Solve Unresolved Past Paper Question: Physics 11/11

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In summary, Daniel was trying to find the power factor for a parallel LCR circuit, but he was having trouble because he made a mistake in the equation. He fixed the problem by remembering that -iwC is the correct symbol for the power factor.
  • #1
FunkyDwarf
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Past paper question - please help, still unresolved! (11/11)

Hey guys

Im studyin for my physics end of year exam (only first year so its prob easy peasy for you guys) but unfortunately i have no answer key as yet so i don't know if what I am doing is right. I get a plausable answer but i just wanted to be sure incase i never get the answers from lecturers (im badgering em heaps don't worry)

Ok this might take a while to set out in latex thing but ill give it a go

The question is about power factor of a parallel LCR ciruit but in a nutshell i need to find C such that the imaginary part of the complex impedance vanishes (ie angle is zero)

Were given that
[tex]L= 5 \\
R = 4 [/tex] and were doing this at mains freq of [tex] \omega = 50Hz[/tex]

Unfortunately i can't upload the image right now but all you need to know is that this parallel circuit has 2 arms, one just has the capacitor the other the resistor and inductor in series.

So, working out impedances:
[tex] Z = \frac{1}{(R+\omega Li)} - \omega Ci[/tex] Now I've always used minus i for the capacitor part because of the angle difference in the phasor, i assume i do that here.

Next i try to get the imaginary parts off the bottom.
[tex] \frac{1}{(R+ \omega Li)} * \frac{(R- \omega Li)}{(R-\omega Li)} = \frac{R- \omega Li}{((R^2)+(\omega L)^2)}[/tex]

Then putting it back in and making it one big happy fraction
[tex]\frac{(R-\omega Li - (\omega Ci)(R^2 +(\omega L)^2)}{(R^2 +(\omega L)^2)}

= \frac{(R-\omega Li - \omega CR^2 i - \omega ^3 L^2Ci)}{(R^2 +(\omega L)^2)}
[/tex]
Now just taking the imaginary bit and equating to zero (getting rid of denominator for obvious reasons)
[tex]
(-\omega L-\omega CR^2-(\omega )^3 L^2C = 0
[/tex] Then just shoving in those numbers and ignoring the minus sign (which has me rather worried) i get C = 7.9 x 10^-5 farad.

Hope this makes sense and I am right!

Cheers for help (please don't flame my stupidity if i made a stupid mistake, first time with latex too)
-G
 
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  • #2
The [tex] tag closes with [ / tex] (without the spaces of course)

Daniel.
 
  • #3
yah just fixing that now cheers :D tryin to get omega to work too :S

hmm some of the changes arent saving. anyway
 
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  • #4
It's \omega and the same for the other symbols and functions.

Daniel.
 
  • #5
yah its not saving the changesEDIT... ah there we go
 
  • #6
Ok i found out what's wrong, its the fact that i used -iwC instead of +iwC, which screws it all up. However i always thought that due to phasors and phase differences etc your supposed to use the minus. Anyone?
 
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Solving unresolved past paper questions in Physics helps students to practice and apply their knowledge of the subject, identify any areas of weakness, and improve their problem-solving skills. It also gives them a better understanding of the types of questions that may appear on their exams.

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