Question

The peak current through an inductor is $10mA$. What is the peak current if

a. The emf frequency is doubled?

b. The emf peak voltage is doubled (at the original frequency)

Short answer

(a) Current is $5mA$

(b) Current is$20mA$

Step-by-step solution

Part (a) Step 1: Given information 

We have been given that,

Current through inductor is$I_0=10mA$

Part (a) Step 2: Simplify

Let Voltage is $V_0$

And Frequency is $f_0$

And Inductance is $L$

And Inductive reactance is localid="1649934635236" $X_L=2{\mathrm{\pi f}}_0\mathrm{L}$

Therefore, Current is

$I_0=\frac{V_0}{2{\mathrm{\pi f}}_0\mathrm{L}}$$I_0=\frac{V_0}{2{\mathrm{\pi f}}_0\mathrm{L}}$

Now, Frequency is doubled i.e. $f=2f_0$

Therefore, Current is,

localid="1649935758169" $$\begin{gathered} I=\frac{V_0}{2\mathrm{\pi fL}} \\ I=\frac{V_0}{2\mathrm{\pi }\times 2f_{0}L} \\ \mathrm{I}=\frac{{\mathrm{I}}_0}{2} \\ \mathrm{I}=\frac{10}{2} \\ \mathrm{I}=5\mathrm{mA} \end{gathered}$$

Hence, Current is $5mA$

Part (b) Step 1: Given information 

We have been given that,

Current through inductor is$I_0=10mA$

Part (b) Step 2: Simplify

Let Voltage is$V_0$

And inductive reactance is$X_L$

Therefore, Current is

$I_0=\frac{V_0}{X_L}$

Now, Voltage is$V=2V_0$

Therefore, Current is,

$$\begin{gathered} I=\frac{V}{X_L} \\ I=\frac{2V_0}{X_L} \\ I=2I_0 \\ I=2\times 10 \\ I=20mA \end{gathered}$$

Hence, current is$20mA$