A) 1.1 mA
B) 1.01 mA
C) 0.01 mA
D) 10 mA
Correct Answer: B
Solution :
Current gain is defined as ratio of collector current to base current. \[Current\,\,gain=\frac{change\,\,in\,\,collector\,\,current}{change\,\,in\,\,ba\operatorname{se}\,\,current}\] \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}\] Also \[{{i}_{E}}={{i}_{B}}+{{i}_{C}}\,\,\,\,\Rightarrow \,\,\Delta {{i}_{E}}=\Delta \,{{i}_{B}}+\Delta {{i}_{C}}\] \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}\] Given, \[\beta \,\,=\,\,100,\,\,\Delta {{i}_{C}}=1mA\] \[\therefore \,\,\,\,\,\,100=\frac{1}{\Delta {{i}_{E}}-1}\] \[\Delta {{i}_{E}}-1=\frac{1}{100}=0.01\] \[{{\Delta }_{i}}E=1+0.01=1.01\,\,mA\]
You need to login to perform this action.
You will be redirected in
3 sec
