MHT CET · Physics · Capacitance
Capacity of a capacitor is \(48 \mu \mathrm{F}\). When it is charged from \(0.1 \mathrm{C}\) to \(0.5 \mathrm{C}\), change in the energy stored is
- A \(2500 \mathrm{~J}\)
- B \(2.5 \times 10^{-3} \mathrm{~J}\)
- C \(2.5 \times 10^{6} \mathrm{~J}\)
- D \(2.42 \times 10^{-2} \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(2500 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Change in energy \(\Delta U=\frac{1}{2}\left[\frac{q_{1}^{2}-q_{2}^{2}}{C}\right]\)
\(
\begin{array}{l}
=\frac{1}{2}\left[\frac{(0.5)^{2}-(0.1)^{2}}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.25-0.01}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.24}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{10^{4}}{2}\right] \\
=2500 \mathrm{~J}
\end{array}
\)
\(
\begin{array}{l}
=\frac{1}{2}\left[\frac{(0.5)^{2}-(0.1)^{2}}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.25-0.01}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.24}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{10^{4}}{2}\right] \\
=2500 \mathrm{~J}
\end{array}
\)
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