Question

In the situation shown in figure, the switch is at position $1$ for long time and it has been shifted from position $1$, to position $2$ at $t=0$. The current in the $3\Omega$ resistor at $t=3\ln 2$ seconds is found to be $\frac{11x}{16}A$. Find the value of $x$.

Answer 1

At $t=0,$ just before flipping switch to position $2$
Charge on capacitor is $q=CV=9\text{C}$


After flipping switch to position 2, circuit is shown above. As we can use cells in parallel formula, we get
$\Rightarrow E_{eff}=\frac{\frac{0}{4}+\frac{12}{12}}{\frac{1}{4}+\frac{1}{12}}=3\text{V}$
$\Rightarrow R_{\text{eq}}=\frac{12\times 4}{12+4}=3\Omega$

So, circuit can be simplified as,

Charge on capacitor at time $t$ is
$Q=Q_0{e}^{-t/RC}+CV(1-e^{-t/RC})$
By differentiating charge, we get
$i=\frac{-Q_0}{RC}{e}^{-t/RC}+\frac{V}{R}{e}^{-t/RC}$
$i=\frac{e^{-t/RC}}{R}(V-\frac{Q_0}{C})=\frac{e^{-\frac{3\ln 2\times 10}{6\times 250}}}{6}[3-36]$
$i={\left(\frac{1}{2}\right)}^2\times \frac{1}{6}\times -\frac{33}{2}$
$|i|=\frac{22}{8}=\frac{11x}{8}\Rightarrow x=2$