KCET · Physics · Capacitance
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
- A \(12 \mu \mathrm{~W}\)
- B \(3 \mu \mathrm{~W}\)
- C Zero
- D \(30 \mu \mathrm{~W}\)
Answer & Solution
Correct Answer
(A) \(12 \mu \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
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,
\(U=\frac{1}{2} C V^2\)
\(\therefore \quad \frac{d U}{d t}=\frac{1}{2} C \cdot 2 V \frac{d V}{d t}=C V \frac{d V}{d t}\)
\(=5 \times 10^{-6} \times 4 \times 0.6\)
\(=12 \times 10^{-6} \mathrm{~W}=12 \mu \mathrm{~W}\)
\(C=5 \mu \mathrm{~F}=5 \times 10^{-6} \mathrm{~F}\)
\(V=4 \mathrm{~V} \Rightarrow d V / d t=0.6 \mathrm{Vs}^{-1}\)
We know that stored energy,
\(U=\frac{1}{2} C V^2\)
\(\therefore \quad \frac{d U}{d t}=\frac{1}{2} C \cdot 2 V \frac{d V}{d t}=C V \frac{d V}{d t}\)
\(=5 \times 10^{-6} \times 4 \times 0.6\)
\(=12 \times 10^{-6} \mathrm{~W}=12 \mu \mathrm{~W}\)
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