Trying to see if I understand Ohms law

AI Thread Summary
Ohm's Law is misunderstood in this discussion, as the correct formula is V = I * R, not V * I = R. To reduce a motor's current from 100 amps to 30 amps at 12 volts, a resistor of 0.28 ohms would be needed, not 360 ohms, and this resistor would need to handle 252 watts. Using a 2.2-ohm resistor in series with a power MOSFET will not achieve 17 amps at 12 volts; the current would be significantly lower. Caution is advised when working with high currents and vehicle batteries due to fire hazards.
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Hello
I'm trying to get my head around ohms law, and have a question.
If I have a motor that draws 12volt 100amp, and I would like to drop the amps down to 30, would I put a 360ohm resister in line.
12 * 30 = 360
V * I = R

Side note can you get a resister that is 480 watt?

Thank you

P.S I've got some nicchrome wire that I could cut to that ohm, but it would glow red hot. Would that happen to all types of reisiters that could do these.
 
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open said:
Hello
I'm trying to get my head around ohms law, and have a question.
If I have a motor that draws 12volt 100amp, and I would like to drop the amps down to 30, would I put a 360ohm resister in line.
12 * 30 = 360
V * I = R

Side note can you get a resister that is 480 watt?

Thank you

No.

The motor has a resistance of 12 / 100 or 0.12 ohms. Assume this is constant.
If you wanted to put 30 amps through this you would have a voltage across the motor of
30 amps * 0.12 ohms or 3.6 volts.

This would leave 12 - 3.6 or 8.4 volts across a series resistor which would have 30 amps flowing in it so it would be 8.4 volts / 30 amps or 0.28 ohms.
The power of the resistor would be 8.4 * 30 or 252 watts.

Note that this would not be a good thing to do to a motor. It may not even rotate with only 3.6 volts across it.

Incidentally, V * I = R is not correct. V * I = power, or V / I = R
 
No.

The motor has a resistance of 12 / 100 or 0.12 ohms. Assume this is constant.
If you wanted to put 30 amps through this you would have a voltage across the motor of
30 amps * 0.12 ohms or 3.6 volts.

This would leave 12 - 3.6 or 8.4 volts across a series resistor which would have 30 amps flowing in it so it would be 8.4 volts / 30 amps or 0.28 ohms.
The power of the resistor would be 8.4 * 30 or 252 watts.

Note that this would not be a good thing to do to a motor. It may not even rotate with only 3.6 volts across it.

Incidentally, V * I = R is not correct. V * I = power, or V / I = R

Thanks for ther quick repley.
If a powermosfet = 50 volt 17 amp, and is in serial with a 2.2ohm reisiter, will it make 17amp at 12 volt, or will a resisiter at .7ohm in serial make 17amp at 12 volt(V/I=R).
 
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open said:
Thanks for ther quick repley.
If a powermosfet = 50 volt 17 amp, and is in serial with a 2.2ohm reisiter, will it make 17amp at 12 volt, or will a resisiter at .7ohm in serial make 17amp at 12 volt(V/I=R).

If you put a 2.2 ohm resistor across a 12 volt supply, it will draw a current of 5.45 amps (12 volts / 2.2 ohms = 5.4545 Amps.)

Putting an extra 0.7 ohms in series would drop the current to 12 /(2.2 + 0.7) or 4.14 amps.

Putting just the 0.7 ohm resistor across the battery would draw a current of 17.14 amps, as you suggest.

A MOSFET does not generate voltage. This one can stand 50 volts when it is not conducting and it can conduct 17 amps when it is acting as a short circuit. The 50 volts and the 17 amps cannot happen at the same time or the MOSFET would be dissipating 850 watts. (50 volts * 17 amps = 850 watts). This is a lot of power for a small MOSFET and may blow it up.
 
OPEN,
Please do not try any of these ideas in practice until you are sure of what you are doing. It seems likely that your task involves vehicle electrics, or a vehicle-type 12V battery.

Be aware that such batteries can pass enormous currents in the case of a short-circuit.
This is a very serious fire hazard, especially when petrol (gasoline) is nearby.
 
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