In the circuit shown in Fig. P30.60, switch S1 has been closed for a long enough time so that the current reads a steady 3.50 A. suddenly, switch S2 is closed and S1 is opened at the same instant.

(a) What is the maximum charge that the capacitor will receive?

(b) What is the current in the inductor at this time?

Principle of conversation of energy

The principle of conservation of energy states that. a system that is isolated from its surroundings, the total energy of the system is conserved.

$E_{Initial}=E_{Final}$

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

$V=IR$

An inductor act as a wire of infinite resistance right after the circuit is closed while it acts as a normal conducting wire after a long time of circuit being closed

A capacitor acts as normal conducting wire when the circuit is just closed while it acts a wire of infinite resistance after a long time of circuit being closed.

Energy stored by the inductor is given by

$E=\frac{1}{2}L{I}^2$

Where L is the inductance of the inductor while I is the current through the system.

Energy of the capacitor is given by

$E=\frac{{Q^2}_{max}}{2C}$

where $Q_{max}$is the maximum charge on the capacitor and is the capacitance of the capacitor.

When the switch $S_1$opened and$S_2$is closed it forms a closed system hence the energy is conserved in the system. Hence entire energy stored by the inductor will be taken up by the capacitor.

Energy in the Inductor is given by$E=\frac{1}{2}L{I}^2$

Where L is the inductance of the inductor while I is the current through the system.

Energy of the capacitor is given by$E=\frac{{Q^2}_{max}}{2C}$

where $Q_{max}$is the maximum charge on the capacitor and $C$is the capacitance of the capacitor.

Equating the two

localid="1664255563744" $$\begin{gathered} \frac{1}{2}L{I}^2=\frac{{Q^2}_{max}}{2C} \\ {Q}_{max}=\sqrt{CL{I}^2} \\ {Q}_{max}=\sqrt{\left(5.0\times {10}^{-6}F\right)\left(2.0\times {10}^{-3}H\right){\left(3.50A\right)}^2} \\ {Q}_{max}=3.5\times {10}^{-4}C \end{gathered}$$

Hence the maximum charge in the capacitor is $Q_{max}=3.5\times {10}^{-4}C$

This state capacitor stores the maximum energy that is all energy of the system is in form of energy of the capacitor which corresponds to zero energy being stored by the inductor

$\frac{1}{2}L{I}^2=0$

$I=0$

Hence the current in the inductor is equal to zero.