A) 100 watt bulb
B) 25 watt bulb
C) None of them
D) Both of them
Correct Answer: B
Solution :
Resistance of 25 W bulb \[=\frac{220\times 220}{25}=1936\,\Omega \] Its safe current \[=\frac{220}{1936}=0.11\,\,amp.\] Resistance of 100 W bulb \[=\frac{220\times 220}{100}=484\,\Omega \] Its safe current \[=\frac{220}{484}=0.48\,\,amp.\] When connected in series to 440 V supply, then the current \[I=\frac{440}{(1936+484)}=0.18\,amp.\] Thus current is greater for 25 W bulb, so it will fuse.
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