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 decrease by $10$%?

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

Answer 1

Answer: (C)

$\tau \ln \frac{10}{9}$

As, we know that $q=q_0{e}^{-t/\tau }$
According to the question, $V_C=\frac{q}{C}=\frac{q_0}{C}{e}^{-t/\tau }$
$\frac{q_0}{10}=\frac{q_0}{C}{e}^{-t/\tau }$
$\Rightarrow e^{t/\tau }=\frac{10}{9}$
$\Rightarrow t=\tau \ln \frac{10}{9}$