Camera power practice exam - Problem 25

[joshtrevino]

Problem Statement:

The design for a transformer secondary calls for #14 gauge copper wire with a 2.7ohm/1000ft resistance according to the NEC. The max current drawn by the electronic bias power drawer is 3amps, powered from the secondary of the transformer. The allowable voltage drop is 7.6v. How far away from the transformer secondary can the drawer be located?

Solution:

R = V/I = (7.6v)/(3A) = 2.53ohms (I get this.)

The solution then states that "the distance, x, is half the distance that the current must travel in the wire (once along each conductor)."

x = (1/2) * 2.53ohms * 1000ft /2.7ohms = 468.52 ft

My question:

Why does the 1/2 come into play in the calculation of the distance?

Please help. Thanks.

 

[Dolphin P.E.]

Problem Statement:The design for a transformer secondary calls for #14 gauge copper wire with a 2.7ohm/1000ft resistance according to the NEC. The max current drawn by the electronic bias power drawer is 3amps, powered from the secondary of the transformer. The allowable voltage drop is 7.6v. How far away from the transformer secondary can the drawer be located?

Solution:

R = V/I = (7.6v)/(3A) = 2.53ohms (I get this.)

The solution then states that "the distance, x, is half the distance that the current must travel in the wire (once along each conductor)."

x = (1/2) * 2.53ohms * 1000ft /2.7ohms = 468.52 ft

My question:

Why does the 1/2 come into play in the calculation of the distance?

Please help. Thanks.

For single phase circuit, if the equipment is located at distance X from the source, then the total length of the wire needed is 2X ( to close the loop).

 

[sam314159]

For single phase circuit, if the equipment is located at distance X from the source, then the total length of the wire needed is 2X ( to close the loop).

Yes, single phase AC and DC voltage drop calculations you need 2X for the return path. Which I guess "single phase" would include DC too, but sometimes that's not obvious.