Engineering Basic RL circuit -- Power example

AI Thread Summary
The discussion revolves around solving a basic RL circuit problem involving a sinusoidal voltage source and power delivery. The participant calculates the power factor angle and current but struggles with the impedance equations and the absence of the imaginary unit 'J' in certain expressions. Clarifications reveal that the total impedance is indeed Z = R + jωL, and its modulus is |Z|^2 = R^2 + ω^2L^2. Misunderstandings about the representation of current and impedance in the solution are addressed, leading to a resolution of the confusion. Ultimately, the participant gains clarity on the power transfer and impedance concepts in RL circuits.
Alex Santos
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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
 
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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
 

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