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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    The electromotive force of primary cell is 2 V. When it is short circuited it gives a current of 4 A. Its internal resistance in ohm is

    A)  0.5                                        

    B)  5.0

    C)  2.0                                        

    D)  8.0

    Correct Answer: A

    Solution :

    EMF  of cell         \[E=I(R+r)\]when it is short circuited \[R=0\] So, internal resistance \[r=\frac{E}{I}=\frac{2}{4}=0.5\Omega \]


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