# Power Triangle

Discussion in 'Homework Help' started by lisa3412, Sep 12, 2014.

1. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
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hello!

this is a question based on Power Triangle.

My answer for Vs= 835.22 @ angle -130.

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I probably did it wrong, but I got Vs= 368.53 volts @ 45°.

3. ### subtech AAC Fanatic!

Nov 21, 2006
123
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Lisa, I think that is off by quite a bit.
I would try figuring all the quantities related to the 30KVA load @ 240 Volts. (VL)
Then do the same with the 40KW load.
Combine the impedances (which are in parallel) of these two loads and insert that value in series with with the 0.1 ohm (resistive) and +j0.1 ohm (X). Then calculate the series circuit values.

Hint:
You can get a ballpark value for Vs by neglecting the angles and just using simple ohmic values.

Last edited: Sep 12, 2014
4. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
0
what I did was that:

We know that P=V*I*Power Factor
V=240 @ angle=0
So we can find out I2.

Is this logic right?
and since Power Factor is lagging, won't Current Angle be negative? that is -cos^-1(0.795)?

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I noticed that it is 240 volts rms. I don't know off hand, but... Should you be using RMS value in your calculations or the Peak (Vpeak) value?

240 volts rms is actually 339.4 volts peak. So if you use 240 volts rms, you will get one set of answers. If you use 339.4 volts peak, you get another set of answers.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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One must use the RMS value in this instance.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Yes you are correct.

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Ok, thank you, then my earlier solution is wrong.

9. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
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So, t_n_k, is my solution correct? please can you verify. .I am having a tough time here!!!

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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One step at a time.
What values for the two individual load currents did you obtain?

11. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
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OK! for I2 i got = 209.643 at an angle of -37.34 degrees

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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That's correct.

And the other.....

13. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
0
154.32 at an angle -25.84

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Wrong magnitude - correct phase angle. How did you work out the magnitude part?

15. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
0
Complex Power S= 30,000 VA
Hence P= 30,000/0.9

Current I1= (30,000/0.9)/(240V*0.9)

16. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
0
Sorry, Magnitude comes out to be 125

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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That's incorrect.

The current magnitude term is simply the total load VA (apparent power) divided by 240V.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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OK we are back in step.

Now work out the total current from the source - i.e. I1 + I2

19. ### lisa3412 Thread Starter Member

Sep 6, 2014
37
0
Ok. So, that comes out to be 125. Then Is= I1+I2
Total Z= o.1+j0.1

Vs= I/Z

Is this correct?

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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You've lost me.

The total current is I1+I2 or 333A at about -33 degrees.

The the line drop is Is*Zline. Add the line drop to 240V to get Vs.