Hello,

Resistance is the property of every material that opposes the current to flow.
Those who have a low resistance are called conductor.
Those who have a high resistance are called insulator.

I am wondering now.

Assuming that my wood has only 1Ω resistance.

Okay, for example I have now a 24 volts battery (Yes, let's upgrade my voltage.). Assuming that my wood is the one I am going to use as the conductor (A wire for example.).

And I have a lamp. Will the lamp be lighted as the wood is the conductor?

I am very confused. Is there any specific formula if how much current does a specific material accept?

It's very hard to explain but I hope you do understand what I mean.

Catatatatatpult.....

 

Try ohm's law.

 

Never put current into your wood. That could hurt.

A lot.

Don't.

 

ErnieM! Play nice.

Cat,
There is a partial list of materials on this page:
http://physics.info/electric-resistance/
Note that most of the materials are some type of metal.
Wood (from trees, Ernie!) when dried actually has a very high resistance; it is a good insulator.

 

Wood is basically an insulator, which some slight potential conductance. You connect a low voltage battery to it you probably won't notice a thing, even if it is wet.

The other guys are worried about higher voltage, but I happen you know you have a pile of battery connectors somewhere.

 

You can get your answers in AAC's free ebook...
http://www.allaboutcircuits.com/vol_1/chpt_1/5.html
And about formulas
http://www.allaboutcircuits.com/vol_1/chpt_2/index.html

Good Luck

 

Based on Ohm's Law, I=E/R (24/1=24). So 24 amperes? I guess Ohm's Law will not answer my question.

 

Ah, yes it would.

Battery specifications: 48 ampere-hour/24 volts
Lamp: 24 amperes/24 volts

Assuming that the wood (My wire in this case) is 2 ohms connected in battery to power up my lamp. Then the lamp will get only 50% of the required amount of ampere.

24/2=12

AHHHHH I DON'T KNOW. HELP ME

 

This calculator will help you figure out how much voltage you will drop across your wiring:
http://www.stealth316.com/2-wire-resistance.htm
For example, 1 Ampere through 5 feet of AWG-20 wire will drop 0.0518 Volts; as that length of AWG-20 copper wire will have 0.0518 Ohms of resistance.

Incandescent lamps have very low resistance when they are cold; when the filament heats up, the resistance increases a large amount.

You will need to calculate your lamps' current by dividing the wattage rating by the voltage rating.

For example:
A 12v lamp rated 24 Watts would be A=P/E; 24W/12V=2 Amperes
You can also calculate the resistance at the normal operating temperature by:
R=E^2 / P; Volts*Volts / Watts; 12*12/24 = 144/24 = 6 Ohms.

 

Would it light? That would depend on the lamp, wouldn't it? 1Ω is quite a low resistance: it's not easy to imagine a piece of wood being that low unless it had been burned to carbon first. If you want to know if a lamp will light when it is in series with a resistor, this requires a little calculation. You need to know the supply voltage (24V in your case), the resistance (1Ω). You also need the normal ratings of the lamp: its voltage and current, or perhaps its wattage, from which we can tell the current rating.

For a filament lamp, these numbers can be used to find an equivalent resistance, although this is not so accurate when the lamp is not at its normal voltage (the lamp resistance is not constant, I'll say more about this later). LEDs are different: they have pretty much constant voltages for a wide range of currents. Let's say you have a 24V 12W filament lamp. What is the normal current for this lamp?

Power = Voltage * Current, so Current = Power/Voltage. I = 12W/24V = 0.5A Resistance = Voltage/Current = 24V/0.5A = 48Ω.

If this lamp is connected in series with a 1Ω resistor, the total resistance will be 1Ω + 48Ω = 49Ω. The current will be 24V/ 49Ω = 0.49A, almost the normal value. The lamp voltage will also be almost normal. You can work this out from the current times the lamp resistance, V = I*R, V = 0.49*48 =23.5V.

A quicker way is to call the series resistor "R1", and the lamp resistor "R2" then use the potential divider formula:

Vout = Vin * R2/(R1+R2) http://en.wikipedia.org/wiki/Voltage_divider

In this case 24V*48Ω/(48Ω+1Ω) = 23.5V A 24V 120W (4.8Ω) lamp would get 24V*4.8Ω/(4.8Ω+1Ω) = 19.9V, so it would be a bit dim.

Your monster 24V 24A lamp (576 Watts!) is only 1Ω equivalent, it would only get half voltage and would be very dim.

Actually, it would be worse than that. Tungsten filament lamp resistances drop as the voltage falls, because at the normal voltage the filament is extremely hot, raising its resistance. At lower voltages the filament is less hot, so the resistance falls. If you ever have access to a resistance meter, try measuring the resistance of an ordinary torch bulb. Then compare this to the resistance calculated from its normal ratings. The cold resistance is likely to be far lower.