A) 100 watt lamp will fuse
B) 40 watt lamp will fuse
C) Both lamps will fuse
D) Neither lamp will fuse
Correct Answer: D
Solution :
Bulb (I) : Rated current \[{{I}_{1}}=\frac{P}{V}=\frac{40}{220}=\frac{2}{11}amp.\] Resistance \[{{R}_{1}}=\frac{{{V}^{2}}}{P}=\frac{{{(220)}^{2}}}{40}=1210\,\Omega \] Bulb (II) : Rated current \[{{I}_{2}}=\frac{100}{220}=\frac{5}{11}amp\] Resistance \[{{R}_{2}}=\frac{{{(220)}^{2}}}{100}=484\,\Omega \] When both are connected in series across 40 V supply Total current through supply \[I=\frac{40}{{{P}_{1}}+{{P}_{2}}}=\frac{40}{1210+484}=\frac{40}{1254}=0.03A\] This current is less than the rated current of each bulb. So neither bulb will fuse. Short Trick : Since VApplied < VRated, neither bulb will fuse.You need to login to perform this action.
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