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I'm not sure if this is the right place to ask or not but I had an electrical question. If you have a strand of mini-lights that are 2.5 V 0.25 watt on a 300 watt strand, it's normally broken into blocks of 50 bulbs in series. If you replace one of the bulbs with a 2.5 V 0.42 watt bulb, then the bulb burns dim and is barely noticeable. I don't understand why a 2.5 V bulb wouldn't burn at its rated watts and would instead burn dim (even dimmer than 0.25 watts that the original bulb burned at).

Any idea what's going on and how I can approach understanding this?

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  • \$\begingroup\$ you either bought a bulb for a 25 light string or you bought a bulb for a 50 light string and you have a 100 light string \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ One thing worth noting when dealing with bulbs (if you’re not already aware) is that they have a cold resistance and a hot resistance. If you attempt to measure the resistance of a bulb that’s off with a meter, you’ll get a completely different reading than when the bulb is under normal operation (glowing and generating significant heat). It’s important to know when doing the math and validating the answers below that model the lamps as resistors. \$\endgroup\$ Commented yesterday

2 Answers 2

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Since both bulbs are rated at 2.5 V, the more powerful bulb must draw more current. (power = voltage × current.)

To draw more current from the same voltage the more powerful lamp filament must have a lower resistance. (power = V2 / R.)

  • 0.25 W lamp: \$ R = \frac {V^2} P = \frac {2.5^2} {0.25} = 25\ \Omega\$
  • 0.42 W lamp: \$ R = \frac {V^2} P = \frac {2.5^2} {0.42} = 15\ \Omega\$ approx.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Top: All 0.25 W lamps with a voltmeter attached to each lamp. Bottom: Lamp 7 has been replaced by a 0.42 W lamp. Note the voltage difference.

When you series connect the lamps the same current flows through all the lamps. The voltage across each lamp will be proportional to its filament resistance.

Note that because the lower lamp chain has lower resistance the current increases slightly from 100 mA to 111.1 mA as indicated on AM1 and AM2.

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  • \$\begingroup\$ Thank you! This makes sense too. The poster above you mentioned that it gets less current but the same voltage but your diagram says it doesn't get the voltage it wants either. I guess they're both right? \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ @Proph brhans does not say that the .42W bulb gets 2.5V, just that it gets 100.8mA. Apply Ohm's law, and you will see.. \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ You're right I just did the math and it works out. I'm so grateful for these answers. What a help in shining light on what's going on here. Thank you all very much and Happy New Year!! \$\endgroup\$ Commented yesterday
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    \$\begingroup\$ OP: @Proph You have the simplified version in the two answers and it should make sense. In reality it's an even more pronounced effect because the bulbs don't behave like simple resistors-- they have a much lower resistance when cold. Also the lumens vs. current curve is nonlinear with much more brightness (and efficiency - but shorter life) at higher currents. Lots of gory details in this set of answers and references. \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ In this example with only 4 bulbs the 250mW ones are also running noticeably above their power rating due to the decrease in the total string resistance — about 309mW each. But in a 50-bulb string, the effect would be much less noticeable (only a couple percent, well within tolerance). \$\endgroup\$ Commented 9 hours ago
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A 1/4W 2.5V bulb has a nominal resistance of 25Ω and will draw 100mA (assuming it's connected to a 2.5V supply). So 50 of those in series will still draw 100mA from a 125V supply and have a nominal 1.25kΩ resistance.
A 0.42W 2.5V bulb will have a nominal resistance of 16.9Ω and will draw 168mA from a 2.5V supply. So 50 of these will draw 168mA from a 125V supply and have a nominal 744Ω resistance.

Now if you take your 1/4W string and replace one of the bulbs with a 0.42W bulb, your string's resistance drops from 1.25kΩ to about 1.24kΩ (virtually no difference) and its operating current increases from 100mA to 100.8mA (also virtually no difference).
So the 49 x 1/4W bulbs still shine at nominally the same brightness since they're getting less than 1% extra current, but your 0.42W bulb is expecting 168mA and only seeing 100.8mA - about 60% of its rated current - and as a result it's going to appear significantly dimmer.

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  • \$\begingroup\$ That makes a lot of sense thank you. \$\endgroup\$ Commented yesterday

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