Question

An LC circuit oscillates at a frequency of $\mathbf{10}\mathbf{.}\mathbf{4}\mathbf{kHz}$. (a) If the capacitance is$\mathbf{340}\mathbf{}\mathbf{\mu F}$, what is the inductance? (b) If the maximum current is$\mathbf{7}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{mA}$, what is the total energy in the circuit? (c) What is the maximum charge on the capacitor?

Short answer

1. The value of inductance, when capacitance $340\mathrm{\mu F}$is,$0.689\mathrm{\mu F}$
2. The total energy in the circuit when the maximum current $720\mathrm{mA}$is,$17.9 \text{pJ}$
3. The maximum charge on the capacitor is$110.3 \text{nC}$.

Step-by-step solution

 Step 1: Identification of the given data

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

Determining the concept

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 $\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }\sqrt{\mathbf{LC}}}$

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

iii) Energy stored in the capacitor is,

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

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

(a) Determining the value of inductance when the capacitance is 340 μF

For LC oscillator, the frequency of oscillation is,

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

Where,

L= Inductance and C= Capacitance

Rearranging the equation for L,

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

Substitute all the value in the above equation.

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

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

(b) Determining the total energy in the circuit when the maximum current is 7.20 mA

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.

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

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

(c) Determining the Maximum charge on the capacitor

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.

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

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.