Physics - EMI/AC Question with Solution | TestHub

Initially, when the switch is closed for a long time, the current in the primary coil reaches a steady state: $I_0=V/R=100\text{V}/10\Omega =10\text{A}$. The secondary coil has no current, and the capacitor is uncharged.

When the switch is opened, the current in the primary coil starts to decay. This change in current induces an EMF in the secondary coil. The maximum energy transferred from the primary to the secondary LC circuit is given by $U_{LC,max}=\frac{1}{2}\frac{M^2}{L_2}{I}_0^2$. This energy is eventually stored in the capacitor as maximum electric energy. Therefore, $\frac{1}{2}\frac{Q_{max}^2}{C}=\frac{1}{2}\frac{M^2}{L_2}{I}_0^2$. Solving for $Q_{max}$: $Q_{max}=M{I}_0\sqrt{\frac{C}{L_2}}$. Substituting the given values: $M=10\times 10^{-3}\text{H}$, $I_0=10\text{A}$, $C=1\times 10^{-6}\text{F}$, and $L_2=40\times 10^{-3}\text{H}$. $Q_{max}=(10\times 10^{-3})\times 10\times \sqrt{\frac{1\times 10^{-6}}{40\times 10^{-3}}}=0.1\times \sqrt{\frac{1}{40\times 10^3}}=0.1\times \sqrt{\frac{1}{40000}}=0.1\times \frac{1}{200}=0.0005\text{C}$. Converting to microcoulombs: $Q_{max}=0.0005\times 10^6\mu \text{C}=500\mu \text{C}$.