JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer                    An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is \[100\Omega \]and the output load resistance is \[10k\Omega .\]The common emitter current gain b is : [JEE Main 10-4-2019 Morning]

    A) \[60\]                           

    B) \[{{10}^{4}}\]

    C) \[6\times {{10}^{2}}\]                     

    D) \[{{10}^{2}}\]

    Correct Answer: D

    Solution :

    \[{{A}_{\text{v}}}\times \beta ={{P}_{gain}}\]           \[60=10{{\log }_{10}}\left( \frac{P}{{{P}_{0}}} \right)\]           \[P={{10}^{6}}={{\beta }^{2}}\times \frac{{{R}_{out}}}{{{R}_{in}}}\]           \[={{\beta }^{2}}\times \frac{{{10}^{4}}}{100}\]           \[{{\beta }^{2}}={{10}^{4}}\] \[\beta =100\]


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