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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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 ................

  1. A \(1 / 2\)
  2. B \(1 / 3\)
  3. C \(1 / 4\)
  4. D \(1 / 6\)
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Answer & Solution

Correct Answer

(D) \(1 / 6\)

Step-by-step Solution

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