Basic RL circuit -- Power example

Click For Summary

Discussion Overview

The discussion revolves around a homework problem involving a simple RL circuit connected to an AC voltage source. Participants explore the calculations related to power, impedance, and the relationships between voltage, current, resistance, and inductance. The focus includes understanding the equations used in the context of power transfer and impedance in RL circuits.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving an RL circuit and expresses confusion about the derivation of certain equations related to current and impedance.
  • Another participant challenges the validity of a formula presented in the solution, asserting that it is incorrect and provides a correction.
  • There is a discussion about the impedance of the inductor, with participants confirming that it should include the imaginary unit 'j' in its representation.
  • Participants clarify that the total impedance is expressed as Z = R + jωL, and its modulus is given by |Z|² = R² + ω²L².
  • One participant expresses gratitude for the clarification, indicating that the confusion was resolved after understanding the corrections made by others.

Areas of Agreement / Disagreement

There is disagreement regarding the correctness of certain formulas related to current and impedance. Some participants correct earlier claims, but no consensus is reached on the initial interpretations of the equations.

Contextual Notes

Participants express uncertainty about the presence of the imaginary unit 'j' in certain equations and the implications of squaring terms in the context of impedance. The discussion highlights the complexity of interpreting circuit equations and the importance of clarity in mathematical representations.

Alex Santos
Messages
14
Reaction score
0

Homework Statement


A simple RL circuit in series connected to a source with V(t) = 30Cos(3t+75°)V and is delivering 14.12 W of power. I'm also given the power factor which is pf=0.857 and is lagging.

Find the values of R and L

Homework Equations



P=Vm*Im/2 * pf for power
theta = Cos^-1(pf)

The Attempt at a Solution



I've tried the problem and I already have the solution which I will link to this post. But I do not understand some part of the problem.
So what I've basically done is I found the angle of the power factor by using the equation above.
Theta = Cos^-1(0.857) = 31.01° ( positive because the powerfactor is lagging)

Then by Ohms law I find the Current which is I = V/Z and Z = R + jwl
so I = V/(R + jwl). We know that the Voltage is sinusoidal and therefor Vm = 30 V .
Then we have I = 30/(R +jwl)

I can also find maximum current by using the equation stated above. Which is
P=Vm * I am / 2 * pf ----> algebra ---> I am = 2*P/(Vm*pf) = 1.098 A

This is how far I could get and I don't know how to get anything else. But in the solution I have it says that
| I | = 30/(R^2 + w^2 * L^2) ... I have absolutely no idea how this is true? But what I find the most disturbing is that I thought that the impedance of the Inductor would be "Jwl" not only "wL", Where is the "J" in that eqation?[/B]

same thing with given solution that (theta)v - (theta)i = avg{R + jwl} = atan(wl/R} ... <-- what is this. Where does this come from and where is the "J"

In the end my solution tries to find two other variables called X and Y. I assume that X is the reactance, and once again I do not know what Y stands for.
The solutions says R^2 + w^2 * L^2 = 30/| I | = 27.3123 = X <--- again where is that J and why is it not there. I need the explanation.
To find Y variable you use this equation R = wl/X and from that w^2 * L^2 / Y ^2 + w^2 L^2 = X
I do not understand what was happening there.

So to sum it all up. I am confused why there is no 'J' in some of the equations and why they are squared in others.
It would be greatly appreciated if I could get some explanation on how this power transfer works.

I will link to pictures with this post. One with my attempt and one of the solution.
Lausn+12_Page_1.png
20141218_215432.jpg
 
Physics news on Phys.org
Alex Santos said:
But in the solution I have it says that
| I | = 30/(R^2 + w^2 * L^2) ... I have absolutely no idea how this is true?
It is not true. Correct formula is:
gif.latex?\mid%20I\mid%20%3D\frac{30}{\sqrt{R^{2}&plus;\omega%20^{2}L^{2}}}.gif

Notice also this is a peak current since peak voltage is 30 V.
 
zoki85 said:
It is not true. Correct formula is:
gif.latex?\mid%20I\mid%20%3D\frac{30}{\sqrt{R^{2}&plus;\omega%20^{2}L^{2}}}.gif

Notice also this is a peak current since peak voltage is 30 V.

Alright thanks for that, but isn't the impedance for inductor Zl = Jwl?
 
Alex Santos said:
Alright thanks for that, but isn't the impedance for inductor Zl = Jwl?
It is. Total impedance is Z=R+jωL, and it's modulus IZI2=R22L2
 
zoki85 said:
It is. Total impedance is Z=R+jωL, and it's modulus IZI2=R22L2

Thank you very much. Problem solved. I just couldn't figure out what the solution was telling me but once you said they were wrong It all came together.

Thanks again for the help :D
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
19
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K