anonymous
  • anonymous
A 120V rms potential at f=1000Hz is applied to a 30 mH inductor, a 40 microF capacitor and a resistor (R). Irms=400mA. What is the value of R? What is the power factor phase angle phi? The only equation I know that I could possibly use to find R is Vrms=IrmsR, when I plugged in my data, I got R=300 ohms. Using R=300 ohms, I rearranged the equation PF = cos phi = R/Z, where Z=sqrt(R^2+[XL-Xc]^2) to solve for R and I got phi=90 degrees. Which makes me suspicious. The data never specified that the circuit was in resonance, so was my original equation invalid?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
300\(\Omega\) is the total impedance Z where \(Z = \sqrt{R^2 + (X_L - X_C)^2)}\) the circuit will resonate when \(X_L = X_C \). i make that about 45 Hz but do it yourself and do not rely on my owlet packet calcs :p
IrishBoy123
  • IrishBoy123
@radar
anonymous
  • anonymous
I'm not sure what you are saying. Is my equation wrong?

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IrishBoy123
  • IrishBoy123
OK you said V = IR I am saying V = IZ Z is as described above make sense?
IrishBoy123
  • IrishBoy123
are you familiar with phasor diagrams? that is how you should "add" the equivalent "resistances" of the resistor inductor and capacitor
anonymous
  • anonymous
So I solve for Z using Vrms and Irms and then solve for R by plugging Z, L, C, and f into the long equation? Where XL = 2 pi f L and Xc = 1/2 pi f C ?
anonymous
  • anonymous
I know what a phasor diagram is but I don't know how to use it in this instance.
IrishBoy123
  • IrishBoy123
Imaginary parts: \(X_L = j \omega L\) and \(X_C = -j /\omega C\) Real part : R use \(w = 2 \pi f\) to get \(\omega\) from 1000Hz frequency |dw:1436290334762:dw| \(X_{net} = X_L - X_C = j \ ( \omega L - 1/ \omega C )\)
anonymous
  • anonymous
I have never taken calculus. I am assuming that you are saying that Z is the hypotenuse of a triangle where R gives the X value and Xnet gives the Y? So I convert f to omega, say XL = omega L and Xc = 1/omega C, subtract Xc from XL to get Xnet and then use pythagorean theorem to get R?
IrishBoy123
  • IrishBoy123
sounds good. |dw:1436290978466:dw|
IrishBoy123
  • IrishBoy123
i would try to be more helpful but i got it in the ear recently for being too helpful so i must tread carefully :p
anonymous
  • anonymous
Thank you, I am just trying to make sure I understand what you are saying. Is my original equation wrong because it is not in resonance?
IrishBoy123
  • IrishBoy123
to be in resonance \(X_L\) and \(X_C\) cancel each other out so \(f_{res} = \frac{1}{2 \pi \sqrt{LC}}\) you can calculate the resonant frequency. i think it's 145Hz but check for yourself
IrishBoy123
  • IrishBoy123
and yes you are right to say that, if it was resonating, you could just say V = IR because at that frequency R = Z.
radar
  • radar
You do not state whether the 120 V was applied to those components connected in series or parallel|dw:1436304451073:dw| Would it make a difference?
anonymous
  • anonymous
I am assuming that they are connected in series.
radar
  • radar
I believe IrishBoy123 in his post has given you the direction you must ttavel. Calculate the reactance at 1,000 Hz if they are equal then consider resonance the mode, if not then the diagram that was provided by IrishBoy123 could be used using the net reactance.
radar
  • radar
Then work under that assumption, what did you get for \[X _{L}\]\[X _{C}\]??
radar
  • radar
The inductive reactance at the 1000 Hz source frequency would be around 188.5 Ohms
anonymous
  • anonymous
This question was not meant to be overly difficult, I don't believe that my instructor would have meant for them to be calculated in parallel.
radar
  • radar
For the capacitance reactance, I calculaten approximately 4 Ohms. The circuit is definitely not resonance. Series is fine. so the circuit is predominately inductive for a series circuit
radar
  • radar
|dw:1436307500401:dw|
IrishBoy123
  • IrishBoy123
|dw:1436307866338:dw|
radar
  • radar
|dw:1436311256873:dw|
radar
  • radar
According to my calculator R=236.56 Ohm, @eabollich What did you get for R?
radar
  • radar
PF =Real Power (PW)/Apparent Power (PA PF = (.4^2)(236.5)/(120)(.4) PF = 37.84/48 = 0.789
radar
  • radar
This value is too much power factor for a lot of industry, many attempt to keep it higher than .8. Correction is usually "capacitor banks" compensation. This is because the biggest contributor is electric motors with their inherent inductive reactance.
anonymous
  • anonymous
That's the resistance I got as well. Thank you both for your help! I figured out the second part of the question as well, it is asking for the phase angle, phi, which is the inverse cos of the PF.
anonymous
  • anonymous
Are you saying that a lower number is more power factor?
radar
  • radar
Power factor of 1 is perfect where the apparent power is all real power. Power factor of 0 is no real power and total reactive power. The reason PF of 1 is desirable is there is no reactive power, while reactive power is simply exchanged between source and load and it would not register on a watt hour meter, the additional currents flowing on the transmission lines, even though the resistance of the lines or low, there will be I^2R losses which the power utility does not like, and will encourage the user to clean up his act by keeping PF close to 1.
anonymous
  • anonymous
Great! I think I understand it better now!
radar
  • radar
Good luck with your studies. PF is also the Cosine of the angle between line voltage and line current. When they both are in phase (0 degrees) Cosine is 1. They will be in phase when the load is pure resistance and no reactance.

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