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 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, \[\alpha =0.9,\,\Delta {{I}_{b}}=4\,mA\] \[\therefore \] \[0.9=\frac{{{I}_{c}}}{4+{{I}_{c}}}\] \[\Rightarrow \] \[0.9(4+{{I}_{c}})={{I}_{c}}\] \[\Rightarrow \] \[3.6+0.9{{I}_{c}}={{I}_{c}}\] \[\Rightarrow \] \[3.6=0.1{{I}_{c}}\] \[\Rightarrow \] \[{{I}_{c}}=36\text{ }mA\]
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