# An inductor has a peak current of 330 mu A when the peak voltage at 44 MHz is 3.6 V. 1) What is...

## Question:

An inductor has a peak current of {eq}330\ \mu A {/eq} when the peak voltage at {eq}44\ MHz {/eq} is {eq}3.6\ V {/eq}.

1) What is the inductance?

2) If the voltage is held constant, what is the peak current at {eq}88\ MHz {/eq} ?

## Electric Current

Electric current is described as the movement of charged particles from one potential point to another in an electric circuit in order to produce the effect of electricity. It is quantified in the conventional units of Ampere.

Given data

• The value of peak current is: {eq}{I_p} = 330\;{\rm{\mu A}} {/eq}.
• The value of the peak voltage is: {eq}{V_p} = 3.6\,{\rm{V}} {/eq}.
• The frequency is: {eq}f = 44\;{\rm{MHz}} {/eq}.

1)

The expression for the inductance is,

{eq}L = \dfrac{{{V_p}}}{{2\pi f \times {I_P}}} {/eq}

Substitute the values.

{eq}\begin{align*} L &= \dfrac{{3.{\rm{6}}\;{\rm{V}}}}{{2\pi \left( {44\;{\rm{MHz}}\left( {\dfrac{{{{10}^6}\;{\rm{Hz}}}}{{1\;{\rm{MHz}}}}} \right)\left( {\dfrac{{1\;{{\rm{s}}^{ - 1}}}}{{1\;{\rm{Hz}}}}} \right)} \right) \times \left( {330\;{\rm{\mu A}}\left( {\dfrac{{{{10}^{ - 6}}\;{\rm{A}}}}{{1\;{\rm{\mu A}}}}} \right)} \right)}}\\ &= 3.945 \times {10^{ - 5}}\;{\rm{V}}{\rm{.s/A}}\left( {\dfrac{{1\;{\rm{H}}}}{{1\;{\rm{V}}{\rm{.s/A}}}}} \right)\\ &= 3.945 \times {10^{ - 5}}{\rm{H}} \end{align*} {/eq}

Thus, the inductance is {eq}3.945 \times {10^{ - 5}}{\rm{H}} {/eq}.

2)

It is provided that the voltage is kept constant. The frequency is {eq}{f_p} = 88\;{\rm{MHz}} {/eq}.

The expression for the peak current is,

{eq}I = \dfrac{{{V_P}}}{{2\pi {f_p} \times L}} {/eq}

Substitute the values.

{eq}\begin{align*} I &= \dfrac{{3.6\;{\rm{V}}}}{{2\pi \left( {88\;{\rm{MHz}}\left( {\dfrac{{{{10}^6}\;{\rm{Hz}}}}{{1\;{\rm{MHz}}}}} \right)\left( {\dfrac{{1\;{{\rm{s}}^{ - 1}}}}{{1\;{\rm{Hz}}}}} \right)} \right) \times \left( {3.945 \times {{10}^{ - 5}}{\rm{H}}\left( {\dfrac{{1\;{\rm{V}}{\rm{.s/A}}}}{{1\;{\rm{H}}}}} \right)} \right)}}\\ &= 1.65 \times {10^{ - 4}}\;{\rm{A}} \end{align*} {/eq}

Thus, the peak current is {eq}1.65 \times {10^{ - 4}}\;{\rm{A}} {/eq}. 