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I had a argument with a friend about whether how can we possibly have too large wire gauge?

Assumption 1. The voltage will never drop over a long distance (200 feet) because the wire is too large. No matter of low voltage is and how little amps source provide.

Assumption 2. The voltage will drop more over a long distance (200 feet) on too large wire (AWG 16 vs AWG 6), if the voltage is low (5V) @ 0.5 Amps source providers.

It sounds like a extreme case but it comes from the phase that "Too large wire never drop current but only hurt your pocket".

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    \$\begingroup\$ Your question is a bit incoherent. I do not understand the purpose of your assumptions. \$\endgroup\$
    – DKNguyen
    Sep 21 '20 at 13:14
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    \$\begingroup\$ A proper understanding of Ohms Law will answer your question. \$\endgroup\$ Sep 21 '20 at 13:27
  • \$\begingroup\$ Be careful of your terminology - AWG6 is normally considered to be larger than AWG6. The size of the wire goes inversely withe the gauge number. \$\endgroup\$ Sep 21 '20 at 14:19
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    \$\begingroup\$ Higher source voltage means lower source current. Lower source current means less voltage loss due to E=IR. This is why long distance power transmission is done by high voltage AC and dropped to low voltage near your home. \$\endgroup\$
    – Kyle B
    Sep 21 '20 at 14:40
  • \$\begingroup\$ @KyleB Actually higher source voltage doesn't mean lower source current. Both current and voltage depend on the design of the system. You can create systems where higher voltage = lower current, and systems where higher voltage = higher current. Usually high-voltage things are designed for lower current but if you make the voltage higher than it's designed for then the current goes up too. \$\endgroup\$
    – user253751
    Sep 21 '20 at 15:17
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More the current flowing through the wire, more will be the voltage drop in the wire. All practical wires have non-zero finite resistance which means you will get a voltage drop equal to current x wire-resistance.

Now, if you need to power up an appliance of lets say 100 W and you are presented with 2 options:

  1. 10 V and 10 amps

  2. 100 V and 1 amp

Lets say wire resistance is 0.1 ohms. Voltage drop in first case will be roughly 1 V which is 10 % of the source voltage. In second case, the voltage drop will be 0.1 V which is 0.1 % of the source voltage.

Thus if both options are feasible, its better to go with second option.

Also, (current x current x resistance) heating loss will be lesser in the second option because of lower current.

That's the reason power plants prefer to send the power over high voltage and lower currents. That reduces wastage.

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400ft (you have to get to load and come back) of #16 AWG = 1.606Ω, so at 0.5A = 0.803V lost to wire and 4.197V at load.

#6 = 0.158Ω and 0.079V voltage drop and 4.921V. So Assumption 1 is incorrect. Wire has resistance (and inductance) and there will be some voltage drop across the feeder.

Area of #6 is 26,250CM and #16 2,582.9CM, so #6 will cost at least 10× #16.

If you don't mind the cost and are willing to use non-solid wire, you can have negligible voltage drop across any distance, but that is not practical. You have to balance acceptable voltage drop to feeder vs wire cost.

"Too large wire never drop current but only hurt your pocket". No wire ever drops current and any size wire drops some voltage. You can make load voltage close to source voltage, but unless it is a superconductor, there will be some voltage drop. So your statement is flawed. Current in = current out.

Assumption 2 does not make sense, but wire resistance is proportional to distance, so the greater the distance at the same current, the greater the voltage drop, which may or may not agree with Assumption 2.

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Assumption 1. The voltage will never drop over a long distance (200 feet) because the wire is too large. No matter of low voltage is and how little amps source provide.

No. This is not true. Any wire, on any distance will have a resistance, thus dropping voltage as the effect of this resistance. This voltage drop can be very important in case of low voltage and high current (lots of amperes) used by the device connected. Or it can be minimal in the case of very low current used by the connected device or high voltage or both combined.

How much current the supply can provide is not related to the voltage drop. Only the current used by powered device, and the length and the gauge of the cable will define how much voltage you lose.

So this assumption contains two mistakes.

Assumption 2. The voltage will drop more over a long distance (200 feet) on too large wire (AWG 16 vs AWG 6), if the voltage is low (5V) @ 0.5 Amps source providers.

As already mentioned in the comments, AWG 16 is thinner and AWG 6 is thicker, So you have to know what you mean by "large". In my definition AWG 6 is larger than 16. So this statement is half correct: An AWG 16 will drop more voltage than AWG 6. But not because it's too large but because it's too thin. Larger wires present lower voltage drop.

It's also true that 5V, 0.5A over 200 feet is not appropriate. It's better to use a higher voltage, 15V, 24V or 36V, whatever, with a normal wire, and then reconvert the current to 5V with a DC/DC converter. 12V is already not enough.

Now if it's just a 5V signal with minimal current, say, 0.02 mA, then it's OK. But here you already have to know what you are doing.

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