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.
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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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