Q & A: bulbs in series

My Friend showed me a simple circuit - just two bulbs conected in series to a power supply. When he put two 60w bulbs in the holders they both shined equally brightly. When he put two 30w bulbs in the holders they both shined equally brightly. But when he put one 60w bulb and one 30w bulb in the holders then the 30w bulb lit up and the other didn’t (well it did but so faintly you could only see if you turned all other lights off!!) Why did this happen? Can anyone help me?
- Anonymous

Sure. The wattage rating for bulbs tells you about how many watts of power they draw if connected directly to a standard 115 V (or so) power supply. The power is equal to the current I times the voltage V. Since I=V/R, where R is the resistance, you can also write the power as V^2/R or as I^2*R. So the higher wattage bulb is the one with lower resistance.
If you put two bulbs in series, their combined resistance is bigger than either one separately. Therefore less current will flow through them than would flow if they were hooked up normally. In this case, the 30 W bulb has twice the resistance of the 60 W bulb, so the total resistance is 1.5 times that of the 30 W bulb and 3 times that of the 60 W bulb. So now the 30 W bulb will get only 2/3 as much current as it does when plugged in all by itself, and the 60 W bulb will get 1/3 of the current it gets when plugged in all by itself. The 30 W bulb will draw 13.3 W of power, and the 60 W bulb will draw only 6.6 W. The 30 W bulb will be a lot less bright than usual, but the 60 W bulb will barely glow at all.

By the way, I've made a somewhat false simplification in this discussion. I've pretended like each bulb has a fixed R regardless of how much current goes through it. Actually, as a bulb heats up its resistance goes way up. However, the general idea is the same as that I described in simple terms.

Another fact that helps to understand what you see is that the visible light that comes out depends very strongly on how hot the filament gets. If it's not hot enough, the light that comes out will mostly be invisible infrared. I bet the barely visible 60 W bulb looks pretty reddish, not white, because the filament isn't hot enough to give off the blue part of the visible spectrum.

Mike W.

(published on 10/22/2007)

Hey Mike, you mentioned in your explanation that both the bulbs will draw different amount of current. Why would they draw different amount of current? They are connected in series. They would definitely have different potential difference across them but I didn't understand the concept of different current. Plus, can you also tell me if the total power of the circuit is 90W or something else? 'Cause if we calculate their resistances (assuming the bulb doesn't heat up!) using R = V^2/P (V could be 110V) and then add both the resistances (they are in series) and calculate total power of the circuit using P = V^2/R(total) then it gives a very small value. Thanks
- Suzanne (age 22)
Edinburgh, UK

 
Nice question. It shows that what I wrote was confusing.

Of course the same current flows through each of the series bulbs. However, that current is a larger fraction of the standard current through the 30W bulb than through the 60W bulb, because in normal operation the 60W bulb has more current through it. That's why the 30 W bulb comes closer to lighting normally.

Mike W.

(published on 03/03/09)