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JCECE Medical JCECE Medical Solved Paper-2005

  • question_answer
    The figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 2.5 V cell. The balance point of the cell in the open circuit is 75 cm. When a resistor of \[10\Omega \]is used in the external circuit of the cell, the balance point shifts to 65 cm length of potentiometer wire. The internal resistance of the cell is:

    A) \[2.5\Omega \]          

    B) \[2.0\,\Omega \]

    C)  \[1.54\,\Omega \]

    D) \[1.0\,\Omega \]

    Correct Answer: C

    Solution :

     For a potentiometer, the internal resistance (r) is given by \[r=R\left( \frac{{{l}_{1}}}{{{l}_{2}}}-1 \right)\] Given, \[R=10\,\Omega ,\,{{l}_{1}}=75\,cm,\,{{l}_{2}}=65\,cm\] \[\therefore \]\[r=10\left( \frac{75}{65}-1 \right)=10\times 0.154\,=1.54\,\Omega \]


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