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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    Two bulbs 40 W and 60 W and rated voltage 240 V are connected in series across a potential difference of 420 V. Which bulb will work at above its rated voltage?

    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.


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