Mike,
If the batteries are connected neg to neg/pos to pos then the lamp voltage is 1.5 volts. If they're connected neg to pos, then the lamp voltage is 3 volts.
Let's assume that the lamp is rated 3 watts. If it's 1.5 volts, the lamp would draw about 2 amps and if it's 3 volts, it would draw about 1 amp.
The simplest way to power your lamp is to put a resistor in series with it. At 1.5 volts and 2 amps you'd need a 5 ohm/20 watt resistor. At 3 volts/1 amp, you'd need a 10 ohm/10 watt resistor.
It would be better to replace the lamp with a standard white diode and a 470 ohm/1 watt resistor. Or, use one of the newer super bright diodes. I'm not sure of the voltage or current ratings of these so I can't comment on the size of the resistor.
Good luck.
Walt
If I am not mistaken, the 12v lighting systems [I have Malibu] use 12v AC rather than DC. If this is correct, it might cause a problem for LEDs, but not incandescent bulbs.
JimC.
Yes, it is normal for the resistor to get hot drawing the load that you are putting on it; and no, it would not be safe to bury it unless there is plenty of air circulation around it because it needs to get rid of the heat into the air or it will burn up.
Using a resistor with AC current is NOT advisable because of the heat problem, and using the Malibu 12 Volt AC power supply to replace two batteries with an output of 3 Volts DC is even more hazardous because you may be powering an LED, which requires polarized DC current. Running an LED from AC current will shorten its life considerably.
Yes it is normal for the resistor to get warm but not hot. Judging from the photo it looks as though you're using a 1/2 or 1 watt resistor. That's much too small. You'll need at least a 10 or 20 watt unit (Radio Shack should have something). And no it's not a good idea to bury it because you need air circulation to dissipate the heat.
Hi Mike,
Judging from the photo It looks as though you have a sophisticated power supply there. Can you measure the current draw? If so, apply 12 volts through that resistor and lamp and record the current. Use the formula I2*R (current squared times the resistance). This will calculate the minimum wattage of the resistor. Example: 1.5 amps *1.5 amps = 2.25 *10 ohms = 22.5 watts. You would need a resistor greater than 22.5 watts, say 25 or 30 watts.
Somehow a campfire fed by AA batteries would have a bulb in the 100 milliamp range or less, not a 1 amp range as the batteries would only last an hour.
And at 9 volts dropped across a resistor, the resistor would get hot.
Possibly the light is already a LED.
For these types of lights, I use a 3 volt 5 amp DC power supply and feed all the dollar store lights I buy as well as the Dept 56 lights.
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