Q130 / 180 · KCET 2024 · Physics

A capacitor of capacitance $5 \mu \mathrm{~F}$ is charged by a battery of emf 10 V . At an instant of time, the potential difference across the capacitors is 4 V and the time rate of change of potential difference across the capacitor is $0.6 \mathrm{Vs}^{-1}$. Then, the time rate at which energy is stored in the capacitor at any instant is

  • Correct answer
    A
    $12 \mu \mathrm{~W}$
  • B
    $3 \mu \mathrm{~W}$
  • C
    Zero
  • D
    $30 \mu \mathrm{~W}$

Solution

Given, battery emf, $e=10 \mathrm{~V}$ $C=5 \mu \mathrm{~F}=5 \times 10^{-6} \mathrm{~F}$ $V=4 \mathrm{~V} \Rightarrow d V / d t=0.6 \mathrm{Vs}^{-1}$ We know that stored energy, $U=\frac{1}{2} C V^2$ $\therefore \quad \frac{d U}{d t}=\frac{1}{2} C \cdot 2 V \frac{d V}{d t}=C V \frac{d V}{d t}$ $=5 \times 10^{-6} \times 4 \times 0.6$ $=12 \times 10^{-6} \mathrm{~W}=12 \mu \mathrm{~W}$

Hence, the correct answer is option: 'A'.

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