gola exam

[kmatthews]

a capacitor of 100uf is connected in parallel with a 20ohm pure resistor. the group is connected to a 200v 50hz supply. calculate the power dissapated in the combination

can someone show me the formula for this

 

[Sintra]

Will have a go at this but could be way wrong.

Impedance = C + R

Impedance (Z) = 120 ohms

Current (I) = V/Z

Current (I) = 200/120

Current (I) = 1.667

Power = VxI

Power = 200x1.667

Power = 333.4 W

 

[pushrod]

a capacitor of 100uf is connected in parallel with a 20ohm pure resistor. the group is connected to a 200v 50hz supply. calculate the power dissapated in the combination

can someone show me the formula for this

mmmh! interesting - i suspect this is a bit of a trick question. A pure capacitor will not disipate any power although it will produce a leading power factor. So you just ignore the capacitor and calculate the power.

You could use P=VxI and since you don't have a current value you could rearrange the formula by substituting in I=V/R to get P=VxV/R or P=V^2/R (V squared divided by R)

so (200x200)/20 = 2000W

(at least that is my take on it after a couple of beers!)

 

[pushrod]

Will have a go at this but could be way wrong.

Impedance = C + R

Impedance (Z) = 120 ohms

Current (I) = V/Z

Current (I) = 200/120

Current (I) = 1.667

Power = VxI

Power = 200x1.667

Power = 333.4 W

I see what you are thinking, but the impedance would have to calculated from Xc which would be 31.8ohms (from 1/(2 pi f c) and then use pythagoras to get Z to equal 37.5 ohms

then I would equal 5.3 A and power would be 1066 watts . I don't think this is the way to go though, but hey its past my bedtime

K.Matthews can you remember any possible answers?
Anyone else got any ideas?

 

[claret73]

Pushrod is pretty much there...

Here's how I did it & been doing a lot of Impedance revising...

First as Pushrod said get Xc

1/2PIfc=Xc
1/2xPIx50x100(to the minus 6)
=31.8

Then find Z (Impedance)
Z= Square Root of R sq + X sq
Z= Square Root of 20sq + 31.8 sq
Z= Square Root of 400 + 1011.24
Z= Square Root of 1411.24
Z= 37.56

Which is what Pushrod got...

P=V Squared/Z (Replacing R)
P= 200v Sq/37.56
P= 1064.96W

I hope that's on the a, b, c or d answer otherwise I'm knackered!

Hope this helps Mate. Got my 301 next Friday

 

[pushrod]

Pushrod is pretty much there...

Here's how I did it & been doing a lot of Impedance revising...

First as Pushrod said get Xc

1/2PIfc=Xc
1/2xPIx50x100(to the minus 6)
=31.8

Then find Z (Impedance)
Z= Square Root of R sq + X sq
Z= Square Root of 20sq + 31.8 sq
Z= Square Root of 400 + 1011.24
Z= Square Root of 1411.24
Z= 37.56

Which is what Pushrod got...

P=V Squared/Z (Replacing R)
P= 200v Sq/37.56
P= 1064.96W

I hope that's on the a, b, c or d answer otherwise I'm knackered!

Hope this helps Mate. Got my 301 next Friday

Sorry to rock the boat but i'm still going with my first answer of 2000watts on the principle that i think watts are dissipated only in the resistive part of the circuit and that inductances and capacitances are"watt-less". The 2nd answer was just expanding on Sintra's thinking.
Now if the question was actually asking for apparent power (VAr) that would be different - you would then be adding currents by pythagorus before doing P =IxV.

I'm not betting my life on it mind!!