Voltage Drop Equation

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shsweet28

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Lets say I am calculating Voltage Drop (Vd) and have a load with a lagging PF, then arcos(PF) gives the angle, I make the angle negative since the load is lagging.

So Vd=Vo-(I<-Angle)*(R+j*X).

But Vd=abs(I)*(R*cos(angle)+sin(angle).

Why do I use the positive angle in the second equation and the negative in the first?

Maybe I am missing something simple.

 
I see yet another error so I will start over.

Lets say I am calculating Voltage Drop (Vd) and have a load with a lagging PF, then arcos(PF) gives the angle, I make the angle negative since the load is lagging.

So,Vd=(I<-Angle)*(R+j*X)

Vd=abs(I)*(R*cos(angle)+X*sin(angle)

Why do I use the positive angle in the second equation and the negative in the first?

 
I found the answer to my question. Please see the image below in case you were curious.

Capture.JPG


 
Is this accurate? I am having a hard time finding anything online to back up the + for lagging, - for leading.

In the NEC Handbook it only shows Rcos(theta)+Xsin(theta) and doesn't state whether the power factor is leading or lagging.

 
I have the same confusion...but i am planning to use only + because nec doesnt say anything about the sign. Can someone help?

 
If the source is a 1 phase center tap line (V LN= 110V and V L-L = 220V), which voltage would you use to calculate voltage at receiving end?

 
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