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brinethery

  • 2 years ago

A circuit is constructed with an AC generator, a resistor, capacitor and inductor as shown. The generator voltage varies in time as ε =Va - Vb = εmsinωt, where εm = 120 V and ω = 225 radians/second. The inductance L = 153 mH. The values for the capacitance C and the resistance R are unkown. What is known is that the current in the circuit leads the voltage across the generator by φ = 48 degrees and the average power delivered to the circuit by the generator is Pavg = 103 W. What is C, the value of the capacitance of the circuit?

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  1. brinethery
    • 2 years ago
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  2. brinethery
    • 2 years ago
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    For answers to the previous questions asked, I have: I_max = 2.566A resistance of circuit, R = 31.29 Ω I'm not sure if those values are needed to calculate the answer to this question, so I'm providing them just in case.

  3. kutabs
    • 2 years ago
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    Well the impedance of inductor is omega*L, for capacitor it is 1/(omega*C). now \[\tan\]=( (X_L - X_C)/R) and \[\cos\]= R/ underroot((X_L-X_C)^2+R^2). Pavg=Vrms*Irms*cos(phi) Now , you have two unknowns (L and C) and the above two equations. Solve it to get the result.

  4. kutabs
    • 2 years ago
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    PS: Vrms= Vmax/underroot(2), Irms=Imax/underroot(2)

  5. brinethery
    • 2 years ago
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    underroot? Do you mean square root?

  6. brinethery
    • 2 years ago
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    Thank you for answering my question, btw. I really appreciate it.

  7. kutabs
    • 2 years ago
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    Yeah, square root. :P And those should be tan(phi) and cos(phi). It's no big a deal, I liked doing it! :D

  8. brinethery
    • 2 years ago
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    Okay I'll work through it and see what I come up with. Smartphysics can be really frustrating! :-P. I liked the mechanics aspect of physics, but electricity and magnetism just is not my thing.

  9. kutabs
    • 2 years ago
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    Okay. I love mechanics! It's so much fun. Electrostatics and magnetics is a lot like mechanics though. But I find optics horrible! I really suck at both geometrical and ray optics. Meanwhile, is Smartphysics a physics forum? I'm googling it now.

  10. brinethery
    • 2 years ago
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    Lol, no. It's the most horrible thing ever invented. It's a site where you submit homework.

  11. kutabs
    • 2 years ago
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    What! You mean to say the internet is used for ACTUAL work! That's crazy talk! I thought it was all about facebook.com and snog.com Hahaha. :P

  12. brinethery
    • 2 years ago
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    No, I mean the questions that are asked. I feel like I'm not learning anything in the class.

  13. brinethery
    • 2 years ago
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    Okay I must be a complete idiot. There are the first two equations you gave: Z = (omega*L) and Z = 1/(omega*C) I guess I'm just missing something, but why do we need the other three equations; the one for tan(phi), cos(phi), and P_avg?

  14. kutabs
    • 2 years ago
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    |dw:1369760684756:dw| Well it works like this: To calculate X_L and X_C you need to find omega*L and 1/omega*C respectively. But X_L andX_C shall be unknown too since C and L are unknown. tan (phi) is given and using X_L and X_C you can form 1 euation containing C and L. Now calculate cos(phi) (equal to= P/(Vrms*Irms) )and cos(phi)=R/ squareroot((X_L-X_C)^2+R^2) <<From the Pythagoras' theorem in the triangle shown. PS: Someone with 300 medals is not an idiot. :)

  15. kutabs
    • 2 years ago
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    In the triangle the angle should be phi, in place of theta.

  16. brinethery
    • 2 years ago
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    So we don't use the "L" value that was given in the description of the problem?

  17. kutabs
    • 2 years ago
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    Oh, I'm so sorry I never saw your reply. :( Of course we do. For calculating the X_L we need it. X_L=omega*L. Similarly, X_C=1/(omega*C) All the values shall be required.

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