Question

In the circuit shown in figure, Find energy stored in $4\mu \text{F}$ capacitor at the steady state.

• A. $0.5\text{mJ}$
• B. $0.8\text{mJ}$
• C. $1.2\text{mJ}$
• D. $1.8\text{mJ}$

Answer 1

The correct option is D $1.8\text{mJ}$

Let currents $I_1$ and $I_2$ in the two loops are as shown in figure. We know that at the steady state no current flows through the branch containing capacitor. So we do not consider any current in the loop containing the capacitor as shown.



Using KVL equation for the loop with $I_1$ current we get,

$-40-5I_1-5I_1-10(I_1-I_2)=0$

$-40-20I_1+10I_2=0$

$\Rightarrow -4-2I_1+I_2=0$

$\Rightarrow 2I_1-I_2=-4........\left(1\right)$

Using KVL equation for the loop with $I_1$ current we get,

$-10I_2-10(I_2-I_1)+20=0$

$\Rightarrow I_1-2I_2=-2........\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get,

$3I_1=-6$

$\Rightarrow I_1=-2\text{A}$ and $I_2=0\text{A}$

Energy stored in capacitor at the steady state is given by,

$U_{4\mu \text{F}}=\frac{1}{2}C{V}_{AB}^2$

Using KVL to the open loop from $\text{A}$ and $\text{B}$ we get,

$\Rightarrow V_A-5(-2)+20=V_B$

$\Rightarrow V_B-V_A=30\text{V}$

$\Rightarrow U_{4\mu \text{F}}=\frac{1}{2}\times 4\times {\left(30\right)}^2\mu \text{J}$

$\Rightarrow U_{4\mu \text{F}}=1800\mu \text{J}=1.8\text{mJ}$

Hence, option (d) is the correct answer.