wire length voltage drops

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shawn73

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Location
Napoleon, Ohio
I just bet my coworker at our factory a 12 pack of beer and won, but I still question the theory behind it all. We pulled 1000' of 16 ga multi-conductor cable in for an alarm circuit. Anyway, he bet me that we would have a voltage drop of 5-6 volts if we put a 24v power supply on one end and checked voltage on the other. I said that there would be less than a volt of drop. It turned out the voltage was the same from one end to the other, 24.4 volts.

Ok, if electricity is related to water (as a lot of people seem to compare), it seems to me that I could have 10,000 foot of wire and as long as there was nothing but my multimeter on the other end there should be no voltage drop. (A garden hose full of water will have the same pressure anywhere within-without any flow.) I believe that if a load is placed on the end there will be a lack of current and a voltage drop because of the distance. Is this correct?

How would there ever be a voltage drop in the situation I described and is there a formula to calculate the drop using only the wire as as the load on the circuit.

Thanks again.

Shawn
 

jim dungar

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Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Re: wire length voltage drops

Voltage drop is actually the voltage lost due to a current flowing through the resistance of the wire.

Because you did not connect any load (other than your meter) you had effectively no current flow and therefore no voltage drop. Try the experiment again using some type of load.
 

physis

Senior Member
Re: wire length voltage drops

If you ignore the intrernal resistance of the voltage souce and there was enough current available there would be no voltage drop if you connected a conductor across the source. Because your measuring the source. But if you move one of the test probes further down the conductor, away from the source terminal, you'll see the voltage dropping as you move along. Of course in the real world this conductor melted a long time ago. When both probes are on the same source terminal the voltage is zero.

If you connect a load at the end of a pair of conductors you'll have three resitances in series. You can measure the voltage across each of the two conductors and across the load. These are the voltage drops.

You'll see that the voltage drop across the conductors are pretty teeny. And the voltage drop across the load is almost the whole supply voltage. That's because the two conductors have very little reistance and the load has almost all the resistance in the circiut. If you connected three equal resistances in the same way the three voltage drops would be equal and would each equal 1/3 of the source voltage.

The formula is ohm's law.
V = Voltage R = Resistance I = Current
These are it's three basic forms
V=IxR I=V/R R=V/I

We'll need an imaginary circuit.

24 Volt source
Two conductors with a resistance of .001 ohm
and a 10 ohm load

To get the voltage drop across one of the coductors you need to know the conuctors resistance and the current going through it. Because V=IxR. We know the resistance is .001

And the current is I=V/R that means we need the total resistance across the voltage source. Series resistances are added so the two conductors are .001 + .001 and the load is 10 so that's 10.002 ohms across the 24 volt source. 24/10.002=2.39952 so the current going through the conductor is 2.39952 amps.

Back to V=IxR 2.39952 x .001 = .00234 volts. That's the voltage drop across the conductor

The current is the same in the load so it's 2.39952 x 10 (it's resistance) = 23.9952 volts.

If you add the voltage drops all together it's only around a hundred millionths of a volt less than it should be. Cause of rounding.

See, nothin to it!

If you double the current by making the load 5 ohms you'll see what happens to the voltage drop across the conductors. Heck put a 1/4 ohm load on it. Make it 120 volts too.
 
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