after testing a metal for resistance (ohms) how can you find out how many watts are resisted?

Question

i have no idea of the voltage or the ampage after it leaves the material so i ask for the wattage or anything else to figure out how much electricity is leaving the resisted material
(the stuff i am working with has a much bigger resistance than average metals)

Answer 1

You really can't unless the metal in question in tested in an actual circuit.

Many metals (nichrome, for instance) increase greatly in resistance as the current heats them up - and, conversely, some decrease in resistance when heated (many semiconductors, for instance).

The equation for watts is P=IE, where P is the power or watts, I is the current or amps, and E is the applied voltage.
Ohms Law (R=E/I) will determine the actual resistance of the metal once the above quantities are known.

Answer 2

Resistance = Voltage/Current

Power = Voltage * Current

So if you apply a voltage of v volts across a resistance of r ohms, it will draw v/r amps.

From the second equation, Power (watts) = v*v/r or v^2 /r

so the power consumed by a given resistance increases as the square of the voltage.

As CanTexan points out, the resistance varies with the temperature and in dissipating the heat, the metal will heat up and the resistance will rise, thus drawing less current for a given PD, consequently consuming less power. There will be an equilibrium point (provided the metal doesn't melt first), which is why a light bulb filament or an electric fire element will heat to a certain point and no further.

Incidentally, that also explains why a light bulb most often blows when switched on. The filament is cool and therefore the current is at its highest. Being old, however, a small part of the filament has a much higher resistance and the power surge at switch on is enough to heat that small part above the melting point of the metal, which generally vaporises explosively!

Answer 3

There is a problem here.

Power (measured in watts) is a function of resistance and current according to the following formula:

P = R * I * I ... Power (P) in watts, resistance (R) in ohms, current (I) in amperes.

However, as current is applied to the resistance, the temperature goes up ... and therefore so does the resistance. This means the power consumed will also go up.

Eventually there will be a thermal equilibrium reached, based on the metal's thermal conductivity and the available surface area (and the temperature of the cooling medium) ... but you have to know what equilibrium temperature is to find the real solution.

Answer 4

Now i hope u are talking of r{internal resistance of metal},due to so many reasons electricians have found out that after passing some voltage in a metal,or some other good conductors some small voltage loss usually occurs.
Reasons:-length of metal or media
-joule's losses due to heat or I^2r losses
( in other machines there are more losses like the eddy current loss in transformers,copper losses ,ect. infact in all static and rotational machines.)
Now this u most take into consideration and then use this formula {i=el/r} where i=current,e=potential defference and l=length of metal.So remember that e will change they by changing i thus giving u your watts change when u use this i^2r formula confirming joule's effect.

Answer 5

Better look at that again.
Yes the formulas people have posted are correct.
But all metals have low resistance. Metals are usually conductors.
The Watts used are determined by how much current is flowing. That will be determined by how much is needed by the device you are using.
You probably want to find how many watts are needed, and how much current will be drawn (by something you use) and then measure the resistance of your metal to see if it is low enough to carry your current.

Answer 6

the two elementary equations for potential are P = V x A and P = I^2R potential = I^2R or potential = Amps x Amps x resistance I^2 = potential / Resistance I = ? P/R = ? 5000/50 = ? one hundred = 10 amps potential = volts x amps so Volts = potential / amps = 5000/10 = 500V P= 5000W V = 500 Volts I = 10 Amps R = 50 Ohms

Answer 7

Depends on the current. P=R*I*I (Power = resistance * current squared). Of course the current depends on the electric potential (voltage) applied across the resistor: V=IR.

Answer 8

you need two known factors to get an answer