Homework Help: Past paper question

1. Nov 10, 2006

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 dunno if what im 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 dont 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 cant 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 ive 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 im right!

Cheers for help (please dont flame my stupidity if i made a stupid mistake, first time with latex too)
-G

Last edited: Nov 10, 2006
2. Nov 10, 2006

dextercioby

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

Daniel.

3. Nov 10, 2006

FunkyDwarf

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: Nov 10, 2006
4. Nov 10, 2006

dextercioby

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

Daniel.

5. Nov 10, 2006

FunkyDwarf

yah its not saving the changes

EDIT... ah there we go

6. Nov 12, 2006

FunkyDwarf

Ok i found out whats 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?

Last edited: Nov 12, 2006