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Punjab Medical Punjab - MET Solved Paper-2011

  • question_answer
    A torch battery consisting of two cells of\[1.45\,\,V\]and an internal resistance\[0.15\Omega \], each cell sending current through the filament of the lamps having resistance\[1.5\Omega \]. The value of current will be

    A) \[16.11\,\,A\]

    B) \[1.611\,\,A\]

    C) \[0.1611\,\,A\]

    D) \[2.6\,\,A\]

    Correct Answer: B

    Solution :

    Emf of two cells\[E=2\times 1.45=2.9\,\,V\] Total internal resistance of two cells                 \[r=2\times 0.15=0.3\Omega \]               \[R=1.5\Omega \] Current, \[i=\frac{E}{R+r}\]                 \[i=\frac{2.9}{1.5+0.3}=1.611\,\,A\]


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