An inductor has a peak current of 250 muA when the peak volt

An inductor has a peak current of 250 muA when the peak voltage at 42 MHz is 3.0 V. You may want to review. For help with math skills, you may want to review: What is the inductance? Express your answer using two significant figures. Part B If the voltage is held constant, what is the peak current at 84 MHz ? Express your answer using two significant figures.

Solution

The impedance of an inductor is 2pi*f*L

Z= 2 * 3.14* 42 * 10^6 *L

V = i*Z

3 = 250*10^-6 * 2* 3.14* 42*10^6 *L

L = 3/65940 = 4.54 * 10^-5 H

(2)

z = 2 * pi * 84 *10^ 6 * 4.54 * 10^-5

V is held constant

Since z is doubled i will become half. So it will be 125 micro amps