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 $BC$ 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

Answer: 16

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
$=4\Delta V_{BC}=4\times 4=16\mu C$
Charge as function of time $=16\left(1-e^{-\frac{t}{16/3}}\right)$

$I_{BC}=\frac{dQ}{dt}=16\left(0+e^{-\frac{1}{16/3}}\times \frac{3}{16}\right)$
$=3\left(e^{-\frac{1}{16/3\mu s}}\right)$