JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    An ac source has an internal resistance of\[{{10}^{4}}\Omega .\] The turn ratio of a transformer so as to match the source to a load of resistance \[10\Omega \] is

    A) \[4.62\times {{10}^{-2}}\]                            

    B) \[2.03\times {{10}^{-2}}\]

    C) \[3.16\times {{10}^{-2}}\]                            

    D) \[5.62\times {{10}^{-2}}\]

    Correct Answer: C

    Solution :

    For primary & secondary circuit, \[{{V}_{s}}{{l}_{s}}={{V}_{p}}{{l}_{p}}\]where symbols have their usual meaning. \[\Rightarrow \]\[\frac{V_{s}^{2}}{{{Z}_{5}}}=\frac{V_{p}^{2}}{{{Z}_{p}}}\]\[\Rightarrow \]\[\frac{V_{s}^{2}}{{{Z}_{5}}}=\sqrt{\frac{Z_{s}^{{}}}{{{Z}_{p}}}}=\sqrt{\frac{10}{{{10}^{4}}}}=\frac{\sqrt{10}}{100}\] So turns ratio, \[\frac{V_{s}^{{}}}{{{V}_{p}}}=\]turn ratio \[=\frac{\sqrt{10}}{100}=3.16\times {{10}^{-2}}\]

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