Thread: resistor calculation

Hi all - I am going to buy a resistor and just wanted to make sure I got my calculations correct. I want to cut the amps it uses in half but keep the same voltage:

Appliance is 12V @ 4 Amps fan, R=V/I = 12/2 = 6 Ohms.

Is this correct?

Thanks!

It's not quite clear what you are doing here. If you put the resistor in series with the fan, then you will have the voltage across the resistor and half across the fan. So the fan will only be getting 6V. Will it still work at that voltage?

In that case you have 6V across the resistor at 2A = 3 ohms.

Note that this means the resistor is dissipating 12W - it will get quite hot.

Originally Posted by Strange

It's not quite clear what you are doing here. If you put the resistor in series with the fan, then you will have the voltage across the resistor and half across the fan. So the fan will only be getting 6V. Will it still work at that voltage?

In that case you have 6V across the resistor at 2A = 3 ohms.

Note that this means the resistor is dissipating 12W - it will get quite hot.

Hey Strange - thanks for the tip. This is where my learning curve comes in!

So if I want it to consume half the power I half the voltage with a resistor...this will mean a fan that usually consumes 4A @ 12V will consume 2A @ 6V...is this correct?

Ah... not quite. You have halved the voltage and the current so the power will be a quarter!

But the motor probably wont have a linear voltage-current relationship.... Maybe you need to hook the fan to a variable supply voltage and see how it behaves.

Originally Posted by Strange

Ah... not quite. You have halved the voltage and the current so the power will be a quarter!

But the motor probably wont have a linear voltage-current relationship.... Maybe you need to hook the fan to a variable supply voltage and see how it behaves.

ah, ok thanks, so if I wanted it cut in half should I aim for around 9V?

I appreciate it there might not be a linear relationship - I was looking into purchasing a potentiometer to test but found them expensive so decided to go down the fixed resistor route.

By cutting "it" in half, I assume you mean power.

It's a somewhat complicated relationship because the resistor cuts the voltage across the fan (because it makes a "voltage divider" circuit) as well as the current through the fan (because it adds to the load).

You would want a 1.25Ω resistor (more accurately, 1.2426407Ω).

Original fan power consumption would be 48W (that is, 12V * 4A = 48W).

Fan resistance is 12V / 4A = 3Ω, and by adding the resistor, total resistance would be 4.25Ω, and current would drop to 2.82A (that is, 12V / 4.25Ω = 2.82A.

The voltage drop across the fan would be 8.47V (that is, 12V * (3Ω/4.25Ω) = 8.47V).

The fan power consumption would be 23.89W (that is, 8.47V * 2.82A = 23.89W), which is about half (49.83%) of the original 48W.

Someone check my work, please.

A 9Ω resistor would be far too drastic. Total resistance = 3Ω + 9Ω = 12Ω, current = 12V / 12Ω = 1A, voltage drop across fan = 12V * (3Ω/12Ω) = 3V, and fan power consumption = 3V * 1A = 3W = 6.25% (or 1/16th) of original 48W.

If we assume that the motor just acts like a resistor in the circuit, then its resistance is 3 ohms. (Power = 12^2/R=48, R=3)

To cut the motor's power in half consider that the power in the motor after the circuit modification is
I^2*R(motor)=24, then
(I^2)*3=24 and I= square root of 8 = 2.83 amps.
To get a current of 2.83 amps at 12 volts requires a total circuit resistance of 12/2.83=4.24 ohms.
The value of the resistor is 4.24-3=1.24 ohms

This agrees with what jrmonroe calculated.