First slide

KVL

Question

The energy stored in the capacitor in the steady state is

Moderate

A

$338 μJ$
B

$196 μ J$
C

$98 μ J$
D

$8 μ J$

Solution

At steady state, capacitor acts as open circuit.
$\begin{array}{l} V_{A} + 3 \times 1 - 14 - 5 \times 3 = V_{B} \\ \Rightarrow V_{A} - V_{B} = 26 V \end{array}$
Energy stored in capacitor,
$U = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 10^{- 6} \times {(26)}^{2} = 338 μJ$

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