Question

A $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{mH}$inductor in an oscillatingLCcircuit stores a maxi- mum energy of $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{J}$. What is the maximum current?

Short answer

The maximum current is$0.115\mathrm{A}$.

Step-by-step solution

Step 1: Given

${\mathrm{U}}_{\max }=100\mathrm{\mu J}=10\times {10}^{-6}\mathrm{J}$

$\mathrm{L}=1.5\mathrm{mH}=1.5\times {10}^{-3}\mathrm{H}$

Determining the concept

By using the formula for maximum energy in the $\mathbf{LC}$circuit and substituting the given values, find the maximum current.

Formulae are as follows:

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

Where,

U = magnetic energy.

L = inductance of solenoid.

I = current.

Determining the maximum current

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

Rearranging the terms,

$$\begin{gathered} {\mathrm{I}}^2=\frac{\text{2}{\mathrm{U}}_{\max }}{\mathrm{L}} \\ \mathrm{I}=\sqrt{\frac{\text{2}{\mathrm{U}}_{\max }}{\mathrm{L}}} \\ \mathrm{I}=\sqrt{\frac{2\times 10\times {10}^{-6}}{1.5\times {10}^{-3}}} \\ \mathrm{I}=0.005\mathrm{A} \end{gathered}$$

Hence,the maximum current is$0.005\mathrm{A}$.

Using the formula for maximum energy stored in the LC circuit, found the maximum current in the circuit.