d t}=16\left(0+e^{-\frac{1}{16 / 3}} \times \frac{3}{16}\right)$
$=3\left(e^{-\frac{1}{16 / 3\, \mu s }}\right)$