# 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