R=V²/P is not working for me. Trying to calculate resistor for alternator charge light.

[rcx132]

I must be making some doofus mistake here.

I plan on using an LED dashboard on my buggy which is a problem for the alternator which relies on an incandescent dashboard battery light to prime it. So I need to put a resister in place of the incandescent bulb. Here is me trying to do GCSE physics and it's not working.

The dashboard battery bulb is 1.1 watts so I calculate the resistance:
R=V²/P = (12 x 12) / 1.1 = 130 Ohms

Then I remembered I have some spare dashboard lights so I put a multimeter to one. I get 13 Ohms. Totally different result.

Confused, I connect the bulb to a power supply. The power supply shows the bulb is 1.1 watts and is drawing 0.089 Amps which according to V=IR gives R = 135 Ohms.

So the multimeter is giving a different result to the power supply and my calculations (the latter two agree with each other).

 

[tigdlo]

Probably the increased temperature. What is the current at 6V? then work out the resistance again, should be less as it won't be as hot.

 

[Paulc20au]

The filament in the bulb has a positive temperature co-efficient. As it heats up, the resistance goes up. The resistance when cold, that you measure is not the operating resistance. You could still just use the incandescent lamp as your resistor, just mount it behind the dash somewhere.

 

[m_c]

To save any calcs, 60-70ohm 5W (minimum) resistor. That'll give you the equivalent of a 2.2W bulb which used to be the common size for 12V alternator warning lamps.

 

[premmington]

To save any calcs, 60-70ohm 5W (minimum) resistor. That'll give you the equivalent of a 2.2W bulb which used to be the common size for 12V alternator warning lamps.

Do calcs at 14.4v

 

[m_c]

Do calcs at 14.4v

There's a wide enough margin, that I've never bothered.
Plus for excitation purposes battery voltage is going to be around 12V, and as per the previous posts, resistance is lower at lower voltages.
IIRC 68ohm is the common resistor value anyway.

However all this is assuming an old alternator.
More modern* ones and reconditioned ones had regulators that removed the need for a big excitation bulb. The warning lamp terminal simply sensed voltage, then the regulator drew power from the main battery terminal for excitation purposes. I discovered that after fitting a replacement alternator on an old split diode setup, and it wouldn't start charging until power was applied to the main terminal. Those vehicles had to get an additional resistor fitted across the split diode so they'd start charging.

*more modern excludes anything newer that uses a LIN bus.

 

[6ply]

As others have said (but using different words) a light bulb is not an Ohmic conductor.

 

[rcx132]

Thanks everyone. It's nice to learn something new. I've bought a mini 10W resistor encased in a heat sink and will screw that to the bracket behind the dash.

 

[aesmith]

The dashboard battery bulb is 1.1 watts so I calculate the resistance:
R=V²/P = (12 x 12) / 1.1 = 130 Ohms

Power is current x voltage, so your calculation should have been 12/1.1 not 12 squared.

 

[Guineafowl]

No, he’s using P = V²/R, which is right.

 

[rcx132]

Power is current x voltage, so your calculation should have been 12/1.1 not 12 squared.

If you take the two basic equations V=IR and P=IV and combine them to get rid of I (current) then you get P = V²/R

 

[m_c]

And since I was needing a break from centripetal force problems, here's the formula juggling required to get P=V²/R