Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

PS: Vrms= Vmax/underroot(2), Irms=Imax/underroot(2)

Answer 4

underroot? Do you mean square root?

Answer 5

Thank you for answering my question, btw. I really appreciate it.

Answer 6

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

Answer 7

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.

Answer 8

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.

Answer 9

Lol, no. It's the most horrible thing ever invented. It's a site where you submit homework.

Answer 10

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

Answer 11

No, I mean the questions that are asked. I feel like I'm not learning anything in the class.

Answer 12

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?

Answer 13

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. :)

Answer 14

In the triangle the angle should be phi, in place of theta.

Answer 15

So we don't use the "L" value that was given in the description of the problem?

Answer 16

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.