A \(20 \mathrm{~F}\) capacitor is charged to \(5 \mathrm{~V}\) and isolated. It is then connected in…

A \(20 \mathrm{~F}\) capacitor is charged to \(5 \mathrm{~V}\) and isolated. It is then connected in parallel with an uncharged \(30 \mathrm{~F}\) capacitor. The decrease in the energy of the system will be
  1. \(25 \mathrm{~J}\)
  2. \(100 \mathrm{~J}\)
  3. \(125 \mathrm{~J}\)
  4. \(150 \mathrm{~J}\)

Solution

$\begin{aligned} & C_s=20 F, V=5 \text{ volt, } \\ & F_x=30 F \end{aligned}$ Decrease in energy, $\begin{aligned} & \Delta U=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_2-V_2\right)^2 \\ & =\frac{1}{2} \times \frac{20 \times 30}{20+30}(5-0)^2 \\ & =\frac{300}{50} \times 25=150 J \end{aligned}$

Asked in: JEE Mains - Capacitance - Test 3

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