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 ................
- A \(1 / 2\)
- B \(1 / 3\)
- C \(1 / 4\)
- D \(1 / 6\)
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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