# NCEES sample exam problem 133



## Ilan (Aug 5, 2007)

A Single-phase transformer serves a 7.2kW resistive load located 200' from the XFR. Each conductor from the XFR to the load has an impedance of (0.78+j0.052) ohm/1000'. If the voltage at the load terminals is 240V, the voltage magnitude (V) at the XFR secondary is most nearly:

I got 245V assuming one conductor from XFR to the load. The solution assumes two conductors and the answer is 250V. :brickwall:

Could some one point out what I am missing here...

Thanks,

Ilan.


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## Flyer_PE (Aug 5, 2007)

Ilan said:


> A Single-phase transformer serves a 7.2kW resistive load located 200' from the XFR. Each conductor from the XFR to the load has an impedance of (0.78+j0.052) ohm/1000'. If the voltage at the load terminals is 240V, the voltage magnitude (V) at the XFR secondary is most nearly:
> I got 245V assuming one conductor from XFR to the load. The solution assumes two conductors and the answer is 250V. :brickwall:
> 
> Could some one point out what I am missing here...
> ...


My first impression, since the you came up with about half of the voltage drop for the correct answer, is that you only used half of the cable. Since the load is 200 ft away, there is actually 400 ft of cable to make the round trip.

I load = 7.2kW/240 V = 30 Amps

Z cable = (2 * 200ft)(0.78+j.052)/1000 ohm = 0.30 + j0.021 ohm

Vcable = 30*(0.30+j0.021) = (9 + j0.624) volts.

The source voltage ends up being 249 and change.

Jim


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## jdd18vm (Aug 5, 2007)

IFR_Pilot said:


> My first impression, since the you came up with about half of the voltage drop for the correct answer, is that you only used half of the cable. Since the load is 200 ft away, there is actually 400 ft of cable to make the round trip.
> I load = 7.2kW/240 V = 30 Amps
> 
> Z cable = (2 * 200ft)(0.78+j.052)/1000 ohm = 0.30 + j0.021 ohm
> ...


Thats exactly what I did wrong too. Using the 200' came up with 4.68 Volts. Slightly different approach:

(400ft/1000ft)(.78+j0.52)=.312+j208 ohms * 30 Amps=9.36 Volts

John


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## Ilan (Aug 6, 2007)

Thanks IFR_Pilot and jdd18vm for the responses.

I understand that the second conductor is the return circuit for single phase circuits. My assumption was that the impedance was given for the entire cable segment. How do we know that how many conductors are there for each circuit/phase.

Thanks,

Ilan.


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## grover (Aug 6, 2007)

Ilan said:


> Thanks IFR_Pilot and jdd18vm for the responses.
> I understand that the second conductor is the return circuit for single phase circuits. My assumption was that the impedance was given for the entire cable segment. How do we know that how many conductors are there for each circuit/phase.
> 
> Thanks,
> ...


Whenever they list the conductor impedence per 1000' of length, they're talking 1-way distance. You just have to know whether or not your system has any current on the neutral that you might have to double (or consider partial) voltage drop for. For balanced 3-phase and 240V 1-phase (EG, just about any balanced shared neutral), there's negligible current (and voltage drop) on the neutral, and voltage drop only has to be calculated 1-way. For simple 1-phase with full current on the neutral, it's double the drop. For asymmatric fault conditions (EG, extremely unbalanced), it's almost always double the drop, or close to it.

I don't think they ask about it on the PE exam, but harmonic current on a 3-phase neutral can throw a real wrench into the calculations, too!


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## Flyer_PE (Aug 6, 2007)

grover said:


> I don't think they ask about it on the PE exam, but harmonic current on a 3-phase neutral can throw a real wrench into the calculations, too!



The only thing I can think that they may hit you with about harmonics in the neutral is that they will typically be 3rd order harmonics (i.e. 180Hz, 540HZ etc.) and that they are caused by non-linear loads. I can't imagine much more depth than that for a six minute sized problem.

Jim


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## Ilan (Aug 6, 2007)

Thanks grover for the clarification.

Ilan.


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