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Basic Electronics » Do resistors in series really add up?

October 17, 2010
by Shanonius
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Hi Nerdkits Forum, Thanks for reading this post. I wanted to test out Ohm's law with regards to how much resistance it would take to stop an LED lighting up. I ran into a finding that has me wondering, though. I used a 12V 0.5A DC wall plug on a circuit comprised of 1 X Red LED and 1 X 47k Ohm resistor. Only the tiniest lighting up was visible from the LED. No problems so far. Then, however, I strung together 5 X 10k Ohm resistors, and the LED lit up fully. I thought to myself, "Surely that adds up to more than the 47k Ohm resistor...why is the LED lighting up fully?" I strung another 2 X 10k Ohm resistors to make 7 in total, but it hardly made a difference. I read somewhere that resistors in series add up, but my experiment seems to suggest otherwise.

Would anyone be able to shed some light on why the above happened? Is there a law I am not aware of to explain the behaviour of resistors in series?

October 17, 2010
by rajabalu21
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Please check whether you are using 10K or 1K. Some times the RED and ORANGE colors are close enough to cause the confusion. If you have a multimeter then try to measure the resistance of the 5 resistors in series.


October 17, 2010
by Ralphxyz
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For that matter if you have a multimeter you should be able to measure the different voltage between the resistors i.e. voltage divider.

I am still thinking about your dim led when you started.


October 17, 2010
by mongo
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LED's light up with a surprisingly light current. I have some that light up at 0.2mA. They don't light up as brightly as they would at 20 mA but the still light up.

The 1X47K resistor should be 1470 ohms. Well within the operating range.

The 5X10K resistor would be 5100 ohms. It would still deliver about 5mA or so, allowing the LED to light up.

My calculations may be a little off...

Maybe the 1X47K is 47000 ohms... that would be a standard value.

Maybe th5 5X10K is 50000 ohms... Not a standard size.

I would suggest checking the values with an ohm meter to be sure.


October 18, 2010
by Rick_S
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if you have (1) 47K resistor with the LED in Series( +12V <---///---|>---> GND ) with a normal 10% resistor you'd have 46350-47470 ohms of resistance.

Vs: five 10K resistors with the LED in Series ( +12V <---///---///---///---///---///---|>---> GND) with the toleraances, you'd have 45000-55000 ohms. Essentially the same circuit.

The only thing that would change would be if like Raja suggested that your resistors are actually 1k or even 100 ohm. That would keep the LED considerably brighter than the 47K would. Or if you have the LED tied in at a midpoint somewhere going to ground. Or if your "series" circuit isn't really in series.

If you don't have a meter, it would be a good investment. Otherwise, if you can take good photo's of the circuit in question, maybe we can help a bit more.


October 18, 2010
by Shanonius
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Thanks for your inights everyone. I have taken your advice and ordered a multimeter online so I can really get to the bottom of things in my experiments. Should arrive in the post in a few days. It's a cheapy for now, but hopefully does the trick. Thanks again!

October 18, 2010
by mongo
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By the way... Resistors in series do indeed add up. Measuring them in parallel is the hard part, unless you are good with reciprocals. An easier way is to make them all the same and then just divide it out. Four 1K resistors in parallel would be 250 ohms whereas in series, they would all add up to 4K.

December 14, 2010
by danuke
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I also recommend ordering a high resistance variable resistor to go along with the multimeter. It is a lot easier than throwing a lot of resistors in series (but still so it to verify the series and parallel resistor rules i.e. calculate and then verify - the Mims book from Radio Shack is a good learn and play book).
If you can get some, order a few Allen-Bradley resistors, they are really easy to read. They are not made anymore, but still around, some in lots or you can single one or two single values for 50 cents to a buck. Beware of scalpers.

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