91Ó°ÊÓ

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.

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

Equivalent Capacitance for series of capacitors is given by

role="math" localid="1664261030111" $\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{i}=1}{\overset{\mathrm{n}}{∑}}\frac{1}{{\mathrm{C}}_{\mathrm{i}}}$

Energy stored by the inductor is given by

role="math" localid="1664261036148" $E=\frac{1}{2}L{I}^2$

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.

In position 1 for the switch the current flows through the resistance$R=125\mathrm{Î\copyright }$only. The inductors act as normal conducting wires, hence use the ohm’s law to get the current in the circuit,

$$\begin{gathered} i=\frac{\mathrm{Î\mu }}{R} \\ i=\frac{75V}{125\mathrm{Î\copyright }} \\ i=0.6A \end{gathered}$$

The two capacitors are in series hence they have the same charge Q.

Find the equivalent capacitance

$\frac{1}{C_{eq}}=\frac{1}{25\mathrm{μ}F}+\frac{1}{35\mathrm{μ}F}$

The equivalent capacitance is$C_{eq}=14.6\mathrm{μ}F$

Use conservation of energy to convert the energy stored in the inductors which will be equal to the energy that can be stored in the capacitors.

$$\begin{gathered} U_c=U_L \\ \frac{Q^2}{2C_{eq}}=\frac{L_{eq}{i}^2}{2} \\ Q=i\sqrt{L_{eq}{C}_{eq}} \end{gathered}$$

Plug the values for all the variables and get$Q=3.24×{10}^{-4}C$

We know that the time period of oscillation is given by

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt{\frac{1}{L_{eq}{C}_{eq}}} \\ \frac{2\mathrm{Ï€}}{T}=\sqrt{\frac{1}{L_{eq}{C}_{eq}}} \end{gathered}$$

Maximum charge in the capacitor is obtained at$T/4$

Use this fact to obtain

$$\begin{gathered} t=\frac{T}{4} \\ t=\frac{2\mathrm{Ï€}\sqrt{{\mathrm{L}}_{\mathrm{eq}}{\mathrm{C}}_{\mathrm{eq}}}}{4} \end{gathered}$$

Solve this for time and get$t=8.5×{10}^{-4}s$