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AFMC AFMC Solved Paper-2002

  • question_answer
    A 100 W 200 V bulb is connected to a 160 V supply. The actual power consumption would be:

    A) 185 W                                   

    B) 100 W

    C) 54 W                                     

    D) 64 W

    Correct Answer: D

    Solution :

    Key Idea: Power is dissipated in inverse proportion of resistance. Power dissipated is given by \[P=\frac{{{V}^{2}}}{R}\] \[\Rightarrow \]  \[R=\frac{{{V}^{2}}}{P}\] Putting the numerical values, we have \[R=\frac{{{(200)}^{2}}}{100}=400\Omega \] Supply voltage \[{{V}_{s}}=160V\] Actual power consumption is given by \[P=\frac{V_{s}^{2}}{R}=\frac{{{(160)}^{2}}}{400}=64\,W\]


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