Question

A part of circuit in steady state along with the currents flowing in the branches, the value of resistances etc., is shown in figure. Calculate the energy stored in the capacitor.

• A. $8\times {10}^{-1}J$
• B. $8\times {10}^{-2}J$
• C. $8\times {10}^{-3}J$
• D. $8\times {10}^{-4}J$

Answer 1

The correct option is D $8\times {10}^{-4}J$

In steady state current from a to b through capacitor is zero.
by applying KCL at node a

$\Rightarrow 2A+1A+0-I_1=0$

$I_1=3A$

KCL at node B

$-2A+1A+I_2=0$

$I_2=1A$
$\Rightarrow V_a-Vb=5I_1+1I_1+2I_2$

$=15+3+2$

$=20$

Energy $=\frac{1}{2}c{v}^2=\frac{1}{2}\times 4\times {10}^{-6}\times 400=8\times {10}^{-4}J$