Question

A capacitor of capacity $6\mu \text{F}$ and initial charge $160\mu \text{C}$ in connected with key S and resistance as shown in figure. Point $M$ is earthed. If key is closed at $t=0$, then the current through resistance $(R=1\Omega$) at $(t=16\mu \text{s})$ is $\frac{4n}{3e}\text{A}$, then find $n$

Answer 1

Initial potential difference across the capacitor = $\frac{160}{60}\text{Volt}$ = $\frac{80}{3}$
When key is closed capacitor starts discharging through a parallel combination of $(3+R)$ = $4\Omega$ and $8\Omega$ both.
Equivalent circuit will be :-
At time $t,$ charge on capacitor will be
$q=Q_{initial}.e^{\frac{-t}{\tau }}$

current through the circuit is
$I=\frac{dq}{dt}=\frac{Q_{initial}}{\tau }.e^{\frac{-t}{\tau }}$

As in a parallel circuit, current is divided inversely proportional to resistance
current through R is $i=\frac{8I}{4+8}$
$i=\frac{2}{3}\frac{Q_{initial}}{\tau }.e^{\frac{-t}{\tau }}$
$i=\frac{1}{3}\frac{160\mu C}{\frac{8}{3}\left(6\mu C\right)}.e^{\frac{-16}{16}}$
$i=\frac{20}{3e}$
Given that $i=\frac{4n}{3e}$
$\Rightarrow n=5$