Question

A capacitor is charged and then made to discharge through a resistance. The time constant is $\tau$. In what time will the potential difference across the capacitor decreases by $10\%$ ?

• A. $\tau \ln \left(0.1\right)$
• B. $\tau \ln \left(0.9\right)$
• C. $\tau \ln \left(\frac{10}{9}\right)$
• D. $\tau \ln \left(\frac{11}{10}\right)$

Answer 1

The correct option is C $\tau \ln \left(\frac{10}{9}\right)$

While discharging, charge on the capacitor after time $t$ is,$$Q=Q_0{e}^{\frac{-t}{\tau }}....\left(1\right)$$We know that potential difference across $C$ is proportional to charge $\left(Q\right)$ on the capacitor.

For the potential difference to fall by $10\%$, $Q$ also must fall by $10\%$

$\therefore Q=0.9Q_0$

Using eq (1) we get,

$0.9Q_0=Q_0{e}^{\frac{-t}{\tau }}$

$\Rightarrow \frac{10}{9}=e^{\frac{t}{\tau }}...\left(2\right)$

Taking $\ln$ on both sides,

$\Rightarrow \ln (\frac{10}{9})=\frac{t}{\tau }$

$\therefore t=\tau \ln (\frac{10}{9})$

Hence, option (c) is the correct answer.