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I am curious about how and why if you take 2 purely resistive incandescent lamps (light bulbs) that are rated at 40 watts and 60 watts, both 120V AC (RMS) ratings, if you put them in series and feed them 240V AC (RMS), what will happen to those 2 lamps? I heard somewhere that the 40 watt light will actually glow brighter than then 60 watt light and I would like to understand why.

I have a 110V to 220V voltage converter so I can actually test this.

I believe if I use 2 same wattage lamps (such as 60W and 60W) and double the input voltage from 120V to 240V, the 2 lamps should burn normal brightness. "Strange" things start to happen when they are different rated wattages. Also I wonder if they are grossly different rated wattages such as 25W and 60W, what might happen.

You can assume that the wires connecting all of these are of negligible resistance.

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    \$\begingroup\$ Google the voltage divider, also notice that 60 watts bulb will have a resistance \$R_{60W} = \frac{120V^2}{60W} = 240 \Omega\$ and 40W one will have \$R_{40W} = \frac{120V^2}{40W} = 360 \Omega\$ \$\endgroup\$
    – G36
    Commented Dec 22, 2019 at 14:56
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    \$\begingroup\$ Also in the real-life resistance of an incandescent lamp is not constant. But will changes with the apply voltage (the resistance of a bulb is nonliear). \$\endgroup\$
    – G36
    Commented Dec 22, 2019 at 15:00
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    \$\begingroup\$ First, calculate the equivalent operating resistance of each lamp using Watt's Law. Then connect them in series in various combinations, use Ohm's Law to calculate the current through the string, and then use Joule's Law to calculate the power dissipated in each resistor. \$\endgroup\$
    – AnalogKid
    Commented Dec 22, 2019 at 15:28
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    \$\begingroup\$ @G36 it was as linear as Edison could make it. Edison had to, at all costs, develop a bulb that worked in constant voltage, and avoid CC mode, since Tesla had the corner on that tech. \$\endgroup\$
    – Harper - Reinstate Monica
    Commented Dec 22, 2019 at 17:24

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Conductors like copper and tungsten have a large +ve Tempco (PTC) such that the R.hot=10x R.cold.

You know that when voltage is applied across 2 series connected R's , the resistor with the largest value sees the most voltage.

40W Bulbs have 50% higher resistance than 60W bulbs so they see 50% more voltage and thus due to the square law \$V^2/R=Pd\$ the rise in temperature accelerates the voltage divider so that the smallest power bulb starts brighter and ends up with 10x more power dissipation than the higher power rated bulb.

The same effect would occur with two identical rated bulbs but one +10% and the other -10%. It just occurs slower and a different power share.

The effect could also be called Thermal Runaway. So if you put two 100W bulbs rated for 120V and used 240V across, the smaller power larger R bulb burns hotter than rated.

Conclusion Verified by simulation.

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    \$\begingroup\$ Press Reset and see peak power settle to >10x Pd ratio with 40W +100W in series Proof tinyurl.com/u2buggd \$\endgroup\$
    – Tony Stewart EE75
    Commented Dec 22, 2019 at 16:14
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    \$\begingroup\$ A 40W+60W in series combined draw 66W but with actual power of 47W+19W tinyurl.com/qls3n89 Now you can compute 25W & 60W and then edit the simulation parts and check. TY @Pete \$\endgroup\$
    – Tony Stewart EE75
    Commented Dec 22, 2019 at 16:19
  • \$\begingroup\$ Related: note that what I have found using low voltage incandescent lamps is that the wattage draw of them is proportional to Vf where Vf is the Voltage factor of the new voltage in question vs some known voltage and wattage combo. For example, if I put 18V thru a 12V 40W rated incandescent light, the voltage factor (Vf) = 18/12=1.5. The predicted wattage of that lamp at 18V (instead of 12V) would then be multiplied by Vf and by square root of Vf. So multiply by 1.5 and also by Sqrt(1.5) = 1.225 so 40 watts * 1.5 * 1.225 = 73.5 watts. If I used 24V, it would be 40 * 2 * 1.414 so 113 watts. \$\endgroup\$
    – David
    Commented Dec 22, 2019 at 22:14
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    \$\begingroup\$ You can use bulbs as constant current sources when the bulbs are less than dim. Also when two 60W bulbs rated for120V are applied across 240V , BUT tolerances are +/-1% the one with higher actual on 120V will be lower in series @ 240V and visa versa for -1% \$\endgroup\$
    – Tony Stewart EE75
    Commented Dec 22, 2019 at 22:26
  • \$\begingroup\$ It is very interesting (and somewhat surprising) that putting a 40W and 60W incandescent lights in series (keeping input voltage at 120VAC RMS), the wattage of the "40W" lamp actually goes UP to 47W and the wattage of the "60W" lamp goes down to 19W. This is something that I would not have guessed because I am adding resistance to the 40W circuit yet that part of the circuit is consuming more wattage that previously (without the "60W" lamp present). \$\endgroup\$
    – David
    Commented Dec 22, 2019 at 22:34

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