Question

A 7.50nF capacitor is charged up to 12.0 V, then disconnected from the power supply and connected in series through a coil. The period of oscillation of the circuit is then measured to be $\mathbf{8}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{s}$. Calculate the maximum current in the circuit.

Short answer

Answer: The maximum current in the circuit is $6.58\times {10}^{-3}\mathrm{A}$.

Step-by-step solution

Energy conservation in LC circuit

The stored energy in capacitor in addition to the stored energy in inductor at any instant equals to the maximum stored energy in the capacitor.

Mathematically it can be given by

$$\begin{gathered} {\mathrm{U}}_{\mathrm{L}}+{\mathrm{U}}_{\mathrm{C}}={\mathrm{U}}_{\mathrm{C}\max } \\ \frac{1}{2}\mathrm{Li}+\frac{{\mathrm{q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}} \end{gathered}$$

Where , ${\mathrm{U}}_{\mathrm{C}\max }$is the maximum energy stored in capacitor and ${\mathrm{U}}_{\mathrm{L}}$and ${\mathrm{U}}_{\mathrm{C}}$are the energy stored in inductor and capacitor respectively.

The definition of time period of oscillation of charge in the LC circuit

It is time at which current or moving charge complete one oscillation in whole LC circuit.

Mathematically it is given

$T=2\mathrm{\pi }\sqrt{\mathrm{LC}}$

Where, $L$is the inductance of inductor ,$C$is the capacitance of capacitor and $T$is time period of oscillation .

Calculation of inductance of inductor

Using

$$\begin{gathered} \mathrm{T}=2\mathrm{\pi }\sqrt{\mathrm{LC}} \\ \mathrm{L}=\frac{{\mathrm{T}}^2}{4{\mathrm{\pi }}^2\mathrm{C}} \end{gathered}$$

Now ,put the values of constants in above expression

$$\begin{gathered} \mathrm{L}=\frac{{\left(8.60\times {10}^{15}\mathrm{s}\right)}^2}{4{\mathrm{\pi }}^2\left(7.50\times {10}^{-9}\mathrm{F}\right)} \\ \mathrm{L}=2.50\times {10}^{-2}\mathrm{H} \end{gathered}$$

Thus, the value of inductance of inductor is $2.50\times {10}^{-2}\mathrm{H}$

Calculation of maximum current  in LC circuit

Using

$\frac{1}{2}\mathrm{Li}+\frac{{\mathrm{q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$

For maximum current in LC circuit stored charge on capacitor must be zero .

So,

role="math" localid="1664168462756" $$\begin{gathered} \frac{1}{2}\mathrm{Li}{\left({\mathrm{i}}_{\max }\right)}^2=\frac{{\mathrm{Q}}^2}{2\mathrm{C}} \\ {\mathrm{i}}_{\max =}\frac{\mathrm{Q}}{\sqrt{\mathrm{LC}}} \\ {\mathrm{i}}_{\max =}\frac{\mathrm{CV}}{\sqrt{\mathrm{LC}}} \end{gathered}$$

Now, put the values of constants in above equation

$$\begin{gathered} {\mathrm{i}}_{\max =}\frac{7.50\times {10}^9\mathrm{F}\times 12.0\mathrm{V}}{\sqrt{2.50\times {10}^{-2}\mathrm{H}\times 7.50\times {10}^9\mathrm{F}}} \\ {\mathrm{i}}_{\max =}6.58\times {10}^{-3}\mathrm{A} \end{gathered}$$

Thus,the maximum current in the circuit is$6.58\times {10}^{-3}\mathrm{A}$