2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • 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. Then the internal resistance of the cell is:

    A)  \[2.5\,\,\Omega \]                        

    B)         \[2.0\,\,\Omega \]        

    C)  \[1.54\,\,\Omega \]                      

    D)  \[1.0\,\,\Omega \]

    E)  \[0.5\,\,\Omega \]

    Correct Answer: C

    Solution :

    \[r=R\left( \frac{{{l}_{1}}}{{{l}_{2}}}-1 \right)=10\left( \frac{75}{65}-1 \right)\] \[=10\times 0.154\] \[=1.54\,\Omega \]


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