Twenty fairy lights, and one shorts.

A string of 20 identical fairy-light bulbs is connected in series across the 220 V mains. One bulb fails by shorting. The power of each of the other 19 bulbs increases by about:

Answer: 11%.

Each bulb's share of the 220 V goes from one twentieth to one nineteenth of it. The resistance of each bulb stays the same, so its power changes as the square of its voltage.

\frac{P'}{P} = {(\frac{20}{19})}^{2} = \frac{400}{361} \approx 1.108

So each remaining bulb takes about 11% more power. Five percent is the straight-line guess, one bulb in twenty, but power goes as the square of the voltage. That is why one shorted bulb shortens the life of the rest of the string.

Original practice question · JEE Physics Decoded

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