A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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+[XLXc]^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?
anonymous
 one year ago
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+[XLXc]^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?

This Question is Closed

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2300\(\Omega\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure what you are saying. Is my equation wrong?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2OK you said V = IR I am saying V = IZ Z is as described above make sense?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2are you familiar with phasor diagrams? that is how you should "add" the equivalent "resistances" of the resistor inductor and capacitor

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0I know what a phasor diagram is but I don't know how to use it in this instance.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2Imaginary 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2sounds good. dw:1436290978466:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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
 one year ago
Best ResponseYou've already chosen the best response.2to 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
 one year ago
Best ResponseYou've already chosen the best response.2and yes you are right to say that, if it was resonating, you could just say V = IR because at that frequency R = Z.

radar
 one year ago
Best ResponseYou've already chosen the best response.0You do not state whether the 120 V was applied to those components connected in series or paralleldw:1436304451073:dw Would it make a difference?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am assuming that they are connected in series.

radar
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Then work under that assumption, what did you get for \[X _{L}\]\[X _{C}\]??

radar
 one year ago
Best ResponseYou've already chosen the best response.0The inductive reactance at the 1000 Hz source frequency would be around 188.5 Ohms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0For 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1436307866338:dw

radar
 one year ago
Best ResponseYou've already chosen the best response.0According to my calculator R=236.56 Ohm, @eabollich What did you get for R?

radar
 one year ago
Best ResponseYou've already chosen the best response.0PF =Real Power (PW)/Apparent Power (PA PF = (.4^2)(236.5)/(120)(.4) PF = 37.84/48 = 0.789

radar
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0That'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
 one year ago
Best ResponseYou've already chosen the best response.0Are you saying that a lower number is more power factor?

radar
 one year ago
Best ResponseYou've already chosen the best response.0Power 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
 one year ago
Best ResponseYou've already chosen the best response.0Great! I think I understand it better now!

radar
 one year ago
Best ResponseYou've already chosen the best response.0Good 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.