Question

In the given electrical circuit, if the switch $S$ is closed then the maximum energy stored in the inductor is

• A. 3 J
• B. 9 J
• C. 12 J
• D. 6 J

Answer 1

Answer: (D)

In figure below, if switch $S$ is open then total energy stored in the capacitors, is suppose $E_0$

Energy in IF capacitor,
$E_1=\frac{1}{2}C{V}^2=\frac{1}{2}C{\left(\frac{Q}{C}\right)}^2$
$=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}\times \frac{4^2}{1}=8J$
Similarly, $E_2=\frac{1}{2}\frac{Q^2}{C}=\frac{2^2}{2\times 2}=1J$
So, the total energy, $E_0=E_1+E_2=8+l=9J$
Now, switch's is closed then the common potential of Capacitors,
$V_{\text{common }}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
$=\frac{1\times 4+2\times 1}{1+2}=\frac{6}{3}=2V$
Hence, now the new arrangement of energy,
$E_1=\frac{1}{2}{C}_1{V}_{\text{common }}^2=\frac{1}{2}\times 1\times 4=2J$
and $E_2=\frac{1}{2}{C}_2{V}_{\text{common }}^2=\frac{1}{2}\times 2\times 4=4J$
Now, from conservation of energy,
Energy stored in the inductor,
$E_L\simeq E_0-\left(E_1+E_2\right)$
$=9-(2+4)=3J$
Hence, the inductor has $3J$ of energy.