NCEES exam problem 129. I am now freaking out about voltage drop

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NEC - Table 9 - Note 2

Ze (for other than .85 PF) = R*PF+Xl*sin(cos^-1(PF))

R(1000ft)=.029

R(250ft)=.029*(250/1000) = .00725

X(1000ft)=.048

X(250ft)=.048*(250/1000)=.012

PF=.8

Ze = .00725*.8+.012*sin(cos^-1(.8))

Ze = .013

Vd = I*Ze = 400*.013 = 5.2V

V = Vs(L-N)-Vd

V=277-5.2 = 271.8

V(LL) = V(LN)*1.732

V(LL) = 271.8*1.732 = 470.77 ≈ 471V
For the bolded part why did you drop to i imaginary term for .012*sin(cos^-1(.8))? with it I get a completely different answer

 
For the bolded part why did you drop to i imaginary term for .012*sin(cos^-1(.8))? with it I get a completely different answer
The formula at the bottom of the table in the NEC and in the NCEES Handbook don't have the j term. The X is just the value of the reactance. You don't include jX, just X. In the end, you just get a number.

 
The formula at the bottom of the table in the NEC and in the NCEES Handbook don't have the j term. The X is just the value of the reactance. You don't include jX, just X. In the end, you just get a number.
Thanks, all this time the numbers have been working out for me except for this question. I feel lucky I caught this mistake before it was too late lol

 
Sorry to revive an older topic but looking at the NEC handbook example we also do not use the complex values  if the PF=.85, is this correct?

So just use magnitudes when doing voltage drops for either .85 PF or =/= .85 PF?

 

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