Capacitors

JEE Main level, numerical answer, medium

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Question 1 of 180

Two identical capacitors each of capacitance C=10 μFC = 10\,\mu\text{F} are connected in parallel and charged to 100 V100\,\text{V}. The total energy stored in the combination is (in J):

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Answer: 0.1

Ceq=2C=20 μFC_{eq} = 2C = 20\,\mu\text{F}. Energy U=12CeqV2=12(20×10−6)(100)2=0.1 JU = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (20 \times 10^{-6}) (100)^2 = 0.1\,\text{J}.

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