Camara Power Sample Test #40

kimball79 · Apr 15, 2010

Anyone work through this question: Camara sample test, power morning #40? See attached

PE_exam.pdf

kimball79 · Apr 15, 2010

kimball79 said:
Anyone work through this question: Camara sample test, power morning #40? See attached

The solution is rather simple but it doesn't make sense.....

Zp= sqrt(R^2+Xl^2 angle (arc tan X/R))

Quantities for R and X are 3 and 4 respectively. First off any idea how they came up with the numerical vaules for R and X?

Rei · Apr 15, 2010

The question asking for PHASE impedance which is resistor=3 and inductor=4. In another word it is impedance Z = 3+j4 and they just convert it to polar form wich is 5<53 ohm.

kimball79 · Apr 15, 2010

Rei said:
The question asking for PHASE impedance which is resistor=3 and inductor=4. In another word it is impedance Z = 3+j4 and they just convert it to polar form wich is 5<53 ohm.

So how did you get R=3 and X=4? Adding up the equivalent resistance/reactance and coverting to actual values given the base of 20kVA?

Rei · Apr 15, 2010

kimball79 said:
So how did you get R=3 and X=4? Adding up the equivalent resistance/reactance and coverting to actual values given the base of 20kVA?

It's the giving value of one leg of the triangle.

I wonder if we are looking at the same question or not. The Camera's #40 that I have is a delta load with giving R=3 and XL=4 for each leg and they are asking for the phase impedance Zp. The diagram that you attached look complicated which make me think we might be talking about two different questions, but the solution that you posted is the same of what I have. It's the conversion between one leg giving impedance 3+j4 to polar form which gives 5<53 deg.