Q135 / 180 · JIPMER 2016 · Physics

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
    $\tau \ln 0.9$
  • Correct answer
    C
    $\tau \ln \frac{10}{9}$
  • D
    $\tau \ln \frac{11}{10}$

Solution

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

Hence, the correct answer is option: 'C'.

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