This may look like a homework, but it's not. Actually it is related to Christmas. My Chrismas lights in front of the house are constructed from segments. Each segment consists of 16 incandescent light bulbs connected in series. Each single light bulb is rated 1.5V 0.07A, so they need 24V power supply. I use hundreds of these.

The problem is that I am unable to obtain identical spare bulbs this year. I can only get 2.4V 0.07A. My question here is simple: What happens if I connect these bulbs together in series?

For simplicity, if I connect 2.4V 0.07A and 1.6V 0.07A together in series, what happens? Which one of them will be brighter or darker? Or will they shine the same? Can I consider such a tiny light bulb as a pure resistive load? May I expect a constant resistance on each bulb, and use Ohm's law to compute the rest?

  • \$\begingroup\$ Its resistance but tungsten is a non-linear PTC with 1:10 ratio from cold to hot. So not an ideal resistor (pure). Result is 2.4V bulb which appears to consume 50% more power with the same current will be brighter than the others but normal. \$\endgroup\$ Dec 8, 2012 at 2:54
  • \$\begingroup\$ My standard old incandescent light bulb seemed to take 5V (6V?). I'm using it as the power indicator and background load for my ATX-to-bench-power rig. Using a 1.5V battery I got no light on it, and 3V (two batteries) I got a dim light. I burnt out one bulb by trying a 12V supply. I don't remember how many were on the string. \$\endgroup\$ Apr 13, 2021 at 16:29

2 Answers 2


Ohm's law applies and those bulbs are indeed just simple resistive loads. To get the series current, first derive the equivalent resistance at rated voltage and current for each lamp type. Add up these resistances and from that and the applied voltage to the entire segment derive the series current.

The bulbs with a lower voltage rating will most likely shine brighter than the higher rated ones. Since both types are in series, the same current passes through them. However, the energy dissipated (in the form of light and heat) by the filament may be different.

From an application note for aerospace lamp components:

Design Volts: The voltage for which the lamp was originally designed. All other ratings (amperes, brightness, and life) are measured at this voltage, and can be changed by rerating the voltage. ... Candlepower is directly proportional to the 3.5 power of the ratio of the applied voltage, and can be increased at the expense of lamp life.

  • 1
    \$\begingroup\$ It that 3.5 power figure applicable only to the filament temperature which is expected to correspond with design voltage, or does it hold over a wide range? I imagine that there's some point below which the relationship breaks down, since many lamps really don't seem to do much of anything at at 1/4 voltage, while 1/128 of a good lamp's output should still be pretty visible. \$\endgroup\$
    – supercat
    Nov 26, 2013 at 16:37
  • \$\begingroup\$ @supercat The power figure is an approximation that works within a range representative of typical usage/tolerance (datasheet needs to be consulted). Beyond that range, like in the example you have provided, the linear relationship assumption doesn't hold. \$\endgroup\$
    – shimofuri
    Nov 27, 2013 at 16:09

Yes, Ohm's law applies regularly. Just recalculate supply voltage or number of bulbs.

  • \$\begingroup\$ I am not going to replace all bulbs. I want to know what happens when I connect different types in series. \$\endgroup\$
    – Al Kepp
    Dec 7, 2012 at 22:18
  • \$\begingroup\$ 2.4V bulb's resistance is 24Ohms, 1.5V's is 21Ohm. So if you will replace half of the bulbs with 2.4V ones — resistance of one segment will increase for 24Ohm. You can compensate this increase by shortening one bulb in segment. Just insert a piece of foil in the socket isntead of a bulb. The current will remain the same. \$\endgroup\$ Dec 7, 2012 at 22:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.