T
tobeeepe
I am posting this as a correction to what I said before, and also for anyone who who needs the proof using the full expression:
Question:
Feeder supplying a balanced load. For a fixed current of 50A, which of the following statements concerning the phase-to neutral voltage drop at the receiving end of the feeder is most correct?
(A) Voltage drop is relatively independent of the power factor.
( Voltage drop will be largest for a unity load power factor.
© Voltage drop will be largest for a zero load power factor
(D) Voltage drop will be larger for a lagging load power factor of 0.707 than either a unity load power factor or a zero load power factor.
The answer given in the book is (D), and proves this using the approximate equation: R*cosphi + X* sin phi
The complete expression for the voltage drop assuming a LAGGING load pf phi is,
Phasor (Vs) - Phasor (Vr) = |I| {(R*cos phi + X* sin phi)+j (X*cos phi - R* sin phi)}
R = X = 0.05 and this simplifies the equation to
Voltage drop Vd = 50*0.05 (cos phi + sin phi) + j (cos phi - sin phi)
For phi = 0, Vd = 2.5 (1+j), |Vd| = 2.5
For phi = 90, Vd = 2.5 (1-j), |Vd| = 2.5
For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535
You will get the same answers if you use the approximate equation too. I was not just comfortable with making the decison based on the approximate equation. That's all.
Question:
Feeder supplying a balanced load. For a fixed current of 50A, which of the following statements concerning the phase-to neutral voltage drop at the receiving end of the feeder is most correct?
(A) Voltage drop is relatively independent of the power factor.
( Voltage drop will be largest for a unity load power factor.
© Voltage drop will be largest for a zero load power factor
(D) Voltage drop will be larger for a lagging load power factor of 0.707 than either a unity load power factor or a zero load power factor.
The answer given in the book is (D), and proves this using the approximate equation: R*cosphi + X* sin phi
The complete expression for the voltage drop assuming a LAGGING load pf phi is,
Phasor (Vs) - Phasor (Vr) = |I| {(R*cos phi + X* sin phi)+j (X*cos phi - R* sin phi)}
R = X = 0.05 and this simplifies the equation to
Voltage drop Vd = 50*0.05 (cos phi + sin phi) + j (cos phi - sin phi)
For phi = 0, Vd = 2.5 (1+j), |Vd| = 2.5
For phi = 90, Vd = 2.5 (1-j), |Vd| = 2.5
For phi = 45, Vd = 2.5 (2*0.707), |Vd| = 3.535
You will get the same answers if you use the approximate equation too. I was not just comfortable with making the decison based on the approximate equation. That's all.