2. Solved Papers
  3. JCECE Engineering

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


You need to login to perform this action.
You will be redirected in 3 sec spinner