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

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Answer & Solution

Correct Answer

(D) \(1 / 6\)

Detailed explanation

In \(t_{1}\) time energy becomes half so charge will become \(\frac{1}{\sqrt{2}}\) time

\(q = Q _{0} e ^{\frac{ T _{1}}{ RC }}=\frac{ Q _{0}}{\sqrt{2}}\)

and \(q = Q _{0} e ^{\frac{ t _{1}}{ RC }}=\frac{ Q _{0}}{8}=\left(\frac{ Q _{0}}{\sqrt{2}}\right)^{6}\)

\(t _{2}=6 t _{1}\)

\(\frac{ t _{1}}{ t _{2}}=\frac{1}{6}\)

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