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2 Laser pointers powered by a single 6v lantern battery

Hi, I need to wire 2 laser pointers to run simultaneously of a single 6v lantern battery. The pointers originally each took three button cell batteries (1.5v) I've measured the current being used by each pointer at 20 milliamps. Is it simply a matter of using two resistors as a voltage divider to get 4.5 volts then wiring the pointers in parallel across that? What resistor values do I need to use for the voltage divider to maximize battery life? (Not hugely important) If it helps, I measured the resistance of the pointers at 1.3K each, though I understand that may change under operating conditions.
What I really need to know is, what is the best way to do this, wasting as little energy from the lantern battery as possible, and preserving the lifespan of the pointers as much as possible. Ideally they would be operating a little below their normal output.
 
all you need is a resistor to limit the current to 20ma or whatever. since you have 6 volts then add them in series it was 1.5v each now if added means 6-3v=3 3v/.02= the resisance of 150 ohms.
 
I don't know where neon got the 3V from, or what kind of series connection he's thinking about.
6V - 4.5V = 1.5V
1.5V / 0.02A = 75 Ohms
So you just need two resistors of 75 ohms each, one in series with each laser.
The lasers (with resistors) are independently connected to the battery.
Current draw will be 40mA total so if the battery is 15Ah it'll last 375 hours.
Using 150 Ohms will (in theory) halve the current and double the battery life.
 
Resqueline appears to be right. Your best bet would be to connect to the two devices in parallel to the 6V battery.

Neon's math seems to have mistaken your 3 1.5V batteries as separate sources for the devices and not the 1.5 volts in series with two other 1.5V batteries that it actually is. I'm pretty sure you can't connect two devices that use 4.5V in series off of a 6V source- so Ohm's Law would dictate that it would have to be a parallel connection.
 
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