Q70 / 90 · KVPY 2012 (SB/SX) · Physics
The capacitor of capacitance $\mathrm{C}$ in the circuit shown in fully charged initially. Resistance is R.-
After the switch $\mathrm{S}$ is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is
After the switch $\mathrm{S}$ is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is
- A$\mathrm{RC} / 2$
- B$2 \mathrm{RC}$ in 2
- C$\mathrm{RC} \ln 2$
- Correct answerD$\frac{\mathrm{RCln} 2}{2}$
Solution
Discharging -
$\mathrm{Q}=\mathrm{Q}_{0} \mathrm{e}^{-t / \mathrm{RC}}, \mathrm{U}^{\prime}=\frac{\mathrm{U}}{2} \Rightarrow \frac{\mathrm{Q}_{0}{ }^{2}}{2 \mathrm{C}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{RC}}=\frac{\mathrm{Q}_{0}{ }^{2}}{2 \mathrm{C}}$
Hence, the correct answer is option: 'D'.