Question

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?

Short answer

1. Inductance is $2.41 \mathrm{\mu H}$.
2. The total energy is 21.4pJ.
3. The maximum charge is 82.2nC.

Step-by-step solution

Identification of the given data

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

Determining the concept

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.

(a) Determining the Inductance

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 $2.41 \mathrm{\mu H}$.

(b) Determining the total energy

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.

(c) Determining the maximum charge

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.