# Impedance problem

#### testing12

Joined Jan 30, 2011
80
Hello Everyone,

I looked through my text and understand the properties reguarding phase difference in current and volatge for capicitors and inductors. I also understand impedance and how its calculated with regards to R L and C. My problem with this particular problem is the solution.

How can we use a triangle to solve this problem? I know in an inductor Current lags voltage by 90 degrees. But the solution above doesn not make sense to me, where does this triangle come from? Immediately I thought of applying KVL, can someone provide some more intuition please?

#### testing12

Joined Jan 30, 2011
80
t n k thank you so much, i have no idea this site offered this information! Its bettern then my text book! You have helped me before thank you very much sir!

#### testing12

Joined Jan 30, 2011
80
Hello,
After some review of the link you sent I do understand impedance and how to obtain it's magnitude from the triangle, I also notice that v=I Z. From there I searched some more and found the following:

I understand how it was done in the solution manual but am still not completely clear as to why the sides of the impedance triangle is replaces with the respective voltages. This is exactly what I'm after.

Thank you very much.

#### neeraj sharma

Joined Jun 20, 2009
5
buddy,
as u said V=IZ,
so if at all u have a triangle with sides as impidence i.e.Z then a triangle with sides V will be a triangle similar to the triangle with sides Z.
A1/A2= B1/B2= C1/C2
thus the pythagorian law applicable to impidence traingle will be applicable to the corresponding voltage triangle as well.

#### bglazierjr

Joined Apr 2, 2011
6
Keep in mind this only will work if you are using it in this form. You must be subtracting real parts from E to find the imaginary parts. Meaning sqrt((Ein^2) - (VResistance^2)) = Vimpedance. Just make sure you sort out your resistance and impedance when doing it in this manner. Notice that the answer doesn't account for the given phase angle of the voltage drop VL. Obviously using Kirchoff's Voltage Law with those answers doesn't work. Your AC sin wave could be at a particular given angle, and the voltage VL most defiantly will be at a particular angle. KVL should appear in this form : (100+10j) - (85+ 0j) - (15+10J) = 0; thats just a generic example not related to this problem. The impedance ZL also varies depending upon at what frequency you are trying to find these voltage drops. The formula is ZL= (2*pi*F*L) ; F = Frequency and L = inductance in Henries