# Solve Unresolved Past Paper Question: Physics 11/11

• FunkyDwarf
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.
FunkyDwarf

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
$$L= 5 \\ R = 4$$ and were doing this at mains freq of $$\omega = 50Hz$$

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:
$$Z = \frac{1}{(R+\omega Li)} - \omega Ci$$ 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.
$$\frac{1}{(R+ \omega Li)} * \frac{(R- \omega Li)}{(R-\omega Li)} = \frac{R- \omega Li}{((R^2)+(\omega L)^2)}$$

Then putting it back in and making it one big happy fraction
$$\frac{(R-\omega Li - (\omega Ci)(R^2 +(\omega L)^2)}{(R^2 +(\omega L)^2)} = \frac{(R-\omega Li - \omega CR^2 i - \omega ^3 L^2Ci)}{(R^2 +(\omega L)^2)}$$
Now just taking the imaginary bit and equating to zero (getting rid of denominator for obvious reasons)
$$(-\omega L-\omega CR^2-(\omega )^3 L^2C = 0$$ 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

Last edited:
The [tex] tag closes with [ / tex] (without the spaces of course)

Daniel.

yah just fixing that now cheers :D tryin to get omega to work too :S

hmm some of the changes arent saving. anyway

Last edited:
It's \omega and the same for the other symbols and functions.

Daniel.

yah its not saving the changesEDIT... ah there we go

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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