Basic RL circuit -- Power example

In summary, the conversation discussed a problem involving a simple RL circuit with a power source and a power factor of 0.857. The goal was to find the values of R and L. The solution provided incorrect equations for finding the current and impedance, but with the correct formulas, the problem can be solved. The impedance for an inductor is Zl = jωL and the total impedance is Z = R + jωL, with a modulus of IZI^2 = R^2 + ω^2L^2.
  • #1
Alex Santos
14
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
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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
 

What is a basic RL circuit?

A basic RL (resistor-inductor) circuit is a type of electrical circuit that contains a resistor and an inductor. The resistor is used to regulate the flow of electric current, while the inductor is used to store energy in the form of a magnetic field.

How does an RL circuit work?

In an RL circuit, when a voltage source is connected, the current begins to flow through the circuit. The resistor limits the amount of current that can flow, while the inductor resists changes in the flow of current. As a result, the current in an RL circuit gradually increases to its maximum value, determined by the resistance and inductance values, and then reaches a steady state.

What is the power in an RL circuit?

The power in an RL circuit is the rate at which energy is converted from electrical energy to other forms, such as heat or light. It is measured in watts and is equal to the product of the voltage and current in the circuit.

How can I calculate the power in an RL circuit?

To calculate the power in an RL circuit, you can use the formula P = I^2 * R, where P is power in watts, I is current in amperes, and R is resistance in ohms. Alternatively, you can use the formula P = V * I, where V is voltage in volts and I is current in amperes.

What are the applications of RL circuits?

RL circuits have a variety of applications, including in power supplies, electric motors, generators, and audio equipment. They are also commonly used in electronic devices to filter out unwanted frequencies and inductors are used in electronic circuits to store energy and create magnetic fields.

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