- #1

FunkyDwarf

- 489

- 0

**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

Last edited: