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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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