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JEE Main & Advanced Physics Semiconducting Devices Question Bank Junction Transistor

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    For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA is [Haryana CET 1991]

    A)            6 mA                                        

    B)            4.8 mA

    C)            24 mA                                     

    D)            8 mA

    Correct Answer: C

    Solution :

                       a = 0.8 \[\Rightarrow \beta =\frac{0.8}{(1-0.8)}\]=4 Also \[\beta =\frac{\Delta {{i}_{c}}}{\Delta {{i}_{b}}}\]\[\Rightarrow \Delta {{i}_{c}}=\beta \times \Delta {{i}_{b}}\]= 4 ´ 6 = 24mA.


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