AC circuit, mystery impedance

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  • Thread starter Enzo
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  • #1
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Homework Statement


http://img159.imageshack.us/img159/1867/es100wi8.th.png [Broken]http://g.imageshack.us/thpix.php [Broken]

a) Write an expression for V1 and V2 both in time domain and phasor domain (Solved)

b) Write an expression for the current I both in time domain and phasor domain (Solved)
c) Calculate the Power factor of the supply and specify whether it is lagging or
leading (Solved)
d) Specify the type of the reactance (XC or XL)
e) Determine the value of X and hence the corresponding value of L or C
f) Calculate the supply average, reactive and apparent power
g) Draw the phasor diagram of the circuit




Homework Equations


f= 250/3


The Attempt at a Solution


a)
V1 = 10sqrt(2)<48 = 9.46+10.51
V2 = 5/sqrt(2)<0 = 3.54
b)
I = v2/r2 = 0.354<0
c)
PF = 0.669, lagging
d)

X is an inductor

e)
VCh2 + Vxl = Vsource
Vxl = 9.46+10.51j - 5/sqrt(2) = 5.93 +10.51j = 12.1<60.6

Zx = Vxl/Is = 12.1<60.6 / 0.354<0 = 16.77+ 29.73j

This tells me that there's a physical impedance associated with the coil, equal in value to 16.77ohms...Meaning that it's not a pure inductor?

To find L:

29.73 = 2*pi*f*L ... 29.73/ (2*pi*250/3) = L = 0.057H

But these results don't add up to the next few parts:

-Active Power-
P(10ohm)=I^2*R=(0.354)^2*10 = 1.25W
P(16.77ohm)=I^2*R=(0.354)^2*16.77 = 2.10W
Total:3.35W

-Reactive Power-
P(XL)=I^2*R=(0.354)^2*29.73 = 3.72VAR

-Apparent Power-
Active+Reactive*j = 3.35+3.72j

which matches:

S=EI=10sqrt(2)<48 * 0.354<0 = 3.35 + 3.72j

Have I done this question correctly? Is it possible that the inductor has a simple resistance component to it?
 
Last edited by a moderator:

Answers and Replies

  • #2
The Electrician
Gold Member
1,285
169
You have apparently chosen the first positive peak of V1 for its phase, but for the phase of V2 you have chosen the first positive going zero crossing. You have to be consistent in these choices.

Looking at the zero crossings of V1, the angle appears to be closer to 45 degrees than 48 degrees.

Since the positive going zero crossing of V1 nearest to the zero time of the graph is at -45 degrees, I would say that V1 = 10sqrt(2)<-45, and then it would be consistent to say that V2 = 5/sqrt(2)<0.

Redo your calculations with this number for V1 and see if you get better results.
 

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