Voltage drop on a parallel circuit

[Anonymous]

How do you figure voltage drop on a parallel circuit? Say you have a 300 foot run, you have 1 1000 watt metal halide @175 feet and anouther @the end of the line. we are using sigle Phase 208 volts. do I use total amperage of both lights on the first half and only The total amperage of 1000 watts on the second half ? I would sure appreciate the help.

[imported_Ron]

Fisrt detrmine the voltage that will remain at the first fixture, then calculate the extension to the second fixture from the first fixture while using the answer from the first calc.
You wouldn't use the total amps, as the amps for the first fixture does not go through the wiring to the second and would give a false answer..

[imported_Ron]

For fun, consider #10 awg copper:
208, 1 phase. 1000W @ 175' and 1000W @ 125' more.
1.94% VD at the first fixture and and additional .69% (or 2.63% cumulative) VD at the second fixture.

[Anonymous]

Ron, thanks for the help. I'm not Quite sure I understand why, but at least I know how now. The reasoning I was going by was If you put an amp meater on the wire below the first light you read the amperage of bothe of the lights at that point. Thank's again Troy

[Wgoodrich]

If you first light is wired parallel to the wires going to the second light then you will only read the load of the second light if measured after the first light. If you measure before the first light the you would read the load of both lights.

The only way the load of the first light would register in a test reading after the first light and after the second light is if it were wired in series. Even in series if tested between the two lights then only the load of the first light would be measured.

Parallel wiring amp load of first light does not carry through to the second light.

House wiring is wired in parallel thus if tested after the first light only the load of the second light will be read.

However if a voltage drop is created by the first light it would affect voltage drop of the second light.

Parallel wiring of two loads on a circuit would have to be done in two calculations as Ron discribed performing the voltage drop of the second light load and distance by starting with the lowered voltage after the first light caused by the load of that first light. Kind of like finding what voltage is left after the first light before you can calculate the voltage drop using that voltage as part of your calculation.

Hope this helps

Wg

[imported_Ron]

Troy,
You're right. from the source to the first light, there will be the total current of both lights. So you calc the first "leg" of the circuit with the total current. The second leg will only have the current of the second part of the circuit (one light only), so the calc for the second leg only has the current for the second light. The second leg will not reflect the current drawn by the first light in the circuit.

[Anonymous]

Ron, Thanks for your help, this helped out a lot with a disagreement I had with a coworker. thanks again Troy