watts, ohms, volts oh my???

[Fourdy]

I will now show my lack of knowledge in the finer points of electricity. I have an older 6 volt defroster fan that draws 2 amps at 6 volts and runs fine.

I bought a voltage reducer from Napa but that didn't help at all. What can I put together that would reduce the voltage. I did find a variable resister that is about as big as a snuff can that I could make work but just too big.

Many thanks in advance. ( I do know how to solder and cut wires. lol)

Fourdy

[Fourdy]

Probably should have said that I need to reduce from 13 volts.

Fourdy

[Dec211975]

V=I x R so say it draws 2 amps. 6 volts = 2 amps x Resistance Resistance = 3 ohms.

The size of resister required will be: Watts = Volts x Amps so 6 x 2 = 12 watts

A 3 ohm 12 watt resistor should work.

Below is a 3 ohm 100 watt Resistor

http://www.ebay.com/itm/3-Ohm-3R-100...item5ae310ada9

[Mike B]

With a resistor inline, your still going to supply 12 volts to the fan motor...just less amperage.

You have to sink the resistor to ground to split the voltage.
I/e 12v------resitor to ground(where resistor is tied to 12v supply)------heater motor.

I've a voltage reducer in my '42...big round ceramic looking thing, and 12v in, one lug to ground and 3-4 outputs.

[Jack E/NJ]

Mike B>>>your still going to supply 12 volts to the fan motor>>>

Sorry to disagree, Mike. With the aforementioned single 12 W resistor is series with the 12 watt motor, half the 12v is dropped across the resistor (6 volts), and half across the motor(6 volts). That is, 6v drop across each.

What you apparently have is the equivalent of a multi-tap potentiometer. The full 12 volts is dropped across the circular length of a very hi-power resistor that has individual taps along its length for various voltages between 0 and 12 volts. The problem with a potentiometer is that it must draw a current far exceeding those currents required by devices connected to the taps in order to get the desired tap voltage.

Jack E/NJ

[Mike B]

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So Jack...your saying if I have 12 volt supply...and in series I put a 2M ohm resistor feeding a light bulb...you think the voltage at the bulb will be less than 12 Volts?

No...it's a series circuit, the resistor will drop the amperage, but the voltage will remain the same.

[41ford1]

Originally Posted by Mike B

So Jack...your saying if I have 12 volt supply...and in series I put a 2M ohm resistor feeding a light bulb...you think the voltage at the bulb will be less than 12 Volts?

No...it's a series circuit, the resistor will drop the amperage, but the voltage will remain the same.

It is a series circuit. There will be a voltage drop across each resistor in the circuit. The total voltage drop for the whole circuit is 12 volts. Your 2M ohm resistor will limit the current to something like .006 milli amps. The resistance of the lamp is a negligible addition to this circuit. The voltage drop across the lamp will be approximately .05 volts. The principle of using an equal value resistor as the load(motor or lamp) halves the voltage and current at the load. The voltages must be measured with the circuit turned on.

[Mike B]

Okay...let's simplify this, or bring me to my senses

12v...1K resistor...device (bulb/motor/etc.)


12v....1k resistor....1 K resistor...Device


12 v...1K resistor...1K resistor...1 K resistor...Device


In each of the above series circuits, how much voltage is applied to the device?

[Marvin/TN]

I bought a voltage reducer from Napa

Very simple and easy to do, Get a blower resistor for a 70's GM. They have 3 or 4 taps out put just find the one closes to 6 Volt. I used one in my 51 F-1 that had 7 volts. It does have to be in the air stream to cool it but will work very well. I used a hole saw to mount the resister in the bottom of my heater so fan motor kept it cool.

[Mike V. Florida]

Originally Posted by Mike B

So Jack...your saying if I have 12 volt supply...and in series I put a 2M ohm resistor feeding a light bulb...you think the voltage at the bulb will be less than 12 Volts?

No...it's a series circuit, the resistor will drop the amperage, but the voltage will remain the same.

Whenever you have a current flowing through a load there is a voltage drop ACROSS that load.

The fan, when working, is the same as a 3 ohm 12 watt resistor. If you place another 3 ohm 12 watt resistor in series, each will have the same voltage drop.

Kirchoff's law states that the sum of the voltage drops, minus the supply voltage will equal zero.

Please note we are talking about ACROSS the resistor and ACROSS the fan.

If you put together the circuit; battery, resistor, fan, and measure across the resistor you will read 6 volts. If you measure across the fan you will measure 6 volts. 6 volts plus 6 volts -12 volts (battery) = zero.

Clear as mud?

[Mike V. Florida]

Originally Posted by Mike B

Okay...let's simplify this, or bring me to my senses

12v...1K resistor...device (bulb/motor/etc.)


12v....1k resistor....1 K resistor...Device


12 v...1K resistor...1K resistor...1 K resistor...Device


In each of the above series circuits, how much voltage is applied to the device?

Not enough info. We need to know more about the device.

If we assume 1k for the device,

12v...1K resistor...device (bulb/motor/etc.) - across each part 6v


12v....1k resistor....1 K resistor...Device - across each part 4 volts


12 v...1K resistor...1K resistor...1 K resistor...Device - across each part 3 volts

[4dFord/SC]

VanWinkle's Law: Wamps + Vohms/Resistors = zzzzzzzzzz

[Planojc]

As the Kingfish once said, "the atom is made up of protons, neutrons, fig newtons and morons."

[rotorwrench]

Fourdy
Randy Rundle has a resistor for fan motors. You might call him and ask about it. I don't know how well it will work or how long it will last but it is a design that should work.
http://www.fifthaveinternetgarage.co...counter_2.html

[Mike B]

Originally Posted by Mike V. Florida

Whenever you have a current flowing through a load there is a voltage drop ACROSS that load.

The fan, when working, is the same as a 3 ohm 12 watt resistor. If you place another 3 ohm 12 watt resistor in series, each will have the same voltage drop.

Kirchoff's law states that the sum of the voltage drops, minus the supply voltage will equal zero.

Please note we are talking about ACROSS the resistor and ACROSS the fan.

If you put together the circuit; battery, resistor, fan, and measure across the resistor you will read 6 volts. If you measure across the fan you will measure 6 volts. 6 volts plus 6 volts -12 volts (battery) = zero.

Clear as mud?

I do believe you're correct and I'm incorrect.

[Jack E/NJ]

Assuming the impedance or resistance of the heater motor doesn't change, a 6v motor will draw 2x the current at 12v and have to dissipate 4x the power or wattage. A 6v motor will definitely run faster at 12v but may overheat and fry. A dropping resistor like that recommended by 12/21/1975 above is recommended to avoid this possibility.

Jack E/NJ

[Bill S]

What DEC211975 said is correct. 3 ohm 12 watt. But that 12 watt resistor would be very hot. Use a wire wound 24 watt min.

[51 MERC-CT]

Is there any reason why the 6 volt heater motor can't be run on 12 volts (just making it run faster)?

[Jack E/NJ]

Post #16 is a response to post #18

Jack E/NJ

[Fourdy]

I need to clarify what kind of defrost fan this is. I hooked up a digital power supply and found that at 6 volts it read 1.9 - 2 amps.

As you can see, there are no ducts of moving air to cool any devices.

Thanks

Fourdy