Help with a Series RL network and Power factor

[Khesahn]

Homework Statement

In a 60 Hz series RL network, the source is supplying 400 W of real power. If R = 15.08 Ω and L = 40 mH, find
a. The power factor. (Indicate if it is leading or lagging)
b. The apparent power supplied.
c. The reactive power supplied.

Homework Equations

None were given

The Attempt at a Solution

Ive been trying to google about power factor and reading up on it for quite some time, but I am still stuck. I know I can get PF by taking cosθ, but I am unsure of how to find θ for this particular problem. I believe it can be done by taking cos^(-1) (Real Power/Apparent Power), but again I was not sure of how to find apparent power.

I guess the biggest problem for me so far is been trying to relate the 60Hz to the rest of the problem. If someone could give me a hint on this I would greatly appreciate it. I had a funeral to attend to so I missed the class discussing most of these things, and its been difficult trying to read up on it online. Any help is appreciated!

 

[gneill]

Homework Statement

In a 60 Hz series RL network, the source is supplying 400 W of real power. If R = 15.08 Ω and L = 40 mH, find
a. The power factor. (Indicate if it is leading or lagging)
b. The apparent power supplied.
c. The reactive power supplied.

Homework Equations

None were given

The Attempt at a Solution

Ive been trying to google about power factor and reading up on it for quite some time, but I am still stuck. I know I can get PF by taking cosθ, but I am unsure of how to find θ for this particular problem. I believe it can be done by taking cos^(-1) (Real Power/Apparent Power), but again I was not sure of how to find apparent power.

I guess the biggest problem for me so far is been trying to relate the 60Hz to the rest of the problem. If someone could give me a hint on this I would greatly appreciate it. I had a funeral to attend to so I missed the class discussing most of these things, and its been difficult trying to read up on it online. Any help is appreciated!

The power factor is the cosine of an angle. This angle is the offset between the current and voltage waveforms. As such, it also happens to be the angle of the impedance vector. If you find the impedance of the series circuit in polar form, the resulting angle will be the one you want. Find its cosine.

If you are as yet unfamiliar with complex valued impedance, then the same can be accomplished using resistance and reactance. Form the impedance triangle with resistance on the base leg and reactance on the vertical leg. The legs make a right angle. The angle of the hypotenuse (the impedance Z) with the resistance leg is the angle you're after.

 

[Khesahn]

I was working on this a bit more after doing some research and this is what I've found.

P=(I^2)R, using my values gives me √(400/15.08)= I, which is I = 5.15A.

Using the equation Z= √(R^2 + (ωL)^2) = Z = √(15.08^2 + (120∏*.040)^2) = Z = 21.33Ω.

I then use Apparent Power = (I^2)Z which gave me App. Power = 565.72 VA.

I used PF = (Real Power)/(App. Power) which gave me .707 and its lagging.

I then used the PF to find θ≈45°.

I used Reactive Power = 400 tan(45) to see Rea. Power = 400VAR.

I believe I was doing this correctly, do you all see anything that might have gotten me to be doing this incorrectly? Again, the equations and methods I am using are based off researching from google. The assignment is due tonight so I don't have a lot of time to make sure I am doing this correctly.

Thanks again!

 

[gneill]

You've found a path to the correct values, so that's fine.

My own approach would have been to use the given resistor and inductor values to calculate the impedance first, and then take the power factor angle directly from that. With the angle in hand, the power triangle is similar to the impedance triangle (in the geometric sense) and so the apparent and reactive powers are simple to find.

 

[rezeew]

Hello, it seems like you are struggling with understanding power factor in a series RL network. Let me try to break it down for you.

First, let's define power factor. Power factor is the ratio of real power (also known as active power) to apparent power. Real power is the power that is actually being used to do work, while apparent power is the total power supplied to the circuit. In a series RL network, the real power is dissipated in the resistor (R) while the reactive power (due to the inductor, L) is not used to do work but rather stored and released back to the circuit.

To find the power factor in this problem, you need to use the formula cosθ = Real Power/Apparent Power. In this case, you are given the real power (400 W) and the values of R and L. To find the apparent power, you can use the formula Apparent Power = √(Real Power^2 + Reactive Power^2). The reactive power can be found using the formula Reactive Power = Voltage^2/X, where X is the reactance of the inductor (XL = 2πfL).

Once you have found the power factor, you can determine if it is leading or lagging by looking at the sign of the cosine value. If the value is positive, the power factor is leading (inductive) and if it is negative, the power factor is lagging (capacitive).

To find the apparent power supplied, you can use the formula Apparent Power = Voltage * Current. You can find the current using Ohm's Law (I = V/R).

Lastly, to find the reactive power supplied, you can use the formula Reactive Power = Voltage^2/X. This will give you the reactive power in volt-amperes reactive (VAR).

I hope this explanation helps you understand power factor in a series RL network better. Remember, practice makes perfect, so keep practicing and don't hesitate to ask for help when needed. Good luck with your homework!