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VIT Engineering VIT Engineering Solved Paper-2015

  • question_answer
    Two sources of equal emf are connected to a resistance R. The internal resistance of these sources are \[{{r}_{1}}\]and \[{{r}_{2}}\]\[\left( {{r}_{1}}>{{r}_{2}} \right).\]If the potential difference across the source having internal Resistance  \[{{r}_{2}}\] is zero, then

    A) \[R=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}}\]                              

    B) \[R={{r}_{2}}\left( \frac{r{{+}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}} \right)\]

    C) \[R=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}+{{r}_{1}}}\]                             

    D)  \[R={{r}_{2}}-{{r}_{1}}\]

    Correct Answer: D

    Solution :

    Let E be the emf of each source. When they are connected in series, then the current in the circuit is given by                \[L=\frac{nh}{2\pi }=\frac{2h}{2\pi }=\frac{h}{\pi }\]                \[\because \] So, potential drop across the cell of intermal resistance \[\lambda =\frac{h}{p}\] is  \[\therefore \] Hence,        \[\lambda \alpha \frac{1}{p}\]                   \[\Rightarrow \] So,             \[\frac{\Delta p}{p}=-\frac{\Delta \lambda }{\lambda }\]


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