91Ó°ÊÓ

1. Frequency of oscillation,$\mathrm{f}=10.4 \text{kHz}=10.4×{10}^3 \text{Hz}$.
2. Capacitance, $\mathrm{C}=340\mathrm{μ¹ó}=340×{10}^{-6}\mathrm{F}$
3. Maximum current,$\mathrm{I}=720\mathrm{mA}=7.20×{10}^{-3}\mathrm{A}$

By using the formula for frequency in the LC oscillator, find the inductance in the circuit. By using the formula of energy stored in the inductor, find the total energy in the circuit for the given maximum current. Find the maximum charge on the capacitor by using the formula for energy stored in the capacitor.

Formulae are as follows:

i) The frequency forLC the oscillator is $\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€}\sqrt{\mathbf{LC}}}$

ii) Energy stored in the inductor is,
$\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{LI}}^{\mathbf{2}}$

$\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{{\mathbf{Q}}^{\mathbf{2}}}{\mathbf{C}}$

Where, U is potential energy, C is capacitance and Q is charge.

For LC oscillator, the frequency of oscillation is,

$\mathrm{f}=\frac{1}{2\mathrm{Ï€}\sqrt{\mathrm{LC}}}$

Where,

L= Inductance and C= Capacitance

Rearranging the equation for L,

$\mathrm{L}=\frac{1}{\mathrm{C}}\left(\frac{1}{4{\mathrm{Ï€}}^2{\mathrm{f}}^2}\right)$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{L}=\frac{1}{340×{10}^{-6} \text{F}}\left(\frac{1}{4×{3.14}^2×{\left(10.4×{10}^3 \text{Hz}\right)}^2}\right) \\ \mathrm{L}=6.89×{10}^{-7} \text{F} \\ =0.689×{10}^{-6} \text{F} \\ \mathrm{L}=0.689\mathrm{μ¹ó} \end{gathered}$$

Hence,the value of inductance, when capacitance is $340\mathrm{μ¹ó}$is,$0.689\mathrm{μ¹ó}$.

The energy stored in the inductor is given as,

$\mathrm{U}=\frac{1}{2}{\mathrm{LI}}^2$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{U}=\frac{1}{2}\left(0.689×{10}^{-6} \text{F}\right){\left(7.20×{10}^{-3} \text{A}\right)}^2 \\ \mathrm{U}=17.9×{10}^{-12} \text{J} \\ \mathrm{U}=17.9 \text{pJ} \end{gathered}$$

Hence, the total energy in the circuit when the maximum current is $7.20\mathrm{mA}$is, $17.9\mathrm{pJ}$.

The energy stored in the capacitor is given as,

$\mathrm{U}=\frac{1}{2}\frac{{\mathrm{Q}}^2}{\mathrm{C}}$

Rearranging the equation for Q,

$\mathrm{Q}=\sqrt{2\mathrm{CU}}$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{Q}=\sqrt{2×340×{10}^{-6} \text{F}×17.9×{10}^{-12} \text{J}} \\ \mathrm{Q}=110.3×{10}^{-9} \text{C} \\ \mathrm{Q}=110.3 \text{nC} \end{gathered}$$

Hence, the maximum charge on the capacitor is $110.3\mathrm{nC}$.

By using formulae for frequency of LC oscillator and energy stored in inductor and capacitor, found the required quantities.