A) \[1.2\,\,mA\]
B) \[12\,\,mA\]
C) \[24\,\,mA\]
D) \[36\,\,mA\]
Correct Answer: D
Solution :
The ratio of collector current \[({{I}_{C}})\] to emitter current \[({{I}_{e}})\] is known as current gain \[(\alpha )\] of a transistor. Therefore, \[\alpha =\frac{\Delta {{I}_{c}}}{\Delta {{I}_{e}}}\] ?.(i) Also, emitter current is equal to sum of change of base current and collector current. Therefore, \[\Delta {{I}_{e}}=\Delta {{I}_{b}}+\Delta {{I}_{c}}\] ?..(ii) From Eqs. (i) and (ii), we get \[\alpha =\frac{{{I}_{c}}}{\Delta {{I}_{b}}+\Delta {{I}_{c}}}\] Given, \[\alpha =0.9,\] \[\Delta {{I}_{b}}=4mA\] \[\therefore \] \[0.9=\frac{{{I}_{c}}}{4+{{I}_{c}}}\] \[\Rightarrow \] \[0.9(4+{{I}_{c}})={{I}_{c}}\] \[\Rightarrow \] \[3.6+0.9{{I}_{c}}={{I}_{c}}\] \[\Rightarrow \] \[3.6=0.1{{I}_{c}}\] \[\Rightarrow \] \[{{I}_{c}}=36mA\]
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