Q144 / 160 · TS EAMCET 2023 (12 May Shift 2) · Physics
If a capacitor of capacitance $100 \mu \mathrm{F}$ is charged at a steady rate of $100 \mu \mathrm{C} \mathrm{s}^{-1}$, then the time taken to produce a potential difference of $100 \mathrm{~V}$ between the capacitor plates is
- A$50 \mathrm{~s}$
- B$200 \mathrm{~s}$
- C$150 \mathrm{~s}$
- Correct answerD$100 \mathrm{~s}$
Solution
Total charge on capacitor,
$$
\begin{aligned}
& Q=C V \\
& =(100 \mu \mathrm{F})(100 \mathrm{~V}) \\
& =100 \times 10^{-6} \times 100=10^{-2} \mathrm{C}
\end{aligned}
$$
Total charge $=Q($ Rate of charge $) \times($ time taken $)$
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{Q}^{\prime} \times \mathrm{t} \\
& \mathrm{t}=\frac{\mathrm{Q}}{\mathrm{Q}^{\prime}}=\frac{10^{-2}}{100 \times 10^{-6}}=\frac{10^{-2}}{10^{-4}}=100 \\
& \mathrm{t}=100 \mathrm{sec}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.