Question

Capacitor in the circuit is in steady state along with the current flowing in the branches. The value of each resisatnce is shown in figure. Calculate the energy stored in the capacitor of capacitance $4\mu \text{F}$.

• A. $6\times {10}^{-4}\text{J}$
• B. $8\times {10}^{-4}\text{J}$
• C. $9\times {10}^{-4}\text{J}$
• D. $12\times {10}^{-4}\text{J}$

Answer 1

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


From the given circuit, using $\text{KCL}$ at the junctions $\text{M}$ and $\text{P}$ we will get, $I_1=3\text{A}$ and $I_2=1\text{A}$ respectively.

Moving along the loop from $\text{MNOP}$ we get,

$V_M-5I_1-I_1-2I_2=V_p$

$\Rightarrow V_M-V_P=6I_1+2I_2=20\text{V}$

Energy stored in the capacitor is,

$\frac{1}{2}C(V_M-V_P{)}^2=\frac{1}{2}\times 4\times {10}^{-6}\times 20\times 20=8\times {10}^{-4}\text{J}$

Hence, option $\left(b\right)$ is correct.