CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    In a common emitter transistor amplifier, the output resistance is 500 k\[\Omega \] and the current gain \[\beta =49\]. If the power gain of the amplifier is \[5\times {{10}^{6}},\] the input resistance is

    A)  325 \[\Omega \]             

    B)         165\[\Omega \]                              

    C)  198 \[\Omega \]             

    D)         225\[\Omega \]

    E)  240 \[\Omega \]

    Correct Answer: E

    Solution :

    Given\[{{R}_{o}}=500\,k\Omega ,\beta =49\]and\[P=5\times {{10}^{6}}\] we have    \[P={{\beta }^{2}}\frac{{{R}_{o}}}{{{R}_{i}}}\]                 \[5\times {{10}^{6}}=\frac{{{(49)}^{2}}\times 500}{{{R}_{i}}}\]                 \[{{R}_{i}}=\frac{{{(49)}^{2}}\times 500}{5\times {{10}^{6}}}\]                 \[=240\,\Omega \]


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