The energy stored in the capacitor in steady state is :

12 μ

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24 μ

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36 μ

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48 μ

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Solution

The correct option is B $24 μ$ J
In steady state, the current flowing through capacitor branch is zero.
$I = \frac{(8 - 3)}{4 + 1} = 1 A$
Potential of point $P = 8 - 4 = 4 V$
Voltage across capacitor $= 4 V$
Energy stored in capacitor $= \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times 3 \times 10^{- 6} \times 16$
$= 24 μ J$

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The energy stored in the capacitor in steady state is :

The correct option is B 24μ JIn steady state, the current flowing through capacitor branch is zero.I=(8−3)4+1=1APotential of point P=8−4=4VVoltage across capacitor =4VEnergy stored in capacitor =12CV2 =12×3×10−6×16 =24μJ