Exam prep practice question18 unit 1

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vjrtapia

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D=(Cmil x Evd)/(2 x12.90 ohms x 16A) Whats the maximum distance that 2, 14 AWG conductors can be run if they carry 16 A and the maximum allowable voltage drop is 10V ?, im not sure how we arrive at 12.90 ohms in the equation?, Thanks
 
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david luchini

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D=(Cmil x Evd)/(2 x12.90 ohms x 16A) Whats the maximum distance that 2, 14 AWG conductors can be run if they carry 16 A and the maximum allowable voltage drop is 10V ?, im not sure how we arrive at 12.90 ohms in the equation?, Thanks

12.9 isn't ohms, exactly. It's a constant equal to (ohms/1000')*cmils. It's generally seen as 12.9 for copper conductors and 21.2 for aluminum conductors.
 

Carultch

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D=(Cmil x Evd)/(2 x12.90 ohms x 16A) Whats the maximum distance that 2, 14 AWG conductors can be run if they carry 16 A and the maximum allowable voltage drop is 10V ?, im not sure how we arrive at 12.90 ohms in the equation?, Thanks

It comes from the resistivity of copper at (typically) 75C, with the dimensional units translated from "square meters per meter" to circular mil per 1000 ft.
 
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