# Light output of different wattage bulbs in series

#### Curlywurly

Joined Mar 16, 2019
8
I have been asked the question: If we put a 5W and a 40W lamp in series and connect them to a 12 V car battery, what will happen?

I stated that the 5W bulb would be brighter than the 40W, but was told it was the other way around...
Could someone explain to me why this is? Cheers!

#### crutschow

Joined Mar 14, 2008
31,110
If they are both 12V incandescent bulbs, than the 5W bulb will have the higher resistance (R = V²/P) so will drop most of the voltage, thus the 5W bulb would be the brightest.
This difference in resistance is increased by the fact that a cold filament has a much lower resistance than a hot filament so as the 5W filament starts to heat up, its resistance will start to increase even more than the 40W filament, which is staying much cooler.

Whoever told you the other way around should not be relied upon as a source for electrical circuit information.

#### Curlywurly

Joined Mar 16, 2019
8
Thank you very much, was hoping I wasn't as silly as I was being told I was!

#### shortbus

Joined Sep 30, 2009
9,754
If they are both 12V incandescent bulbs, than the 5W bulb will have the higher resistance (R = V²/P) so will drop most of the voltage, thus the 5W bulb would be the brightest.
This difference in resistance is increased by the fact that a cold filament has a much lower resistance than a hot filament so as the 5W filament starts to heat up, its resistance will start to increase even more than the 40W filament, which is staying much cooler.

Whoever told you the other way around should not be relied upon as a source for electrical circuit information.
Are you sure about that? It doesn't match up what other places selling bulbs have to say -
• 40 Watt incandescent lamp produces only 380-460 lumens and uses 40 Watts of energy per hour.

#### Ylli

Joined Nov 13, 2015
1,060
Shortbus, the question was what happens when you put the bulbs in series, not which is the brightest at nominal voltage. As crutschow says, the current through both will be the same, but more voltage will be dropped across the lower wattage (higher resistance) bulb. Therefore the lower wattage bulb will dissipate more power and be brighter.

#### wayneh

Joined Sep 9, 2010
17,189
Correct. In the extreme consider a 5W bulb in series with a 500W bulb. All the voltage will drop across the 5W and it will illuminate just as it would without the 500W bulb in series. The 500W bulb might as well be a piece of wire. The current passing through the 500W bulb is limited by the 5W bulb. It won't be enough to make the filament warm and no light will be produced.

#### shortbus

Joined Sep 30, 2009
9,754
Shortbus, the question was what happens when you put the bulbs in series, not which is the brightest at nominal voltage. As crutschow says, the current through both will be the same, but more voltage will be dropped across the lower wattage (higher resistance) bulb. Therefore the lower wattage bulb will dissipate more power and be brighter.
Totally missed that. Saw it as which made the most light. My bad.

#### WBahn

Joined Mar 31, 2012
27,859

#### shteii01

Joined Feb 19, 2010
4,644
Ptotal=5+40=45 W
Itotal (this is series circuit)=45/12=3.75 A

5/3.75=1.33 V
40/3.75=10.65 V

The 40 W bulb is closer do its designed operating voltage (of 12 V), therefore I expect 40 W bulb to be brighter.

#### WBahn

Joined Mar 31, 2012
27,859
Shortbus, the question was what happens when you put the bulbs in series, not which is the brightest at nominal voltage. As crutschow says, the current through both will be the same, but more voltage will be dropped across the lower wattage (higher resistance) bulb. Therefore the lower wattage bulb will dissipate more power and be brighter.
Yes and no. With a big difference like 5 W and 40 W, this is likely the case. But if the wattages were closer together, then other factors come into play. For instance, the number of lumens per watt generally increases with rated power, with high wattage bulbs being several times as efficient as lower wattage ones. Then there's the issue that not all bulbs of the same wattage put out the same amount of light for a variety of reasons, so it's entirely possible to find a lower wattage bulb that, at it's rated operating point, actually puts out more light than a somewhat higher wattage bulb that is intended for a different application. But while you might find something like that with a 60 W bulb and a 75 W bulb, very unlikely to find it with a 5 W and a 40 W.

To be safe, these kinds of issues can be swept aside (in case it's a trick question, if nothing else), by stating that the answer assumes that, other than power, the bulbs are identical and that they furthermore have the same luminous efficiency.

