As there has not been any activity in here lately...

[D Skelton]

As it is a parallel circuit you need to use current to work out power factor. You have the resistive current (34A), work out the reactive current in the same fashion. Current through the inductor and current through the capacitors.

 

[Simonslimline]

I may have gone quite wrong here ...

xl current 99.25A

xc current 18.72A

 

[D Skelton]

Your xc current is wrong. Remember, xc = 1 / (c x pi x 2 x Hz)

 

[Simonslimline]

xc current 56.2A , making the power factor of the circuit 0.62.

 

[D Skelton]

Good stuff, so you have that value. Now work out what the reactive current would be if the circuit had a hypothetical power factor of 0.9.

 

[Simonslimline]

Not 100% sure on how to do this, but am i close with 38A

 

[D Skelton]

What was your reactive current with a circuit P.F. of 0.62? You must have a value of 42.69A?

If you divide your resistive current by 0.62 what do you get? Well take this logic and divide your resistive current by 0.90 and with this figure, you can then work out your reactive current if the circuit had a P.F. of 0.90 yes.

You are off the mark with 38A but think it through again.

 

[Simonslimline]

IxL - Ixc = 42.64A ( slightly different value, i suppose to do with rounding off each calculation)

Reactive current with a power factor of 0.90 = 37.77A hopefully!

 

[D Skelton]

Nope lol.

Right, this is the sum you need to do:

34A / 0.9 = 37.78A

then...

sqrt of (37.78[SUP]2[/SUP] - 34[SUP]2[/SUP])

... This gives you your reactive current with a circuit power factor of 0.9

Work this out then we will move on to the next step

 

[Simonslimline]

Ok so i get a reactive current of 16.47A. so √(34²+16.47²) = 37.78A



34÷37.78=0.90

I appreciate your patience by the way, maths never was my strong point.

 

[D Skelton]

Bingo, now we have to scrub the difference between the reactive current of 42.69A at a power factor of 0.62 and the hypothetical reactive current of 16.47A at a power factor 0.90.

Subtract the 16.47A from the 42.69A, you then have a value of reactance to reduce. Is it inductive or capacative? Is it an inductor you need to place at 'A' to correct a leading current or a capacitor to place at 'A' to correct a lagging current?

 

[Simonslimline]

Just to be sure , the 42.69a was worked out by subtracting xc from xl?


If so then the (42.69-16.47)=26.22A is inductive and Point A needs a capacitor.

 

[D Skelton]

Just to be sure , the 42.69a was worked out by subtracting xc from xl?

It was worked out from subtracting I[SUB]Xc[/SUB] from I[SUB]Xl[/SUB]

I[SUB]Xc[/SUB] being the sum of current passing through the capacitors and I[SUB]Xl[/SUB] being the sum of current passing through the inductor

 

[D Skelton]

Good, so you've worked out the current to rectify, you have worked out that you need to rectify it with a capacitor. What next? What size capacitor do you need and how do you work this out?

 

[Simonslimline]

Sorry yeah thats what i meant. I have edited my post. The remaining value 26.22A is inductive and point A needs a capacitor.