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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    When a resistance of 2\[\Omega \] is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5\[\Omega ,\] the current is 0.25 A. The emf of the cell is

    A)  1.0 V                                    

    B)  1.5V

    C)  2.0V                                     

    D)  2.5V

    Correct Answer: B

    Solution :

    EMF of the cell is given by \[E=I(R+r)\] For \[2\Omega \] resistance \[E=0.5(2+r)\]         ... (i) For \[5\Omega \] resistance  \[E=0.25(5+r)\]       ... (ii) Solving these two equations, we get                 \[r=1\Omega \] Now, from Eq. (i), we get                 \[E=1.5V\]


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