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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Physics
Chapter
Capacitance

JEE Main 2012 · Offline · Physics · Capacitance

The figure shows an experimental plot for discharging of a...

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: The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant of this

Options

  1. A150 sec150 \mathrm{~sec} and 200 sec200 \mathrm{~sec}
  2. B00 and 50 sec50 \mathrm{~sec}
  3. C50 sec50 \mathrm{~sec} and 100 sec100 \mathrm{~sec}
  4. D100 sec100 \mathrm{~sec} and 150 sec150 \mathrm{~sec}

Correct answer

D. 100 sec100 \mathrm{~sec} and 150 sec150 \mathrm{~sec}

Step-by-step solution

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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