practice test questions

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royta

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I took the Mike Holt 25 question practice test HERE. There are a few questions I got wrong, but on many, I am able to figure out why I got them wrong. This thread is for those I am unable to figure out.

Q14. An existing junction box is located 65 ft from the panelboard and contains 4 THHN aluminum conductors. What size copper conductor can be used to extend this circuit 85 ft and supply a 50A, 208V load? Note: Apply the NEC recommended voltage-drop limits.

A. 8 THHN
B. 6 THHN
C. 4 THHN
D. 10 THHN

I chose B, but the answer is C. The original 65 ft length of wire only has a 3.3V voltage drop, which is 1.6% of 208V. To arrive at my answer, I simply did a voltage drop calc using 6.24V which is 3% (this is the NEC recommended voltage-drop limit, isn't it?) of 208V and using the 85 ft length. I arrived at 17572 kcmil which is #6. However, since the answer is #4, I'm guessing I should have added the 65 ft and 85 ft together for a total of 150 ft. When I do that, I arrive at a kcmil of 31009, which bumps me up to #4. Is it correct that I am to add up the original length of wire and the new length of wire for my total voltage drop distance? Thanks.

Roy

more to follow I'm sure. :)
 
im with you on the 3.3 v on alum wire and 6.4 total allowed . subtracting the 3.3 leaves 2.94 v drop for the cu wire. dividing the 2.94v by 50 a is resistance of .0588 ohms. the 2 way distance on cu is 170 ft change than to thousand ft is .170 k . divide the .0588r by .170 gives .345 ohms /k ft. # 4 copper is slightly less at .308 ohms /k ft
 
To explain simply, when you place two resistances in series, you can't rely on the voltage at the end of the first one to remain as it was, and calculate the second portion as if the end of the first is a source.

You have to start over at the source, and use the total resistance to calculate the total voltage drop. If you wish, you can then assign each portion of the total drop to its respective resistance segment.
 
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