Voltage drop

A 3 phase 12 kV feeder supplies a load of 30 A at 0.8 power factor lagging. The line impedance from the 12 kV source to the load is 5 + j15 ohm. The approximate voltage drop (V) per phase is given by:

(A)  1170

(B)   195

(C)  780

(D)  390

D

For a three phase circuit, the voltage drop per phase is given by:

VD = I (RCosφ + XSinφ) = 30(5*0.8 + 15*0.6) = 390V

Why isn't answer (5+15j) * 30 < -36.86 = 474.34volts?  All they are asking for is voltage drop per phase. The voltage drop at the load shouldn't matter. 

While doing VD problems, my instinct will say to use (5+15j) * 30 < -36.86 = 474.34volts which i think is the most accurate one. And choose the most near one option. I know the difference is too much but no other way. If got extra time in exam to check then will apply VD = I (RCosφ + XSinφ) at the end. I wish in actual exam we don't have to see both answers as option choice and hope the answer is near to our calculation.

 What i conclude to do is if the question asks for NEC. use NEC: VD = I (RCosφ + XSinφ). If not then use Ohms law.

I'm interested in knowing the logic though. The two methods would take the same time, it just that the method given with the problem is difficult to remember compare to ohms law. I understand the voltage drop per NEC, those types of problems are worded differently. They give you conductor size and pf of the load etc and and if the pf of the load is different than 0.85 table values then you use the formula VD = I (RCosφ + XSinφ)

In this case they have given the pf of the amps draw but they are not using it. You would think complex impedance times Complex current should give you everything you need including proper angle if needed.

May be some like Zach can throw some light here.   

roy167 said:

A 3 phase 12 kV feeder supplies a load of 30 A at 0.8 power factor lagging. The line impedance from the 12 kV source to the load is 5 + j15 ohm. The approximate voltage drop (V) per phase is given by:

(A)  1170

(B)   195

(C)  780

(D)  390

D

For a three phase circuit, the voltage drop per phase is given by:

VD = I (RCosφ + XSinφ) = 30(5*0.8 + 15*0.6) = 390V

Why isn't answer (5+15j) * 30 < -36.86 = 474.34volts?  All they are asking for is voltage drop per phase. The voltage drop at the load shouldn't matter. 

Click to expand...

Hi @roy167, where did you find this problem? It appears to me like the author made a mistake if the red text in your original post is the author's solution. The problem already gives the line impedance in ohm's (5 + j15 ohm), and not the conductor characteristics plus circuit length. The following formula is only used to calculate the effective impedance from the R and XL NEC® Chapter 9 Table 9 values:

     RCosφ + XSinφ

We are not given the conductor characteristics and the circuit length, so we would not be calculating effective impedance. 

I get the same answer as you, here is how I work it out:

     V_drop_1ø = IL•Zline

          ϴi = cos-1(0.80)
          ϴi = -37º when Vln is at a reference of 0º

     V_drop_1ø = (30A<-37º)/(5 + j15Ω)
     V_drop_1ø = 474.3V<35º

474 Volts, rounded to the nearest volt, is dropped across each phase conductor assuming a balanced system. 

I think the author made a mistake. 

roy167 said:

I'm interested in knowing the logic though. The two methods would take the same time, it just that the method given with the problem is difficult to remember compare to ohms law. I understand the voltage drop per NEC, those types of problems are worded differently. They give you conductor size and pf of the load etc and and if the pf of the load is different than 0.85 table values then you use the formula VD = I (RCosφ + XSinφ)

In this case they have given the pf of the amps draw but they are not using it. You would think complex impedance times Complex current should give you everything you need including proper angle if needed.

May be some like Zach can throw some light here.   

Click to expand...

You know I can't resist 

Zach,

What was in red font was by the author. Not sure about the source. We both get the same answer.  

  V_drop_1ø = IL/Zline

          ϴi = cos-1(0.80)
          ϴi = -37º when Vln is at a reference of 0º

     V_drop_1ø = (30A<-37º)/(5 + j15Ω)
     V_drop_1ø = 474.3V<35º

I think this should be V_drop_1ø = IL * Zline and not division

roy167 said:

Zach,

What was in red font was by the author. Not sure about the source. We both get the same answer.  

  V_drop_1ø = IL/Zline

          ϴi = cos-1(0.80)
          ϴi = -37º when Vln is at a reference of 0º

     V_drop_1ø = (30A<-37º)/(5 + j15Ω)
     V_drop_1ø = 474.3V<35º

I think this should be V_drop_1ø = IL * Zline and not division

Click to expand...

Good catch @roy167 , thanks. I've corrected my original post.