# Formula for resistance to drop to a particular voltage

Discussion in 'Electronics Resources' started by driley, Sep 2, 2016.

1. ### driley Thread Starter New Member

Sep 1, 2010
2
0
What is the formula one needs to use in order to figure out what size resistor I need (in DC) for a particular voltage in (12 volts) to the desired voltage out (6 volts). The application is this: My wife purchased these luminara candles that take two C batteries (3 volts) from QVC. There are 12 of these candles. To put batteries in these ALL THE TIME to keep them lit would cost a small fortune. I have a 12 volt transformer (120v AC to 12v DC). I don't mind putting two together in series, but 4 seems a bit much. I don't want there to be anymore than one other unit to be a factor should something fail. Otherwise, this will start to seem like trying to figure out which light bulb went out in Christmas tree light strings! ...how do you spell PAINFUL!

To be sure, this is probably already on here. I've been searching for a bit, but with the amount of posts in all the forums... it's just exhausting to try and find. If that sounds lazy, I apologize. I just have better things to do with my time and I'll find this information elsewhere. (Not upset/bitter, just stating facts.) Thanks ahead of time should anyone wish to respond to this presumably easy question based upon all the other things that are on here! GREAT site, by the way!!! I can only hope to understand HALF of the stuff on it by the time I die! Blessings.

2. ### DGElder Member

Apr 3, 2016
344
85
I would just put them in series: 3 branches, 4 bulbs each - assuming your transformer has the power. These are LEDs? How often do you expect them to burn out and how much trouble is it to check 3 bulbs? Putting a drop down resistor in series just generates a lot of heat and wastes electricity.

If you are determined then you need to find P, the power rating ( in Watts), of the bulb. I assume you are putting one resistor in series with 6 parallel branches, and each branch has two 3V bulbs in series.

If P is the power rating for each bulb - in watts, then your resistor needs to be 3/P ohms. The power rating of your resistor needs to be at least somewhat over 12P and your transformer needs to be at least 24P.

That is one huge resistor or more practically several large resistors in parallel to burn off all that power. A more practical and cost effective approach is to find a 6V transformer rated at 12P.

Something like this...

Last edited: Sep 2, 2016
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3. ### hp1729 Well-Known Member

Nov 23, 2015
1,950
219
Where do you buy your batteries? Buy in bulk. Fry's. Dollar Tree.

4. ### AnalogKid Distinguished Member

Aug 1, 2013
4,531
1,248
Insert a current meter between the batteries and one candle. Measure the current. Use Ohm's Law for the resistance, Watts Law for the power. Use a resistor rated for at least twice the calculated power.

ak

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5. ### #12 Expert

Nov 30, 2010
16,277
6,788
It's really dead simple, right after you know how, but we need 2 numbers. Voltage is 3? What's the current. With that all the other numbers fall out of the equations like shaking a colander full of sugar.

Can't measure the current? Try Harbor Freight for a \$4 meter.

That's a good answer, too!

6. ### dl324 Distinguished Member

Mar 30, 2015
3,242
619
The problem with taking this approach is that the voltage dropped by the resistor depends on the current through it (V = IR).

Assuming you want to connect multiple or all of the candles to the same wall wart, the voltage to the candles will depend on how many of them are turned on.

A better solution would be to build a simple voltage regulator to drop the 12V down to 6V. Since the candles will probably operate on weak batteries, you could try a fixed 5V regulator like LM7805. In a TO-22o package, it can handle around an amp. Measure the current draw of one of the candles to see if it can handle all 12.

An even simpler solution is to go to Goodwill and look for a 5-6V wall wart.