Question

A part of a circuit in steady state along with the current flowing in the branches, with value of each resistance is shown in figure. What will be the energy stored in the capacitor $C_0$

• A. $6\times {10}^{-4}J$
• B. $8\times {10}^{-4}J$
• C. $16\times {10}^{-4}J$
• D. Zero

Answer 1

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

Applying Kirchhoff’s first law at junctions A and B respectively we have $2+1-i_1=0$ i.e., $i_1=3A$ and $i_2+1-2=0$ i.e., $i_2=1A$
Now applying Kirchhoff’s second law to the mesh ADCBA treating capacitor as a seat of emf V in open circuit
$-3\times 5-3\times 1-1\times 2+V=0$ i.e $V(=V_A-V_B)=20V$
So, energy stored in the capacitor $U=\frac{1}{2}(4\times {10}^{-6})\times (20{)}^2=8\times {10}^{-4}J$