Question

An inductor has a peak current of $330\mu A$ when the peak voltage at$45MHz$is $2.2V$

a. What is the inductance?

b. If the peak voltage is held constant, what is the peak current at $90MHz$?

Short answer

(a) We get value of Inductance as$L=0.23mH$

(b) When we held peak voltage constant we get peak current,$I=0.0169mA$

Step-by-step solution

Part (a) Step 1 : Given information 

We have been given that

Peak current $\left({\mathrm{I}}_{\circ }\right)=330\mu A=33\times {10}^{-5}A$

Peak voltage $\left(V_{\circ }\right)=2.2V$

Frequency localid="1650176753268" $\left(f\right)=45MHz=45\times {10}^{6}Hz$

We need to find out the inductance when peak current and voltage at a particular frequency are given to us.

Part(a) Step 2 : Explanation

As we know,

$$\begin{gathered} X_L=\omega L=2\pi fL \\ \Rightarrow \frac{V}{I}=2\pi fL \\ \Rightarrow L=\frac{V}{2\pi fI}=\frac{2.2}{2\times 3.14\times 45\times {10}^6\times 33\times {10}^{-5}} \\ \Rightarrow L=2.3\times {10}^{-4}H=0.23mH \end{gathered}$$

Part (b) Step 1 : Given Information 

We have been given that

Peak voltage $\left(V_{\circ }\right)=2.2V$

Frequency$\left(f\right)=90MHz=90\times {10}^{6}Hz$

We need to find out the peak current value at 90 MHz

Part(b) Step 2 : Explanation

We Know ,

$$\begin{gathered} V=I{X}_L=I\left(\omega L\right) \\ \Rightarrow 2.2=I(2\pi \times 90\times {10}^6\times L) \\ \Rightarrow I=169\times {10}^{-7}A=0.0169mA \end{gathered}$$