What defines the brightness of a bulb?


Under assumption that three bulbs are connected to constant voltage, brightness actually changes. Brightness is very loosely proportional to power $P = U I = R I^2 = \frac{U^2}{R}$, so it is necessary to calculate the change of current/voltage through the remaining two bulbs, after the first breaks.

Considering your very case, if all three bulbs are the same and under assumption that resistivity of the bulb does not change with bulb's temperature (typical textbook assumption, which is actually not true), before the bulb breaks, the two paired bulbs have smaller brightness than the sole one. This is because the voltage splits in ratio 2:1 in favor of sole bulb. After the break, both remaining bulbs have the same brightness, because voltage splits 1:1.


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Updated on April 10, 2020


  • Cyrus
    Cyrus over 3 years

    So I have a question. There are three identical bulbs, 2 of them are connected in parallel and the third is basically in series, on the same circuit. If the one of the lamps in parallel breaks, what happens to the brightness of the other two? I don't know how to work this out, and what affects the brightness because I know its power but in this case we have to consider voltage or current.

  • Cyrus
    Cyrus over 11 years
    Ah therefore the bulb which used to be in series brightness will decrease and the one which used to be in parallel, now in series, brightness would increase correct? That is the right answer, I just didnt know how to get it. Also why does the voltage favour the sole bulb? If volatge is shared equally in parallel the two parallel bulbs would have same brightness, and since the bulbs are identical the one in series should have the same brightness as the others since it they have the same resistance. Am I on the right lines?
  • Pygmalion
    Pygmalion over 11 years
    First part you are right. The second part you must replace all three bulbs with resistors. If you have two resistors in parallel and one in series, you can easily show that the two resistors in parallel get less voltage.