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WB JEE Medical WB JEE Medical Solved Paper-2010

  • question_answer
    In a common emitter configuration, a transistor has \[\beta =50\] and input resistance \[\text{1}\,\text{k }\!\!\Omega\!\!\text{ }\text{.}\] If the peak value of AC input is 0.01 V, then the peak value of collector current is

    A)  \[0.01\,\mu A\]                              

    B)  \[0.25\,\mu A\]

    C)  \[100\,\mu A\]                

    D)  \[500\,\mu A\]

    Correct Answer: D

    Solution :

                     Given that the current gain of the transistor is \[\beta =50.\] Input resistance \[=1\,k\Omega ,\]input voltage = 0.01 V Hence,     \[50=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}\Rightarrow \Delta {{i}_{C}}=50\Delta {{i}_{B}}\]     Also the change in base current is \[\Delta {{i}_{B}}=\frac{\text{input}\,\text{voltage}}{\text{input}\,\text{resistance}}=\frac{0.01}{1\times {{10}^{3}}}={{10}^{-5}}A\] So, \[\Delta {{i}_{C}}=50\times {{10}^{-5}}=500\mu A\]


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