What is the Impedance of a Particular Inductor with Appreciable Resistance?

[Kris1120]

Homework Statement

A particular inductor has appreciable resis-
tance. When the inductor is connected to a
19.7 V battery, the current in the inductor is
4.38 A. When it is connected to AC source
with an rms output of 19.7 V and a frequency
of 58 Hz, the current drops to 1.65 A.
Find the impedance at 58 Hz.
Answer in units of
.

Homework Equations

Z=sqrt(R^2 +(XL-XC)^2)

XL=V/I

R=V/I

The Attempt at a Solution

XL=19.7 V / 4.38 A = 4.49772 ohms

R = 19.7 V / 1.65 A = 11.9394 ohms

Z= sqrt( 11.9394^2 + 4.49772^2) = 12.7585 ohms

 

[nasu]

I am afraid you mixed them up.
The higher current is when you have only resistance. (I1=V/R)
The lower current is I2 = V/Z and not V/XL.

 

[Kris1120]

Ok I understand my mistake now. Thank you for clearing that up for me!