If 2 lamps of same power are connected in series, the first one connected glows better and the next one gets dimmer..then Why doesnt the serial lights( Christmas lights ) get dimmer even though they are connected in series and share the same voltage..??

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    \$\begingroup\$ You might find some useful information here that explains some of your doubts. \$\endgroup\$ Feb 4, 2015 at 13:11
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    \$\begingroup\$ Who says the first get brighter and the second gets dimmer? If they're the same impedance/resistance, then they should be equally sharing the voltage. Thus, equally rated lights will both become dimmer. On a side note, I am not sure if 'serial' is the proper term to use, usually that deals with communication. Series is often the preferred term. \$\endgroup\$ Feb 4, 2015 at 13:33
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    \$\begingroup\$ Note: The following is not criticism - being wrong is an excellent way to learn. | This claim -> "... If 2 lamps of same power are connected in series, the first one connected glows better and the next one gets dimmer ..." is incorrect, and so the question based on it is meaningless. It would be useful if you explained why you thought it was true. \$\endgroup\$
    – Russell McMahon
    Feb 4, 2015 at 13:39
  • \$\begingroup\$ Barry perfectly explains it. Based on your observation, it can be concluded that the two bulbs you mentioned are not of "the same power". You might have them rated for the same voltage, but if their filaments are not the same (of equal resistance), they will not seem to glow equally. If you meant to say that when you connect the second bulb, the first one becomes dimmer, but the second one is of the same light intensity as the first one then they simply share half of the voltage each - in this case, refer to the Andy Aka's answer. \$\endgroup\$
    – Nazar
    Feb 4, 2015 at 15:01
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    \$\begingroup\$ See the comment below by Underdetermined to Barry's answer. It completely explains it. The bulbs glow with a certain brightness. If you added more bulbs in SERIES, they would all glow with the same brightness again, but a lower brightness. When you plug in another string, you are adding another string in parallel, so the second string is still operating under the same voltage drop as the first. The second string, plugged in in parallel, will glow just as brightly as the first. This is why house wiring is in parallel, and why we need circuit breakers. \$\endgroup\$
    – abalter
    Feb 4, 2015 at 21:04

3 Answers 3


Christmas lights don't exactly "share" the same voltage on each lamp - they happen to have approximately the same voltage but that is because they are all in series and have roughly the same impedance to current when glowing.

If your supply is 240V ac and you have 20 lamps in series, each one will have approximately 12V across the lamp terminals. It makes sense for each lamp therefore to be rated for 12V operation and of course if you tried to put that across 240V it would instantly fail and blow.

If you took a 240V lamp and put it in series with another 240V lamp, each lamp would be receiving 120V and therefore the brightness would approximately reduce by a factor of four.

  • \$\begingroup\$ That is what happens, but perhaps you could explain why the "first" lamp doesn't take up all 240V and operate at full brightness with the second getting 0V and being off. \$\endgroup\$
    – Craig
    Feb 5, 2015 at 0:17
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    \$\begingroup\$ @Craig Since the lamps are connected in series, the same amount of current must goe through both: it comes out from the wall socket, in through the first bulb, out into the second, and then back into the wall. If the two bulbs are essentially the same electrically, since the current through both are the same their voltages will both be the same, i.e. half of the total voltage. \$\endgroup\$ Feb 5, 2015 at 2:19
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    \$\begingroup\$ @Craig - a lamp is a filament of wire. This filament is maybe 10mm long. It has resistance and this dictates how much current passes thru the filament for a given voltage across it. Would you expect the (upper) half of the filament that is connected to the higher voltage to glow more than the "lower" half? \$\endgroup\$
    – Andy aka
    Feb 5, 2015 at 9:25
  • \$\begingroup\$ Note that, when you take out a single lamp, you could receive a shock of more than 12V (theoretically up to the full 230Vac), since your body impedance may be much higher compared to a (cold) christmas light. \$\endgroup\$
    – Sanchises
    Feb 5, 2015 at 9:27
  • \$\begingroup\$ @sanchises - is this relevant? If so who is it aimed at? \$\endgroup\$
    – Andy aka
    Feb 5, 2015 at 9:28

The answer to your second question is related to your first. If 2 lamps rated for the same voltage and power are connected in series, then the same current must flow through each one. Since they are rated for the same voltage and power, the resistance of their filaments must be the same. Thus if the current through them is the same, the voltage will also be the same. Hence, the power, which is the product of voltage and current, will also be the same. Thus they will glow equally. The same is true of a string of Christmas lights. It doesn't matter how many; they will still glow equally.

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    \$\begingroup\$ One crucial point missing in this answer but mentioned by Andy is that if you added more and more lights they will all be dimmer, but all by an equal ammount (if they are the same lights). \$\endgroup\$ Feb 4, 2015 at 17:01

As was suggested in a comment by @JarrodChristman on the question, if the two lights claim to be rated the same, but they are not the same brightness when connected in series, then the lights do not actually have the same specs. Perhaps the filament on one bulb was extruded a little thinner than the other; even though they are both approximately the same rating, one has a slightly higher resistance, and burns a little hotter (and brighter, and will burn out sooner).

In a strand of Christmas lights, each bulb dissipates so little power (1A/125V on a 50 light string is ~2.5W per bulb) that a 5% manufacturing variance on the filaments means the bulbs are only going to vary by ±0.125W. With a 40W household bulb, a 5% manufacturing variance means the bulbs could vary by as much as ±2W.

I'd bet that on that string of Christmas lights you consider to be "equal" brightness, if you were to measure the luminance of each bulb individually (with total isolation from all other light sources) with a precise instrument, you would indeed find that some are not quite as bright as others.

  • \$\begingroup\$ It is entirely possible for two lamps which will both pass 100mA with 3.0 volts across them and give off the same amount of light will pass differing amounts of current with a lower voltage across them. One might burn slightly hotter at 3.0 volts, but have some of its light output obstructed by a support structure which helps improve longevity [absent such a structure, the hotter bulb would give off more light but burn out sooner]. If both lamps have equal resistance at normal operating temperature, the one which operates hotter will likely have... \$\endgroup\$
    – supercat
    Feb 4, 2015 at 18:16
  • \$\begingroup\$ ...a lower resistance at room temperature. If the two bulbs are operated in series from a 3.0-volt supply, their temperature will likely be closer to room temperature than to normal operating temperature, so the bulb which was hotter during normal operation would likely have a lower resistance and thus receive a smaller share of the power. \$\endgroup\$
    – supercat
    Feb 4, 2015 at 18:19

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