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A fully charged parallel plate capacitor of $1 \mu \mathrm{F}$ capacity is discharging through a resistor. If the energy of the capacitor reduces to half in one second then the value of resistance will be
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The correct answer is:
$\frac{2 \times 10^6}{\ln 2} \Omega$
As $Q=Q_0 e^{-t / \tau}$. When energy is $50 \%$ then $Q=\frac{Q_0}{\sqrt{2}}$
$\begin{aligned} & \frac{Q_0}{\sqrt{2}}=Q_0 e^{-t / \tau} \\ & e^{t / \tau}=\sqrt{2} \\ & \frac{t}{\tau}=\ln (\sqrt{2}) \quad \tau=\frac{t}{\ln (\sqrt{2})} \\ & R C=\frac{2}{\ln 2} \\ & R=\frac{2}{C \ln 2}=\frac{2}{10^{-6} \times \ln 2}=\frac{2 \times 10^6}{\ln 2} \Omega\end{aligned}$
$\begin{aligned} & \frac{Q_0}{\sqrt{2}}=Q_0 e^{-t / \tau} \\ & e^{t / \tau}=\sqrt{2} \\ & \frac{t}{\tau}=\ln (\sqrt{2}) \quad \tau=\frac{t}{\ln (\sqrt{2})} \\ & R C=\frac{2}{\ln 2} \\ & R=\frac{2}{C \ln 2}=\frac{2}{10^{-6} \times \ln 2}=\frac{2 \times 10^6}{\ln 2} \Omega\end{aligned}$
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