# What wattage resistor is needed?

#### Scavenger

Joined Jan 5, 2005
19
Hi All.

Last year I bought some LED string lights (photo attached) last year that run on 3 AA batteries. Each string has a 30 ohm resistor in series with the lights. I'd like to replace the batteries with a USB connector. I'd like to connect three strings in parallel to minimize the number of USB connectors needed.

When I take one string and connect to a USB plug (5.08V) I get a current of 68mA. If I'm applying Ohm's Law correctly (P=IV), then 5.08V x 0.068A = 0.344W. When using the batteries (4.85V) I get a current of 60mA, or 0.291W.

If I run three strings in parallel, through one resistor, do I need a 1W resistor? If I'm judging the current resistor (photo attached) it's a 1/4W resistor. Isn't it too small a wattage for the original circuit since the power exceeds 1/4W? I think I could just leave each of the three resistors in series with each string, and then connect all three in parallel, but that seems like it would be bulky.

I'm a newbie at this kind of stuff, so maybe I've read something incorrectly.

Thanks for any help.

Richard

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#### Dodgydave

Joined Jun 22, 2012
9,848
Yes you're right with the wattage 345mW, I would use 3 separate 1/2W resistors one for each chain.

#### BobTPH

Joined Jun 5, 2013
2,729
You want to leave the 3 separate resistors on the strings. You could add a resistor before the 3 strings to reduce the voltage from the 5V of the USB to the 4.5V of 3AA cells. You would calculate this as follows. Add the 3 currents from the strings, which comes out to 180mA. Now you need a resistor that drops 0.5V at 180mA which comes out to 2.7 Ohms. It needs to dissipate

0.5 ^2 / 2.7 = 0.09W

so pretty much any resistor would do.

Bob

#### Wolframore

Joined Jan 21, 2019
2,183
Your total power calculation is correct, however the power dissipated at the resistor and it's power rating is incorrect.

There is voltage dropped across the LED(s) therefore you do not dissipate the full voltage at the resistor. If after powering the LED's you have 0.4V left, your power dissipated by that resistor should be recalculated. 0.4V* 60mA = 24 mW which is why it does not get too hot to touch.

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#### Audioguru again

Joined Oct 21, 2019
2,374
With 5.08V and a 30 ohm resistor the current is measured to be 68mA. Then the power dissipated by the resistor is simply 68mA x 68mA x 30 ohms= 139mW. The LEDs release light and heat for the remaining power provided by the power supply.

A 1/4W resistor will get very warm with 139mW. Three strings of LEDs in parallel and with a 10 ohms series resistor will heat the resistor with 416mA which needs a 1W resistor.

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#### Scavenger

Joined Jan 5, 2005
19
Thanks for all the replies.
After seeing the post from 'Audioguru again' I was much more comfortable leaving the original 1/4W resistors in place. I wired it up and don't feel any significant heat from the resistors.

Thanks again.

Richard