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Initially the potential difference of a 8 μF capacitor is 30 V. Then it is changed to 40 V. What is the increase in energy?(a) 28 × 10^-4 J(b) 18 × 10^-4 J(c) 8 × 10^-4 J(d) 24 × 10^-4 JThe question was posed to me during an internship interview.This intriguing question comes from Energy Stored in a Capacitor in division Electrostatic Potential and Capacitance of Physics – Class 12 |
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Answer» The CORRECT OPTION is (a) 28 × 10^-4 J |
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