hello!

this is a question based on Power Triangle.

My answer for Vs= 835.22 @ angle -130.

Can somebody please verify this?

 

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

 

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.

 

what I did was that:

For the 40kw load:
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)?

 

what I did was that:

For the 40kw load:
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)?

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.

 

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.

One must use the RMS value in this instance.

 

what I did was that:

For the 40kw load:
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)?

Yes you are correct.

 

One must use the RMS value in this instance.

Ok, thank you, then my earlier solution is wrong.

 

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

 

One step at a time.
What values for the two individual load currents did you obtain?

 

OK! for I2 i got = 209.643 at an angle of -37.34 degrees

 

That's correct.

And the other.....

 

154.32 at an angle -25.84

 

Wrong magnitude - correct phase angle. How did you work out the magnitude part?

 

Complex Power S= 30,000 VA
Hence P= 30,000/0.9

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

 

Sorry, Magnitude comes out to be 125

 

That's incorrect.

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

 

OK we are back in step.

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

 

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

Vs= I/Z

Is this correct?

 

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.