A) 60 W bulb
B) 40 W bulb
C) Both will work
D) None of these
Correct Answer: B
Solution :
Resistance of\[40W\]bulb\[=\frac{240\times 240}{40}\] \[=1440\Omega \] Its safe current\[=\frac{240}{1440}=0.167\,\,A\] Resistance of\[60W\]bulb\[=\frac{240\times 240}{40}\] \[=960\Omega \] Its safe current\[=\frac{240}{960}=0.25\,\,A\] When connected in series to\[420\,\,V\]supply, then the current \[i=\frac{420}{1440+960}=\frac{420}{2400}\] Thus, current is greater for\[40W\]bulb, so it will fuse.You need to login to perform this action.
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