# NCEES #512 explanation



## ElecPwrPEOct11 (Oct 17, 2011)

This is the first voltage drop problem that I've seen where there are multiple sets of parallel conductors. I sat there thinking about it for a good while but eventually determined that the the impedance for this problem would be half that for a single run of conductors, but that was purely by logic.

Looking at the NCEES solution they don't explain it at all but stick a (2) in the denominator of the Z conversion formula. Can anyone explain the math behind this solution? What would happen with 3 parallel sets of conductors? Thanks!


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## DK PE (Oct 17, 2011)

ElecPwrPEOct11 said:


> This is the first voltage drop problem that I've seen where there are multiple sets of parallel conductors. I sat there thinking about it for a good while but eventually determined that the the impedance for this problem would be half that for a single run of conductors, but that was purely by logic.
> 
> Looking at the NCEES solution they don't explain it at all but stick a (2) in the denominator of the Z conversion formula. Can anyone explain the math behind this solution? What would happen with 3 parallel sets of conductors? Thanks!


It is simply R1 || R2 with two equal resistors.

For three the same thing that would happen if I drew three equal resistors in parallel and asked you to solve for an new equivalent resistor (or inductor)...


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## scotieb24 (Oct 17, 2011)

So why doesn't the power factor matter in this case. I thought Ze = R*PF + XL*sin(arccosin(PF)) for power factors other that .85


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## ElecPwrPEOct11 (Oct 18, 2011)

DK PE- that's what I eventually reasoned for the problem- two parallel impedances so they add like resistors in parallel. I was confused by the NCEES solution though. Thanks for confirming my logic.

scotieb24- The problem asks for the conductor impedance, not effective Z. The conductor doesn't care what sort of load it is attached to, it will have the same impedance regardless. You could use Ze if you were finding the voltage drop.


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## rick.conner (Oct 24, 2011)

if the conductors were rated at a different temp would the calc be the same?


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## ElecPwrPEOct11 (Oct 24, 2011)

^ Rick- The table values are accurate "only at 75oC, but are representative for 600-volt wire types operating at 60Hz."

I don't think the impedance of the wire changes appreciably when the temperature rating changes. I've never seen a voltage drop problem where the wire temp rating mattered.


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## rick.conner (Oct 24, 2011)

Thanks.

Guys I did the calculation as r1*r1/(r1+r1) and of course it is an answer though wrong according to the ncees. they just find the impedance and divide by 2.


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## vdubEE (Oct 24, 2011)

Rick,

If both impedances/resistances, that are in parallel, are equal, the effective impedance/resistance will be exactly half the original value of one.

You need to take into account the reactance as well as the resistance. You equation should be Z1*Z2/(Z1+Z2) where Z1=Z2.


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## rick.conner (Oct 24, 2011)

yea my brain is beat - sorry. forgot to * 500' and divide by 1000' which is /2.

thanks


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## brad9m (Jan 24, 2012)

I agree with what you guys are saying here about the parallel paths. That is how I finally reasoned it out. But if you look in the 2011 handbook, Calculation Example 2 for Table 9, it doesn't agree. That example uses 3 wires and they never divide by 3. On a side note, I have been staring alternately between my monitor and NCEES for the past 6 hours, so it is quite possible that I am just losing my mind.


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## brad9m (Jan 24, 2012)

brad9m said:


> I agree with what you guys are saying here about the parallel paths. That is how I finally reasoned it out. But if you look in the 2011 handbook, Calculation Example 2 for Table 9, it doesn't agree. That example uses 3 wires and they never divide by 3. On a side note, I have been staring alternately between my monitor and NCEES for the past 6 hours, so it is quite possible that I am just losing my mind.


After taking a break and re-reading everything, I think I understand it. I just misinterpreted what I was reading.


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## tkaauwai (Apr 9, 2013)

I look at this problem a different way. The problem states that the "motor circuit consists of two sets of [copper conductors] in steel conduits with a total circuit length of 500 ft." I took this to mean there are two conduits -- one can be thought of as going TO the motor and the other one is coming FROM the motor, and the "total circuit length" is 500 ft. If you assume that the total circuit is made up of just the two conduits and that the conduits are equal in length, then the distance used to find the complex impedance for the cable "*to* the motor" (and not the one coming back) would be 250/1000 (or 500/(1000*2) as they have in the answer).

Having said that, the assumption for using Table 9 is that (among other things) there are 3 single conductors in a raceway/conduit, so I personally have a hard time visualizing how anything in this problem is parallel. This problem is either worded poorly or is meant to be tricky. Either way, I hope to never see something like this actually show up on the exam.


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