A Problem of Brightness

It’s been a while since I posted anything in the Cute Problems folder, so here’s a nice one:

Two physics students are studying at desks in the same room. The two outlets of the wall socket to which their desk lamps are attached have inadvertently been wired in series rather than in parallel. The only bulbs available for the lamps are designed for the nominal mains voltage (240V) : Student A chooses a 200W bulb for his lamp; Student B opts for a 50W bulb for hers.

Which is the brighter student?

Hint: Assume that the filament in each bulb obeys Ohm’s Law.

Answers through the comments box please, preferably with working…..

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11 Responses to “A Problem of Brightness”

  1. Anton Garrett Says:

    I am presuming that the two students don’t know the sockets are in series. P = V^2 /R so, for bulbs rated at the same voltage, the 50W bulb has 4 times the resistance of the 200W bulb. When they are connected in series, 4/5 of the voltage is therefore across the 50W bulb and, since P = VI and I is the same, 4/5 of the power is dissipated in the 50W bulb and 1/5 in the 200W bulb.

    What (Watt?) if the students DO have the wiring diagram? Then it becomes a game-theoretical situation. (I can’t tell if Peter meant this, depending on the pun about ‘brighter’.) What bulb should they buy (a) if they can confer; (b) if they can’t?

    • Anton Garrett Says:

      Suppose that student 1 buys a bulb rated at P_1 watts, and analogously for student 2. Then the power output P’_1 of student 1’s bulb in the series setup is

      P_1 P_2^2 / (P_1 + P_2)^2

      and analogously for student 2. Now for the game theory. Because

      P’_1 / P’_2 = P_2 / P_1

      each student wishes to undercut the other by getting a lower-rated bulb. But the lower the ratings, the higher the total resistance, and the less power that both get…

    • telescoper Says:

      The pun was deliberate, in the sense that I assumed the dimmer student would assume that the bulb with the highest power would produce more light…

      In the example as given the 50W bulb delivers 32W and the 200W bulb only 8W.

  2. stallphill Says:

    Let A be Aaron (m) and B be Betty (f). Simply, it is my experience that the woman is (just about always) brighter.

  3. Ignoring the physical reality that the filament resistance rises rapidly with temperature, so Ohm’s law doesn’t apply (and also ignoring Stefan’s law, and assuming that light output is proportional to electrical power dissipation)…
    The 50W bulb has 4 times the resistance of the 200W bulb. When they are in series the current is the same in both bulbs so the power dissipation is 4 times that of the 200W bulb (and is (16/25)*50W). The 50W bulb is brighter.

    • telescoper Says:

      Yes, Ohm’s law doesn’t really apply to light filaments but there you go. I’m not sure why you need to know Stefan’s Law, however. It seems irrelevant to me.

      It’s a much tougher question if you allow the resistance to vary with temperature, but it’s soluble if you have model for R(T).

      • “I’m not sure why you need to know Stefan’s Law, however. It seems irrelevant to me.”

        Without it, you have to assume that the amount of visible amount emitted is proportional to the electrical power. More voltage here means a higher temperature which will shift the peak of the spectrum. Yes, the higher the temperature, the higher the intensity at any wavelength, but what is seen is the integral over the visible range. Imagine a cool bulb radiating mainly in the infrared, or a very hot one in the ultraviolet.

      • telescoper Says:

        That’s Wien’s Displacement Law rather than Stefan’s law…

      • Yes, but Stefan’s claim that the power is proportional to the fourth power of the temperature refers to the bolometric luminosity, whereas here we have the visual luminosity, thus Wien’s displacement law interacts with Stefan’s law here.

  4. Agreed!

  5. andyinkuwait Says:

    Very common AS Level question that I give to my students quite regularly.

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