Topic: Energy stored in a capacitor· Electrostatics

A capacitor of 2\ μ F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

A capacitor of \(2\ \mu F\) is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

circuit with battery V, switch S (positions 1,2), 2 uF and 8 uF capacitors

1. 0%

2. 20%

3. 75%

4. 80%

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Answer: 4. 80%

Key idea: charge is conserved; it is shared between the capacitors until the potentials are equal, and the difference in energy is lost.

Initially \(U_i=\tfrac{Q^2}{2C_1}\) with \(C_1=2\ \mu F\).
After connecting the \(8\ \mu F\) capacitor: \(U_f=\tfrac{Q^2}{2(C_1+C_2)}=\tfrac{Q^2}{2\times 10}\).
\(U_f/U_i=\tfrac{2}{10}=20\%\), so \(80\%\) of the energy is dissipated.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A capacitor of 2\ μ F is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

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