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The capacitance of a variable capacitor joined with a battery of 100 V is changed from 2 μF to 10 μF. What is the change in the energy stored in it ?(A) 2 X 10⁻² J (B) 2.5 X 10⁻² J (C) 6.5 X 10⁻² J (D) 4 X 10⁻² J |
| Answer» DENTS,◆ Answer - (D)∆E = 4×10^-2 J◆ Explaination -# Given -V = 100 VC1 = 2 μF = 2×10^-6 FC2 = 10 μF = 10×10^-6 F# SOLUTION -CHANGE in energy stored in CAPACITOR is given by -∆E = 1/2 (C2-C1)V^2∆E = 1/2 × (10×10^-6 - 2×10^-6) × 100^2∆E = 4×10^-2 JTherefore, the change in the energy stored in capacitor is 4×10^-2 J.Thanks dear... | |