An inductor has a peak current of 270 muA when the peak volt

An inductor has a peak current of 270 muA when the peak voltage at 44MHz is 2.5 V. What is the inductance? Express your answer using two significant figures.

Solution

1.

i = 270*10^-6 Amp.

V = 2.5 V

Z = V/i = 2.5/(270*10^-6)

Z = 9259.259 ohm

Z = XL = wL

w = 2*pi*f

f = 44*10^6 Hz

w = 2*3.14*44*10^6 = 2.76*10^8 rad/sec

2.

XL = wL

L = XL/w = 9259.259)/(2.76*10^8) = 0.000033548 = 33.54*10^-6 H

L = 33.54 uH

C.

if f2 = 88*10^6

w2 = 2*w1 = 2*2.76*10^8 = 5.52*10^8

XL = 5.52*10^8*33.54*10^-6 = 18514.08 ohm

i = V/R = (2.5/18514.08) = 1.35*10^-4 Amp.