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FCC Test

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hutch1980

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I just joined here. I've been looking for a place to get some help on some things. I'm trying to get my GROL test done and there are some things on element 3 that I don't know how to do. I can memorize the answers but I would like to actually know how to get that answer. One problem is with power factors. example: What does the power factor equal in an R-L circuit having a 60 degree phase angle between the voltage and the current? The answer is 0.5. The other section I was having problems with was the impedance networks. Example: What is the impedance of a network composed of a 100 picofarad capacitor in parallel with a 4000 ohm resistor, at 500 Khz? specify your answer in polar coordinates. The answer is 2490 ohms,/-51.5 degrees. No clue how to get that though. Any help with this would be greatly appreciated. Thanks!
 
The first one is easy enough. PF = cos (the angle)

I'm not getting 2490 ohms for the second one.
 
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I thought the power factor had some thing to do with the cosine but I can't get that answer no matter which way I do it. Did you get it to work out on a calculator? The polar coordinate question I have no clue on. questions 3-17B1 through 3-17B6 on element 3 are all questions like this. I just know the answer because the FCC website gives the questions and answers. I just got the questions I printed out from there and verified thats their answer. 2490 ohms,/-51.5 degrees. I've asked every one I know into electronics including an electrical engineer. He is looking into it because he says he remembers covering it but hasn't used it in years. I was hoping I might get lucky on here.
 
Calculator: put in 60, press "cos".
There are still a lot of people besides me here. You aren't stopped yet.
 
To combine parallel impedances you use the formula 1/Zp = 1/Z1 + 1/Z2+ ...+1/Zn. This has to be done in vector (phasor) format so the algebra can get rather messy. Here is a site that shows the equations and has a calculator to generate the answer.

If you plug in the value for the resistor and the capacitor impedance at 500kHz you will see the answer for the parallel calculation (for the capacitor leave the real impedance at 0, and for the resistor leave the imaginary [j] impedance at 0). Edit: (You enter the capacitor impedance with a minus sign to get the correct sign on the phase angle.
 
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#12, For some reason when I do the cosine on my iphone it was giving me a different answer. I grabbed a calculator and it came out right. I had tried on my phone before but kept getting a different answer. As for that second problem i'm going to look at the above site and see if I can figure it out now. Thanks for the help.
 
What answer does your iphone give you for the cosine of 60 degrees? Edit: Is it .588?
 
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I'm sure I'm doing something dumb but it's giving me -0.952.......... Not sure where the mistake is. I just started using my calculator for all of those problems and it's working fine now. I'm still having some problems on the polar coordinate questions but I'm working on those.
 
Your problem with the first question is that your iphone is calculation cos(60) in radians, not degrees.

As for the second question, if you still have not gotten it

the capacitors reactance is 1/(2*pi*f*c*j) which is -3183.098j Ω or 3183.098 angle 90 Ω

then you add them in parallel: (1/4000 + 1/-3183.098j)^-1 = 1550.9065 -1948.9266j Ω = 2490.707 angle - 51.48811 degrees
 
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