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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    In an\[~n-p-n\] transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current je and base current \[IE\]are given by:

    A)  \[{{I}_{E}}=9\,mA,\,{{I}_{B}}=-1\,mA\]

    B)  \[{{I}_{E}}=-1\,mA,\,{{I}_{B}}=9\,mA\]

    C) \[{{I}_{E}}=1\,mA,\,{{I}_{B}}=11\,mA\]

    D)  \[{{I}_{E}}=11\,mA,{{I}_{B}}=1\,mA\]

    Correct Answer: D

    Solution :

    Collector current \[{{I}_{C}}=10\,mA=0.9\,{{I}_{E}}\] (where \[{{I}_{E}}\]the emitter current) Thus      \[{{I}_{E}}=\frac{10}{0.9}=11\,mA\] For \[a\text{ }n-p-n\]transistor base current is given by \[{{I}_{B}}={{I}_{E}}-{{I}_{C}}=11-10=1\,mA\]


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