A capacitor of capacitance C is charged to voltage V. The energy stored is:
A. CV
B. ½C/V
C. ½CV²
D. CV²
Answer: Option C
Solution (By JKSSB Mock Tests)
Energy stored in capacitor U = ½CV² = ½QV = Q²/(2C). This derives from work done to charge the capacitor against increasing voltage. Option A has wrong dimensions (energy should be joules, CV is coulomb-volt = joule, but missing factor ½); C and D are dimensionally incorrect. Memory tip: 'Capacitor energy = ½CV², analogous to spring energy ½kx²'. This formula-based question tests electrostatics knowledge, frequently appearing in competitive exams. Always recall the three equivalent forms and choose based on given quantities.
Explanation:
Work done to stretch spring from 0 to x: W₁ = ½kx². Work done from 0 to 2x: W₂ = ½k(2x)² = 2kx². Thus work for additional stretch from x to 2x: ΔW = W₂ - W₁ = 2kx² - ½kx² = ³/₂kx². Spring force is variable (F=kx), so work is integral of F·dx, yielding parabolic energy storage. Memory aid: Elastic potential energy U = ½kx²; always calculate difference for incremental work. This problem tests understanding of work done by variable forces, a common theme in energy conservation questions in competitive exams.
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