variable capacitor is connected to a 200 V battery. If its capacitance is changed from 2 µF to X µF, the decrease in energy of the capacitor is 2 X 10⁻² J. The value of X is
The energy stored in a capacitor is given by U = (1/2) C V². The decrease in energy is given as 2 × 10⁻² J. Substituting initial capacitance 2 µF, voltage 200 V, and solving for X, we get X = 1 µF. Correct answer: (a) 1 µF.
Class 12th Science Physics NCERT MCQ Questions
NCERT Books MCQ Questions Session 2024-2025.
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The correct answer is (a) 1 µF.
The energy stored in a capacitor is given by
U = 1/2CV². The decrease in energy is:
ΔU = 1/2(2 × 10⁻⁶ – X) (200)² = 2 × 10⁻²J
Solving for X, we get 1 µF.
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