Question

Find the energy stored in the circuit containing four capacitors at the steady state.

• A. $99\mu \text{J}$
• B. $49.5\mu \text{J}$
• C. $62\mu \text{J}$
• D. $75\mu \text{J}$

Answer 1

The correct option is B $49.5\mu \text{J}$

At steady state there will be no current through branch containing capacitors.

Current will flow as shown below.

Current in the loop $\text{JCDI}$ is,

$i_1=\frac{V}{R_{eq}}=\frac{6}{3}=2\text{A}$

Current in the loop $\text{IDEH}$ is,

$i_2=\frac{6}{3+3}=1\text{A}$

Potential difference across $1\Omega ,2\Omega ,3\Omega$ resistances are,

$V_1=i_1\times 1=2\times 1=2\text{V}$

$V_2=i_1\times 2=2\times 2=4\text{V}$

$V_3=i_2\times 3=1\times 3=3\text{V}$

Since, all the capacitors are connected across resistors in parallel combination, therefore the potential difference across the capacitors will be same as potential difference across the resistors.

Thus, total energy stored in circuit,

$U=U_1+U_2+U_3+U_4$

$U=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2+\frac{1}{2}{C}_3{V}_3^2+\frac{1}{2}{C}_4{V}_4^2$

$U=[\frac{1}{2}\times 1\times 4]+[\frac{1}{2}\times 2\times 16]+[\frac{1}{2}\times 3\times 9]+[\frac{1}{2}\times 4\times 9]$

$U=2+16+13.5+18$

$\therefore U=49.5\mu \text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(b\right)$ is correct.