# Another Ohm's Law Question

Discussion in 'General Electronics Chat' started by c0de3, Apr 30, 2013.

1. ### c0de3 Thread Starter Active Member

May 1, 2009
50
1
So why is it some things are so simple to understand when you read about it in a book... then when you attempt to apply them they become so difficult?

I have a circuit with a 12V power source. A 1KOhm Pot and a Fan. I'm trying to calculate the max wattage I would need the resistor to handle as to not overheat in the circuit.

The fan is rated as 12V .35A. I'm afraid I don't have enough knowledge to understand how to calculate the wattage across the resistor in this circuit. I would think the wattage would be greatest when the Pot is set at the full 1Kohm? So I've tried several calculation base on 1Kohm, but nothing seems to come out correctly.

Thanks for any tips/pointers on getting this one figured out.

Last edited: Apr 30, 2013
2. ### toffee_pie Active Member

Oct 31, 2009
162
7
what is the power rating of your variable resistor?

that rating is the maximum it can handle before burning out.

ie, a 200mw pot can safely carry 200mW through it.

3. ### MrChips Moderator

Oct 2, 2009
12,227
3,279
You have problems which are easily solved by applying Ohm's Law.

The fan requires 0.35A @ 12V. What is the DC resistance of the fan?

To reduce the voltage and current to the fan by one-half, what would be the required resistance to put in series with the fan?

What would be the wattage consumed by such a resistor?

4. ### square wave New Member

Apr 25, 2013
6
1
My guess:

Power = Voltage^2/Resistance = Voltage*Current

The fan is going to try and draw .35A at 12V, so to prevent burnout your pot needs to be rated to handle at least

12V * .35A = 4.2 Watts

actually I think I'm missing something here. 4.2 watts seems huge.

Last edited: Apr 30, 2013
5. ### c0de3 Thread Starter Active Member

May 1, 2009
50
1
The resistance would 35 ohms?
So 1/2 would be 6V. So I would need series resistance to drop 6V?

I can't get my head around this... I guess I'd need 1/2 of 12V and 1/2 of .35A... So 6/.175=35 ohms? Guess that makes sense, the fan and the resistor become a voltage divider. So when they are equal in resistance they each get 1/2 the voltage. But I don't think I'm right.

If I'm right, which I don't think I am it would be 6v*.175A=1.05W?

6. ### #12 Expert

Nov 30, 2010
16,020
6,533
The 34.28 ohm number is right...until you cut the voltage to the fan, then the resistance of the fan gets unreliable, but this is a good exercise. The not obvious point is that the power affecting the resistor is highest when it is equal to the power absorbed by the load.

ps, Squarewave, you're scope is turned up too bright!

7. ### crutschow Expert

Mar 14, 2008
12,519
3,064
That's correct if the load was a constant resistance, but the motor is not. And you need to consider the maximum current the pot can carry as well as the power. Since the pot has to carry close to .35A when it is near its minimum resistance, the pot should really be sized for that, giving a power rating of .35A$^{2}$ x 1k = 122.5W! Obviously not a practical size.

A better way would be to use a transistor to absorb the power. Then you can use a small pot to control the transistor which then varies the motor voltage and current. Use the 1kΩ pot with the wiper connected to the base of a power transistor, such as a 2N3055, with the emitter to the motor and the collector to +12V. Connect the ends of the pot between 12V and common. That will reduce the maximum motor voltage by about 0.7V, but that should not have a significant effect on its performance.

8. ### c0de3 Thread Starter Active Member

May 1, 2009
50
1

Hmm. Not sure I understand what you are saying. I think you're saying when the resistor is at 0 ohms (or 1 or whatever the lowest setting is on this thing) that is when the most current is traveling through because the fan is "pulling" the most volts&amps? Sorry I know I use all the wrong terms...

I guess I didn't think this was right because with almost no resistance the pot doesn't have to "burn-up" any current it just lets it pass by? To me as I add resistance I'm asking the resistor to limit the current which means change it into heat?

9. ### c0de3 Thread Starter Active Member

May 1, 2009
50
1
Good point. I hinted at this in my reply to #12. However, something seems odd as I have this very circuit running on my bench and my pot isn't even warm...

That's why I thought my math or understanding was wrong. Also I'm surprised by your formula. I would have though the max was 12v*.35A=4.2W?

10. ### #12 Expert

Nov 30, 2010
16,020
6,533
When the voltage drop across the resistor is the same as the voltage drop across the load, the power on the resistor is maximum for this setup. The resistor will not use up 6 volts at one ohm. It will use up 6 volts when it is (allegedly) 35 ohms...except the motor will change its current load when you lower the voltage to it. That is the starting point to math this out. Then you measure to see if the motor wimps out when you lower its voltage or if it allows more and more current as it tries desperately to keep its speed up. Depends on which kind of motor you have.

11. ### toffee_pie Active Member

Oct 31, 2009
162
7
122watt resitor

nice...!

12. ### #12 Expert

Nov 30, 2010
16,020
6,533
The problem with the 122 watt thing is that 12 volts can never cause 122 watts through a 1000 ohm resistor. P = E^2/R, max. That's .144 watts. Crutschow was just showing a theoretical. When the pot "is near its minimum resistance", it won't be 1000 ohms. When the pot is equal to whatever resistance drops 6 volts, that will be the maximum heat on the pot. Depending on the motor, it might be 35 ohms and it might not.

13. ### toffee_pie Active Member

Oct 31, 2009
162
7
rather interesting.

so in a case like this, assuming a fixed 12v supply and a fixed fan current of say .2 or .3 amps @ 12v dc also.

how to properly size a wattage resistor for maximum power transfer?

for max transfer we have Rs = Rl ?

do we match the resistor with the internal impedance of the fan?