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Question

In the circuit shown in figure, Find energy stored in 4 μF capacitor at the steady state.


A
0.5 mJ
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B
0.8 mJ
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C
1.2 mJ
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D
1.8 mJ
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Solution

The correct option is D 1.8 mJ
Let currents I1 and I2 in the two loops are as shown in figure. We know that at the steady state no current flows through the branch containing capacitor. So we do not consider any current in the loop containing the capacitor as shown.



Using KVL equation for the loop with I1 current we get,

405I15I110(I1I2)=0

4020I1+10I2=0

42I1+I2=0

2I1I2=4 ........(1)

Using KVL equation for the loop with I1 current we get,

10I210(I2I1)+20=0

I12I2=2 ........(2)

Solving (1) and (2) we get,

3I1=6

I1=2 A and I2=0 A

Energy stored in capacitor at the steady state is given by,

U4 μ F=12CV2AB

Using KVL to the open loop from A and B we get,

VA5(2)+20=VB

VBVA=30 V

U4 μ F=12×4×(30)2 μJ

U4 μ F=1800 μ J=1.8 mJ

Hence, option (d) is the correct answer.

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