A) 40 W bulb
B) 60 W bulb
C) Both will work
D) None of the above
Correct Answer: B
Solution :
Resitance \[{{427}^{o}}C\] Resistance of 40 W bulb \[{{27}^{o}}C\] \[0.52\] Its safe current \[0.31\] Resistance of 60 W bulb \[0.81\] \[0.42\] Its safe current \[{{r}_{A}}\] When connected in series to 420 V supply, then the current \[{{r}_{B}}\] \[{{v}_{A}}\] Thus, current is greater for 40 W bulb, so it will fuse.
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