I hav a twelve volt battery, I connect it to a 10 Ohm resistor. Obviously I need a high wattage resistor. So i get 12/10 = 1.2 amps. so power = 14.4watt. I use a 20 watt resistor, when i do that , the resistor get very hot, I can barely touchit with my finger, juts a fraction of a second. Is this normal ??
ken
Sure. Fourteen watts is a lot of power dissipation. Even a one watt resistor gets very uncomfortable when it is dissipating one watt.
Chuck
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A 20 watt resistor is probably rated such that it can safely dissipate
20 watts, but that doesn't mean to say it's safe to touch :)An old definition of power ratings was that 'at the rated power, the internal temperature is 150C at an ambient of 25C' - I seem to recall that was for carbon comp. Without knowing the precise part you are using, it's not possible to tell what the internal temperature is, but I wouldn't touch anything small dissipating more than a few milliwatts.
Cheers
PeteS
Note: A resistor's job is to convert electricity into heat. Resistors don't *produce* power--they *consume* it; the Subject line should have said "Resistor dissipation".
You've know this all your life, but I'll remind you: When things get *too* hot, they smell funny. For comparison:
If you want it to run cooler, and it is the rectangle sand type ceramic - strap it to a heat sink with a little heat grease between it and the alluminum. Or you can devide it as to say (2) 20 ohm 10 watt resistors in parralel. If space permits, get 10- 100 ohm 5 watt. etc.
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You built a 14 watt heater..........all electric heating elements are nothing but wirewound resistors......
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Except the PTC ceramics. And the microwave ovens.
John
to get max power
source resistance should be equal to load .
Nope. Think about this circuit, and apply ohm's law:
+--------------------------+ | | [SourceResistance] | | | Battery [LoadResistance] | | +--------------------------+
Okay, but lets use some simple values to make the math easy. Battery is 1 volt, internal resistance is 1 ohm.
If load resistance is 1 ohm, the load current is 1/2 amp, and the power into the load is .25 watt.
Raise the load resistor to 1.2 ohm, and the current falls to .4545 amp, and the power falls to .248 watt.
Lower the load to .8 ohm and the current rises to .5555 amp, but the power is .247 watt.
Find me a value for the load resistor that causes it to absorb more than 1/4 watt.
it won't get you max usable energy, (unless a warm battery is useful) but it will get the most power.
Bye. Jasen
Yes, but that's not often done in power applications, as it implies a power transfer efficiency of 50%. Generators and batteries don't like that.
John
Simple. He said "source resistance should be equal to the load".
So consider your source resistance at .1 ohms. Yields ~ .909 watts with the 1 ohm load.
Using his thinking, he would add a .9 ohm resistor to make the source resistance equal to the load, and get .25 watts.
In the battery/resistor circuit, you want the source resistance as low as possible for maximum power - you do not want it equal to the load.
Ed
To get the maximum power out of the source.
A 1 volt source with .1 ohms resistance feeding a 1 ohm load would produce a current of .909 amps, but that would put .826 watts into the
1 ohm load.
No, it means that if you changed the load to .1 ohm, you would get a current of 5 amperes and a load power of 2.5 watts.
I don't see that. Normally you want the load to be a much higher resistance than the source resistance, but not because it produces the highest output power.
Obviously .909 watts overall.
His statement says the source resistance "should be". It does not say the load resistance "should be". That indicates changing the source resistance to match the load. You would not increase your battery source resistance to match the resistance of your headlights. That would reduce the power you got out of the source and the power dissipated in the headlights.
I believe you keep thinking of power transfer the right way, instead of thinking about what he said, which was wrong. For a given source resistance, maximum power transfer will occur when Rs = Rl. Maximum power transfer will *NOT* occur for a given *load* resistance, when Rs = Rl. In the circuit we are discussing - a resistor across a battery - the source resistance changes with the state of charge of the battery. The load resistance does not change. It's illustrated by the real world auto headlamp circuit, where the source resistance most definitely should *NOT* be equal to the load resistance of a headlamp for maximum power transfer. (That is true whether the bulb is cold or hot - the fact that the bulb resistance increases when hot does not invalidate the point.)
If you would not add resistance to the source to make a bulb connected across it glow brighter, you do see it.
Ed
(snip)
You win. I didn't catch the nuance of the "should".
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