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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

  1. A \(2500 \mathrm{~J}\)
  2. B \(2.5 \times 10^{-3} \mathrm{~J}\)
  3. C \(2.5 \times 10^{6} \mathrm{~J}\)
  4. D \(2.42 \times 10^{-2} \mathrm{~J}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2500 \mathrm{~J}\)

Step-by-step Solution

Detailed explanation

Change in energy \(\Delta U=\frac{1}{2}\left[\frac{q_{1}^{2}-q_{2}^{2}}{C}\right]\)
\(
\begin{array}{l}
=\frac{1}{2}\left[\frac{(0.5)^{2}-(0.1)^{2}}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.25-0.01}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{0.24}{48 \times 10^{-6}}\right] \\
=\frac{1}{2}\left[\frac{10^{4}}{2}\right] \\
=2500 \mathrm{~J}
\end{array}
\)