The work done in charging a capacitor to charge Q is stored as:
A. Heat energy
B. Electrostatic potential energy
C. Chemical energy
D. Magnetic energy
Answer: Option B
Solution (By JKSSB Mock Tests)
Work done to charge a capacitor is stored as electrostatic potential energy in the electric field between plates: U = ½QV = ½CV² = Q²/(2C). This energy can be recovered when capacitor discharges. Heat (A) is dissipated in resistance during charging, but ideal capacitor stores energy electrostatically. Memory aid: 'Capacitor energy = ½CV², stored in electric field'. This electrostatics concept is frequently tested in competitive exams. Always distinguish ideal capacitor (no resistance) from real circuits where some energy is lost as heat during charging.
Explanation:
Resistance R = V²/P = 220²/100 = 484 Ω. At 110 V, P = V²/R = 110²/484 = 12100/484 = 25 W. Assuming resistance constant. Power reduces to quarter when voltage halved (P ∝ V² for fixed R).
Explanation:
Heat travels through spoon from hot end to cold end by conduction (molecular vibration and free electrons). Handle gets hot. Metal good conductor.
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