2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-1997

  • question_answer
    The emf of a Daniel cell is 1.08 V. When the terminals of the cells are connected to a resistance of 3\[\Omega \], the potential difference across the terminals is found to be 0.6 V. Then the internal resistance of the cell is:

    A)  1.8\[\Omega \]                               

    B)  2.4\[\Omega \]

    C)  3.24\[\Omega \]                             

    D)  0.2\[\Omega \]

    Correct Answer: B

    Solution :

                     The emf of cell is\[E=V+ir\]Where\[V=0.6V,\,\] P.D. between terminal of the cell connected to a resistance \[R=3\Omega \] \[E=V+\left( \frac{V}{R} \right)r\] \[1.08=0.6+\left( \frac{0.6}{3} \right)r\]Where V = 0.6 V PD between terminal of  the cell conected a resistance \[R=3\Omega \] \[0.2r=1.08-0.6=0.48\] \[r=\frac{0.48}{0.2}=2.4\Omega \]


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