Say you have a 12v car battery and you want to power a light bulb in two scenarios.

1) 12v battery connected to bulb with 5 feet of wire
2) 12v battery connected to bulb with 1000 feet of wire

I know the second option the bulb will receive less voltage, my question is, would all that extra wire add load/drain the battery faster?

I think it all has to do with the gauge of the wire not the length.

ok. I guess I was just wondering if voltage loss counts as energy consumed. I'm building a small solar setup and need to account for every watt.

voltage = current * resistance

power (watts) = current * voltage

don't listen to the shithead who said that length doesn't matter. the wire length increases the resistance of your circuit.

Great this confirmed my fear.. the wire going to the camera from the solar panel will actually use a lot more power than the camera itself.

This is a little tricky to answer. The short answer, if the gauge of the wire is the same in both cases, then the 1000' feet of wire will provide additional resistance and therefore flow less current. It would drain the battery slower.

Think about the extreme case, say at 1000 ohm resistor across the battery instead of your 1000' of wire and bulb. The 1000 ohm resistor would draw less current. Of even more extreme, air across the terminals -- in the megaohms. How quickly does a battery discharge just sitting there?

The "right" answer is that the battery is a voltage source. Since it isn't an ideal voltage source you would model it as an an ideal voltage source in series with some internal resistance. The internal resistance of a lead acid battery (12V) is probably fairly low compared to your load so just approximate it as 0.

If instead you had an ideal current source, it would drive the specified current across the given load. Adding more wire would cause the voltage across the wire and bulb to increase so that the current across is the same.

I don't think it's been stated explicitly in this thread, but there's no voltage drop from the long wire. The wire does act like a resistor. It heats up and uses power from the battery, and this power doesn't get to whatever device you're trying to run. If you hook an incandescent bulb up to a short wire or a long wire, the bulb on the long wire will be dimmer. The battery will drain in the same time. You'd just be getting less useful work out of the circuit with the long wire.

When dealing with electricity you consider two factors; Volts and amps.

Amps is the AMOUNT of electricity and voltage is the PRESSURE the electricity is moving.

Think of a wire as a hose for electricity. The gauge of the wire is the width of the hose.

http://www.helcohi.com/sse/body/hp.html

Volts don't mean shit! Amps is what will light a bulb and power a cellphone.

Question Nark Wrote:
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> When dealing with electricity you consider two
> factors; Volts and amps.
>
> Amps is the AMOUNT of electricity and voltage is
> the PRESSURE the electricity is moving.
>
> Think of a wire as a hose for electricity. The
> gauge of the wire is the width of the hose.
>
> http://www.helcohi.com/sse/body/hp.html
>
> Volts don't mean shit! Amps is what will light a
> bulb and power a cellphone.

Also wrong. What is of final concern in electricity is power, which is measured in watts. Volts x amps gives you the watts. You can have very high voltage, and very low amps, and get a lot of power through a wire. Long distance transmissions line use very high voltage and low amps, and carry a lot of power.

The long wire will act as a voltage divider with the bulb (load). Usually it's not a big deal if your wires are short. With longer wires, there can be significant voltage drop through the wires leading to less voltage at the load. The NEC (this is from memory) allows a 2-3% drop (loss) in the wires. If you get more loss, you need to increase the wire gauge. There are online calculators for this.

As for transmission lines, power is I^2*R. Since wire has a finite resistance, you want to run the least current possible over a transmission line. So your only choice is to raise the voltage. So it might a few amps flowing on the lines but it will kill you dead while your 12V won't even tickle your finger. Again, this is because you are dealing with a voltage source. You body is a couple 100k ohms. 12V/500k = practically 0. 69,000V/500k = 0.14 amps. I^2*R for that one, about 10,000W. You practically glow.

FmGUD Wrote:
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> Say you have a 12v car battery and you want to
> power a light bulb in two scenarios.
>
> 1) 12v battery connected to bulb with 5 feet of
> wire
> 2) 12v battery connected to bulb with 1000 feet of
> wire
>
> I know the second option the bulb will receive
> less voltage, my question is, would all that extra
> wire add load/drain the battery faster?

Have you thought about a power inverter?

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13 4826 0948 82695 25847. Yes.

Ok I found another site that had a better calculator and it looks like it's not going to take near as much power as I thought. This calculator says for the 143m 24vdc run between the solar panel and cctv camera, I will lose only .13w. I thought it was going to be 20w or something, whew.
Attachments:

Link to calculator
http://photovoltaic-software.com/DC_AC_drop_voltage_energy_losses_calculator.php

4CKUM Wrote:
-------------------------------------------------------
> This is a little tricky to answer. The short
> answer, if the gauge of the wire is the same in
> both cases, then the 1000' feet of wire will
> provide additional resistance and therefore flow
> less current. It would drain the battery slower.
>
> Think about the extreme case, say at 1000 ohm
> resistor across the battery instead of your 1000'
> of wire and bulb. The 1000 ohm resistor would
> draw less current. Of even more extreme, air
> across the terminals -- in the megaohms. How
> quickly does a battery discharge just sitting
> there?
>
> The "right" answer is that the battery is a
> voltage source. Since it isn't an ideal voltage
> source you would model it as an an ideal voltage
> source in series with some internal resistance.
> The internal resistance of a lead acid battery
> (12V) is probably fairly low compared to your load
> so just approximate it as 0.
>
> If instead you had an ideal current source, it
> would drive the specified current across the given
> load. Adding more wire would cause the voltage
> across the wire and bulb to increase so that the
> current across is the same.

Blessed are the murderous.
Attachments:

We like to call it engineering.

Be careful with calculators. Sometimes they don't count the wire distance correctly. For instance, if you are 143m away, you have 286m of wire. Some calculators want the distance of wire, some what the distance of the run.

FmGUD Wrote:
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> 2) 12v battery connected to bulb with 1000 feet of


Why not move the car closer to light bulb #2 and save yourself the trouble?

FmGUD Wrote:
-------------------------------------------------------
> Say you have a 12v car battery and you want to
> power a light bulb in two scenarios.
>
> 1) 12v battery connected to bulb with 5 feet of
> wire
> 2) 12v battery connected to bulb with 1000 feet of
> wire
>
> I know the second option the bulb will receive
> less voltage, my question is, would all that extra
> wire add load/drain the battery faster?


The wire acts like a resistor. The longer it is the greater the resistance.

The greater the distanc between power source and load(light bulb) the dimmer the light bulb.