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Q.
The energy stored in the capacitor in the steady state is
Current Electricity
Solution:
At steady state, capacitor acts as open circuit.
$V_{A}+3 \times 1-14-5 \times 3=V_{B}$
$\Rightarrow V_{A}-V_{B}=26 \, V$
Energy stored in capacitor,
$U=\frac{1}{2} C V^{2}=\frac{1}{2} \times 10^{-6} \times(26)^{2}=338 \, \mu J$