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Question

The energy stored in the capacitor in steady state is :

12567_31e19044c69c45f99257cd6e148523fa.png

A
12μ J
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B
24μ J
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C
36μJ
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D
48μ J
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Solution

The correct option is B 24μ J
In steady state, the current flowing through capacitor branch is zero.
I=(83)4+1=1A
Potential of point P=84=4V
Voltage across capacitor =4V
Energy stored in capacitor =12CV2
=12×3×106×16
=24μJ
66004_12567_ans_42c65fcfde23414ea5c1138658f18cd7.png

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