AC circuit, mystery impedance

[Enzo]

Homework Statement

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

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?

 

[The Electrician]

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.

 

[piercas]

Your calculations seem correct, but your interpretation of the results may be causing some confusion. Let's break down each part:

d) The type of reactance is in fact an inductor (XL), as you correctly stated. However, this does not mean that there is no resistance associated with the inductor. In fact, all real inductors have some amount of resistance, which is known as the winding resistance. This resistance is usually small compared to the reactance, but it is still present.

e) You correctly calculated the value of X, which represents the reactance of the inductor. This value includes both the inductive reactance and the winding resistance (X = XL + R). From this, you can calculate the inductance by using the formula you provided (L = X / (2*pi*f)). This will give you the total inductance, which includes the inductance of the coil as well as any parasitic inductances.

f) Your calculations for the active, reactive, and apparent power seem correct. The apparent power (S) is the vector sum of the active and reactive power, so it makes sense that it matches the total power calculation (P = S * cos(phi)). Keep in mind that the reactive power is not consumed by the circuit, but it is necessary for the operation of the inductor.

g) The phasor diagram for this circuit will have a voltage phasor V1 at an angle of 48 degrees (since V1 is 10sqrt(2) < 48), a current phasor I at an angle of 0 degrees (since I is 0.354 < 0), and a voltage phasor V2 at an angle of 0 degrees (since V2 is 5/sqrt(2) < 0). The phasor diagram will also show the inductive reactance (X) at an angle of 60.6 degrees (since X is 16.77 + 29.73j). This should give you a visual representation of the circuit and how the different components are related to each other.

Overall, it seems like you have done the calculations correctly. The discrepancy between the active power calculation and the total power calculation is most likely due to the winding resistance of the inductor, which is present in all real inductors. Keep in mind that even though the winding resistance may seem small, it can still have an impact on