Current Electricity

JEE Main level, single correct, medium

266 of 380

Question 266 of 380

Two electric bulbs marked 25 W−220 V25 \mathrm{~W}-220 \mathrm{~V} and 100 W−220 V100 \mathrm{~W}-220 \mathrm{~V} are connected in series to a 440 V440 \mathrm{~V} supply.
Which of the bulbs will fuse?

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Answer: (C) 25 W25 \mathrm{~W}

Resistances of both the bulbs are
R1=V2P1=220225\mathrm{R}_1=\frac{\mathrm{V}^2}{\mathrm{P}_1}=\frac{220^2}{25}
R2=V2P2=2202100\mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}=\frac{220^2}{100}
Hence R1>R2R_1\gt R_2
When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440 V440\mathrm{~V} devides in such a way that voltage across 25 w25 \mathrm{~w} bulb will be more than 220 V220 \mathrm{~V}.
Hence 25 w25 \mathrm{~w} bulb will fuse.

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