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

Question

Aicharya · Accepted Answer

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

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

Solution