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  • A capacitor is discharging through a resistor R. Consider in time , the energy stored in the ca

A capacitor is discharging through a resistor R. Consider in time \mathrm{\mathrm{t}_{1}}, the energy stored in the capacitor reduces to half of its initial value and in time \mathrm{t_{2},} the charge store reduces to one eighth of its initial value. The ratio \mathrm{t_{1} / t_{2}} will be

Option: 1

\mathrm{1 / 2}


Option: 2

\mathrm{1 / 3}


Option: 3

\mathrm{1 / 4}


Option: 4

\mathrm{1 / 6}


Answers (1)

best_answer

For discharging cpapaitor

\mathrm{q =q_{0} e^{\frac{-t}{R c}}} \\

\mathrm{u=\frac{q^{2}}{2 c} =\frac{q_{0}^{2}}{2 c} e^{-\frac{2 t}{R c}} }

If energy become half, the charge becomes \mathrm{\frac{1}{\sqrt{2}}} times

\mathrm{\frac{q _{0}}{\sqrt{2}}=q_{0} e^{-t / \tau }} \\

\mathrm{e^{-t /\tau }=\frac{1}{\sqrt{2}} }                   ..........(1)

\mathrm{e^{-t_{1} / \tau }=\left(\frac{1}{2}\right)^{\frac{1}{2}}} \\

similarly at t2 charge becomes \frac{1}{8} times

\mathrm{e^{-t_{2} / \tau }=\left(\frac{1}{2}\right)^{3}} \\

\mathrm{\frac{t_{1}}{t_{2}}=\frac{1}{6}}

Hence the correct option is 4

Posted by

Ritika Harsh

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