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Old 18th October 2010, 12:51 AM   #1
wdcw is offline wdcw  Australia
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Default True Power-- Math Question

I have attached a picture of a parallel RLC circuit that I am trying to calculate the Power Factor.

Branch 1 is a Resistor of 10 Ohms
Branch 2 is an Inductor of .02H & has 4 Ohms DC resistance.
Branch 3 is a 6 0hm Resistor with a 200uF Capacitor in Series.

I have calculated the Branch Impedances, Inductive Reactance, Capacitive Reactance, Currents etc & also the Apparent Power kVA.

I am trying to work out how to get the Power Factor of this circuit.

PF = True Power/ Apparent Power P/S =?
I am not sure how to Find the True Power?

I have looked at different formula's but none of them seem to work for my pea sized brain, can someone help me out with this, I thought that the equation for Apparent power was for True power but obviously not?.

Cheers.
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File Type: gif True Power.gif (18.1 KB, 94 views)

Last edited by wdcw; 18th October 2010 at 12:57 AM.
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Old 18th October 2010, 02:08 AM   #2
Irakli is offline Irakli  United States
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true power is actual consumed power. In your circuit only resistances are dissipating (consuming) power. So, calculate power dissipation by R1, R2, R3, sum them up and you get a true power (calculate currents through resistors and use formula I^2*R)

Apparent Power = total current through circuit * applied Voltage (do not forget you have reactive elements in the circuit)

PF=True Power/apparent power
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Old 18th October 2010, 02:19 AM   #3
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I would use AC analysis (not transient) as its a linear network. This then will give you the current from the source and its phase. Then true power is calculated by taking the cosine of the phase angle (the PF) and multiplying this by the apparent power. I think, I haven't tried it....
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Old 18th October 2010, 03:33 AM   #4
wdcw is offline wdcw  Australia
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Thanks for the replies,


Irakli,

I am a little confused with this.

So I just use the Resistance values of R1, R2, R3 & not the Z Impedance total value of each branch?

I calculated things using the Parallel resistance calculation & then the current for each branch but things don't seem correct.

So do I just Add R1 + R2 + R3 which gives 20 Ohms?

230V/20R = 11.5A

True Power = I^2 * R = 132.25 * 20 = 2.645kW

PF = P/S True Power/ Apparent Power
= 2.645kW/6.624kVA
PF = .4

Is this correct, I thought I may have to use the Z impedance total for each brach in parallel & then work things out from there?
I have to say I am a little confused.

Cheers

Last edited by wdcw; 18th October 2010 at 03:38 AM.
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Old 18th October 2010, 03:38 AM   #5
Irakli is offline Irakli  United States
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no, not like that.

Calculate the power dissipated in each resistor and add them up afterwards.
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Old 18th October 2010, 03:55 AM   #6
wdcw is offline wdcw  Australia
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Hi Irakli,

I did that before but something doesn't seem right.

R1 = 230/10 =23A
R2 = 230/4 =57.5A
R3 = 230/6 =38.3A

IT = 118A

Where to from here as things don't seem right?

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Old 18th October 2010, 04:34 AM   #7
Irakli is offline Irakli  United States
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OK, let' do it by book, with some compex numbers:

1. Let's calculate total complex impedance of the circuit.

1/Ztot=1/Z1+1/Z2+1/Z3

Z1=R1=10ohm
Z2=R2+j*2*Pi*f*L=4+J*6.28*50*0.02=4+j*6.28 (4 is a real part and 6.28 is imaginary part of the Z2 impedance)
Z3=R3+j/(2*Pi*f*C)=6-j*15.92 (please doublecheck the math yourself).

Thus, current I=V/Z=V*(1/Z)=230*(1/Z)=44.36-j*13.61
As you see current has real and imaginary parts.
Now we need to calculate angle between voltage and current.

tg(a)=Im(I)/Re(I)=-13.61/44.36=-0.31

"a" is an angle how much current is shifted vs. voltage.

Power Factor is simply PF=|cos(a)|. Knowing tg(a) you find |cos(a)|

|cos(a)|=sqrt(1/(1+tg(a)*tg(a)))=0.96 (again, please check math to be sure)
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Old 18th October 2010, 04:48 AM   #8
wdcw is offline wdcw  Australia
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Hi Irakli,

I'm sure your math is correct & thank you very much for that, unfortunately I do not understand complex numbers at this stage as I am only learning.

Could you possibly help me with standard math numbers.

Thank You

Cheers
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Old 18th October 2010, 05:02 AM   #9
Irakli is offline Irakli  United States
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Here is another way to calculate;

1. determine |Z1|, |Z2| and |Z3|

e.g. |Z2|=sqrt (R2*R2+wL*wL) where w=2*Pi*f

2. calculate currents through each branch (your simulator already shows them)

|I1|=23A, |I2|=30.9 , |I3|=13.52

calculate power dissipated by ACTIVE elements (resistors) in each branch

true power (power which is converted into heat) P=P1+P2+P3=R1*I1^2+R2*I2^2+R3*I3^2=5290+3812+1097= 10199=10.20KW (thats a lot of power,lol)

Now we need to calculate apparent power which is S=|V|*|Itot|=230*|Itot|

Itot is not simply a sum of |I1|+|I2|+|I3|. I think thats where you have a problem. You need to sum currents as vectors and then find the magnitude.

as I mentioned in previous post, Itot=44.36-j*13.61. So, |itot|=sqrt(44.36^2+13.61^2)=46.06A (not 118A as in your calculations)

Apparent power S=230*46.06=10594W

power Factpr PF=10.20/10.59=0.96

Have fun
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Old 18th October 2010, 05:07 AM   #10
wdcw is offline wdcw  Australia
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Thanks Irakli,

I now see where I was going wrong & you were correct where you mentioned it:

"Itot is not simply a sum of |I1|+|I2|+|I3|. I think thats where you have a problem. You need to sum currents as vectors and then find the magnitude.

as I mentioned in previous post, Itot=44.36-j*13.61. So, |itot|=sqrt(44.36^2+13.61^2)=46.06A (not 118A as in your calculations)"

This is where I was stumbling but now I see how you have worked it out.

Thank You very much for your time, greatly appreciated!

Cheers

Last edited by wdcw; 18th October 2010 at 05:10 AM.
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