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CMC Medical CMC-Medical VELLORE Solved Paper-2015

  • question_answer
    A cell of emf X is connected across a resistor R. The potential difference across the wire is measured as Y. The internal resistance of the cell should be

    A)  \[\frac{R-X}{2Y}\]                           

    B)  \[\frac{X-R}{Y+R}\]

    C)  \[\frac{Y}{X}+R\]                            

    D)  \[\frac{X-R}{R}\]

    E)  \[\left( \frac{X-Y}{Y} \right)R\]

    Correct Answer: E

    Solution :

                    Let r be internal resistance of the cell. The current through the circuit \[I=\frac{E}{(R+r)}\] Substituting \[E=x,\]we get                 \[I=\frac{X}{R+r}\] Potential difference across \[R\] is \[V=IR\] when \[V=Y,\] then \[I=\frac{X}{R+r}\] \[\Rightarrow \]                               \[Y=\frac{XR}{R+r}\] which on solving, gives \[r=\left( \frac{X-Y}{Y} \right)R\]


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