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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    The transfer characteristic curve of a transistor, having input and output resistance \[100\Omega \] and \[100k\Omega \]respectively, is shown in the figure. The Voltage and Power gain, are respectively :                          [JEE Main Held on 12-4-2019 Morning]

    A) \[5\times {{10}^{4}},5\times {{10}^{5}}\]

    B) \[5\times {{10}^{4}},5\times {{10}^{6}}\]

    C) \[5\times {{10}^{4}},2.5\times {{10}^{6}}\]

    D) \[2.5\times {{10}^{4}},2.5\times {{10}^{6}}\]

    Correct Answer: C

    Solution :

    \[{{V}_{gain}}=\left( \frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}} \right)\frac{{{R}_{out}}}{{{R}_{in}}}\]                   \[=\left( \frac{5\times {{10}^{-3}}}{100\times {{10}^{-6}}} \right)\times {{10}^{3}}\]           \[=\frac{1}{20}\times {{10}^{6}}=5\times {{10}^{4}}\]           \[{{P}_{gain}}=\left( \frac{\Delta {{I}_{C}}}{\Delta {{I}_{b}}} \right)({{V}_{gain}})\] \[=\left( \frac{5\times {{10}^{-3}}}{100\times {{10}^{-6}}} \right)(5\times {{10}^{4}})\]            \[=2.5\times {{10}^{6}}\]


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