Capacitors

JEE Main level, single correct, medium

180 of 180

Question 180 of 180

The figure shows an experimental plot for discharging of a capacitor in an R−CR-C circuit. The time constant τ\tau of this circuit lies between:

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Answer: (D) 100 sec100 \mathrm{~sec} and 150 sec150 \mathrm{~sec}

For discharging of an RC circuit,
V=V0e−t/τV=V_0 e^{-t / \tau}
So, when
V=V02V=\frac{V_0}{2}
V02=V0e−t/τ\frac{\mathrm{V}_0}{2}=\mathrm{V}_0 \mathrm{e}^{-\mathrm{t} / \tau}
ln⁡12=−tτ⇒τ=tln⁡2\ln \frac{1}{2}=-\frac{\mathrm{t}}{\tau} \Rightarrow \tau=\frac{\mathrm{t}}{\ln 2}
From graph when V=V02,t=100 s∴τ=100ln⁡2=144.3 sec\mathrm{V}=\frac{\mathrm{V}_0}{2}, \mathrm{t}=100 \mathrm{~s} \quad \therefore \tau=\frac{100}{\ln 2}=144.3 \mathrm{~sec}

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