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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 $10 \%$ ?
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Verified Answer
The correct answer is:
$\tau \ln \frac{10}{9}$
As, we know that $q=q_0 e^{-t / \tau}$
According to the question, $V_C=\frac{q}{C}=\frac{q_0}{C} e^{-t / \tau}$
$\begin{aligned} \Rightarrow & & \frac{q_0}{10} & =\frac{q_0}{C} e^{-t / \tau} \\ \Rightarrow & & e^{t / \tau} & =\frac{10}{9} \\ \Rightarrow & & t & =\tau \ln \frac{10}{9}\end{aligned}$
According to the question, $V_C=\frac{q}{C}=\frac{q_0}{C} e^{-t / \tau}$
$\begin{aligned} \Rightarrow & & \frac{q_0}{10} & =\frac{q_0}{C} e^{-t / \tau} \\ \Rightarrow & & e^{t / \tau} & =\frac{10}{9} \\ \Rightarrow & & t & =\tau \ln \frac{10}{9}\end{aligned}$
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