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Question

Figure shows part of a circuit. Energy stored in 2μF capacitor at steady state is:

A
100 μJ
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B
50 μJ
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C
200 μJ
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D
80 μJ
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Solution

The correct option is A 100 μJ
By junction law, we can see the current through 5 Ω is 2A.
So, potential difference across 5 Ω or 2μF is V=5×2=10V
Hence, energy stored in the capacitor is12CV2=12×2×102=100 μJ

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