Capacitors

JEE Main level, single correct, medium

162 of 180

Question 162 of 180

A capacitor is discharging through a resistor RR. Consider in time t1{t}_{1}, the energy stored in the capacitor reduces to half of its initial value and in time t2{t}_{2}, the charge stored reduces to one eighth of its initial value. The ratio t1t2\dfrac{{t}_{1}}{{t}_{2}} will be

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Answer: (D) 16\dfrac{1}{6}

We know that in the discharging circuit the charge is given by,
q=q0e−tRCq={q}_{0}{e}^{\dfrac{-t}{RC}} ⇒tRC=ln⁡(q0q)\Rightarrow \dfrac{t}{RC}=\ln \left(\dfrac{{q}_{0}}{q}\right)
In time t1{t}_{1} the energy stored reduces to half. Hence, q122C=12×q022C⇒q0q1=2\dfrac{{{q}_{1}}^{2}}{2C}=\dfrac{1}{2}\times \dfrac{{{q}_{0}}^{2}}{2C}\Rightarrow \dfrac{{q}_{0}}{{q}_{1}}=\sqrt{2}. Therefore,
t1RC=ln⁡(2)=12ln⁡2\dfrac{{t}_{1}}{RC}=\ln \left(\sqrt{2}\right)=\dfrac{1}{2}\ln 2.
In time t2{t}_{2} the charge stored reduces to 18th{\dfrac{1}{8}}^{th} of initial value. Hence, q2q0=18\dfrac{{q}_{2}}{{q}_{0}}=\dfrac{1}{8}. Therefore,
t2RC=ln⁡(8)=3ln⁡2\dfrac{{t}_{2}}{RC}=\ln \left(8\right)=3\ln 2.
Hence,
t1t2=12ln⁡23ln⁡2=16\dfrac{{t}_{1}}{{t}_{2}}=\dfrac{\dfrac{1}{2}\ln 2}{3\ln 2}=\dfrac{1}{6}

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