Figure shows as uncharged capacitor of capacitance $C$, two resistors of resistances $R_{1}$ and $R_{2}$ a battery of voltage $V$ and a two-way switch $S$. Terminal 1 of $S$ is connected to terminal 2 . The time required for the charge on the capacitor to reach half of the final charge is

Question

Aicharya · Accepted Answer

The charge on the capacitor at time $t$ is given by $Q=Q_{0}\left(1-e^{-t / \tau_{1}}\right)$ where time constant $\tau_{1}=R_{1} C$ and $Q_{0}$ is the final charge, i.e., the charge when the capacitor is fully charged to the voltage of the battery. For $Q$ to become equal to $\frac{Q_{0}}{2}$, the time $t$ required is given by $\begin{aligned} \frac{Q_{0}}{2} &=Q_{0}\left(1-e^{-t / \tau_{1}}\right) \ \Rightarrow 1-e^{-t / \tau_{1}} &=\frac{1}{2} \end{aligned}$ $\Rightarrow$$e^{-t / \tau_{1}}=\frac{1}{2}$ $\Rightarrow \quad e^{-t / \tau_{1}}=2$ $\Rightarrow \quad \frac{t}{\tau_{1}}=\ln (2)$ $\Rightarrow \quad t=\tau_{1} \ln (2)=R_{1} C \ln (2)$ So the correct choice is $(\mathrm{c})$. ,

Figure shows as uncharged capacitor of capacitance \(C\), two resistors of resistances \(R_{1}\) and…

Solution