#### crutschow

Joined Mar 14, 2008
31,110
Ptotal=5+40=45 W
That's not correct.
You add the resistances, not the wattages.

#### ElectricSpidey

Joined Dec 2, 2017
2,179
Ptotal=5+40=45 W
Itotal (this is series circuit)=45/12=3.75 A

5/3.75=1.33 V
40/3.75=10.65 V

The 40 W bulb is closer do its designed operating voltage (of 12 V), therefore I expect 40 W bulb to be brighter.
Unfortunately the wattage ratings you are using in your calculations would only be valid if both bulbs were receiving 12 volts, but this is not the case with them in series.

In case anybody really cares I did an experiment using a 2.5 and a 25 watt automotive bulbs and the higher wattage bulb didn’t even light.

Joined Feb 20, 2016
4,251
The 5W lamp will light and the 40W will just look cold as has been stated a few times above. It is complicated as the lamps have positive temperature coefficients so it is pretty hard to figure out the actual resistances.

But even if you go by the hot resistance...
5W at 12V = 28.8 Ohms, and 417mA.
40W at 12V = 3.6 Ohms, and 3.3Amps.
If you put these in series, you get 28.8 + 3.6 = 32.4 Ohms.
The current now will be 12/32.4 = 370mA.
That is pretty close to the 5W lamp full power, (3.9W) but nowhere near the 40W lamp (0.5W).
In actual fact, the 40W lamp will not be hot enough to raise the resistance to its operating level, so even more current will flow so the 5W lamp will be closer to full power than that.

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#### crutschow

Joined Mar 14, 2008
31,110

#### Danko

Joined Nov 22, 2017
1,489
But even if you go by the hot resistance...
5W at 12V = 28.8 Ohms, and 417mA.
40W at 12V = 3.6 Ohms, and 3.3Amps.
If you put these in series, you get 28.8 + 3.6 = 32.4 Ohms.
The current now will be 12/32.4 = 370mA.
That is pretty close to the 5W lamp full power, (3.9W) but nowhere near the 40W lamp (0.5W).
Even more:
Resistance of cold lamp is about 10...15 times lower than in working state,
so resistance of in series connected lamps is 28.8 + (3.6 / 10) = 29.16 Ohm
and current: 12 / 29.16 = 411 mA, then
power_5W lamp is 4.86 W,
power_40W lamp is 0.06 W.
EDIT:

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#### mvas

Joined Jun 19, 2017
539
If these are 12 Volt LED Lights then ...
There is possibility that neither LED Light will illuminate when connected in series.

#### Tonyr1084

Joined Sep 24, 2015
7,194
Yes and no. With a big difference like 5 W and 40 W, this is likely the case. But if the wattages were closer together, then other factors come into play. For instance, the number of lumens per watt generally increases with rated power, with high wattage bulbs being several times as efficient as lower wattage ones. Then there's the issue that not all bulbs of the same wattage put out the same amount of light for a variety of reasons, so it's entirely possible to find a lower wattage bulb that, at it's rated operating point, actually puts out more light than a somewhat higher wattage bulb that is intended for a different application. But while you might find something like that with a 60 W bulb and a 75 W bulb, very unlikely to find it with a 5 W and a 40 W.
Some manufacturers would rate their wattages at 103 volts whereas others might rate their wattages at 120 volts.

On the subject of resistance: When a bulb is fully lit -
A 5 watt bulb at 12 volts sees 417 mA. 12 ÷ 417 = 28.8Ω (when fully lit).
A 40 watt bulb at 12 volts sees 3333 mA (3 1/3 amps). 12 ÷ 3.333 = 3.6Ω (again fully lit)
The total wattage (if both lights were fully lit) would be 12v÷(3.6Ω + 28.8Ω)=370mA. 370mAx12v=4.444••• watts. If any bulb is going to light it's going to be the one closest to the power of the circuit. In this case the 5 watt bulb will be brightest as @crutschow said.

As for calculating wattage in series (assuming full brightness on each account)
1 ÷ ( 1 ÷ 5w + 1 ÷ 40w ) = 1 ÷ ( 0.2 + 0.025 ) = 1 ÷ 0.225 = 4.444•••w

#### shteii01

Joined Feb 19, 2010
4,644
Which brings us back to the original message. OP was informed that 40 W bulb will be brighter.