A capacitor of capacitance$\mathbf{158}\mathbf{ }\mathbf{\mu F}$and an inductor form an LC circuit that oscillates at 8.15 kHz, with a current amplitude of 4.21mA. What are (a) the inductance, (b) the total energy in the circuit, and (c) the maximum charge on the capacitor?

1. Capacitor is $158 \mathrm{\mu F}$.
2. Frequency is 81.5kHz.
3. Current is 4.21mA.

Use the formula for angular frequency for LC circuit in terms of inductance and capacitor. To find the total energy, use inductance and current, and use current and angular frequency to find the charge.

The formulae are as follows:

$\mathit{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$ …(i)

Here, $\mathit{\omega }$ is the angular frequency, L is the inductance, C is capacitance.

$\mathit{\omega }\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{f}$ …(ii)

Here, $\mathit{\omega }$ is the angular frequency, f is the frequency.

$\mathit{Q}\mathbf{=}\frac{\mathbf{I}}{\mathbf{\omega }}$ …(iii)

Here, $\mathit{\omega }$is the angular frequency, Q is the charge.

$\mathbf{U}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{LI}}^{\mathbf{2}}$ …(iv)

Here, U is energy, L is inductance, and I is current.

Use the following formula to find inductance:

$\begin{array}{c}\omega =\frac{1}{\sqrt{LC}}\\ 2\pi f=\frac{1}{\sqrt{LC}}\\ f=\frac{1}{2\pi }\frac{1}{\sqrt{LC}}\end{array}$

Substitute all the value in the above equation.

role="math" localid="1663246656941" $\begin{array}{c}8.15\times {10}^3 \mathrm{Hz}=\frac{1}{2\pi }\times \frac{1}{\sqrt{L\times 158\times {10}^{-6} \mathrm{F}}}\\ L=2.41\times {10}^{-6} \mathrm{H}\\ L=2.41 \mathrm{\mu H}\end{array}$

Hence, Inductance is $2.41 \mathrm{\mu H}$.

Use the following formula to find the total energy.

$U=0.5\times L{I}^2$

Substitute all the value in the above equation.

$\begin{array}{c}U=0.5\times 2.41\times {10}^{-6} \mathrm{H}\times {\left(4.21\times {10}^{-3} \mathrm{A}\right)}^2\\ U=21.4\times {10}^{-12} \mathrm{J}\\ U=21.4 \mathrm{pJ}\end{array}$

Hence, the total energy is 21.4pJ.

Use the following formula to find the charge.

$Q=\frac{I}{\omega }$

Substitute all the values in the above equation.

$\begin{array}{c}Q=\frac{4.21\times {10}^{-3} \mathrm{A}}{2\times \pi \times 8.15\times {10}^3 \mathrm{H}}\\ Q=82.2\times {10}^{-9} \mathrm{C}\\ Q=82.2 \mathrm{nC}\end{array}$

Hence, the maximum charge is 82.2 nC.