A) 100 watt lamp will fuse
B) 40 watt lamp will fuse
C) Both lamps will fuse
D) Neither lamp will fuse
Correct Answer: B
Solution :
[b] Maximum capacity of currents in 40 W \[=\frac{40}{220}=\frac{2}{11}=1.82A\] Maximum capacity of current in 100 W \[=\frac{110}{220}=\frac{5}{11}=0.454A\] Resistances of bulbs are \[\frac{220\times 220}{40}=120\Omega \] & \[\frac{220\times 220}{100}=484\Omega \] Now when they are joined in series with 400 volt. Current \[=\frac{400}{1210+484}=0.236\] As current 0.236 A is more than max. capacity of 40 W bulb so, this bulb will fuse.
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