A) \[\alpha =0.9,\beta =9.0\]
B) the base current is 10 mA
C) the emitter current is 1 mA
D) \[\alpha =9.0,\text{ }\beta =0.9\]
E) \[\alpha =0.99,\text{ }\beta =99.0\]
Correct Answer: A
Solution :
Let\[{{I}_{e}}\]be the emitter current. \[\therefore \] \[\frac{90}{100}\times {{I}_{e}}={{I}_{c}}\] \[\Rightarrow \] \[{{I}_{e}}=\frac{100{{I}_{c}}}{90}=\frac{10}{9}{{I}_{c}}\] \[\therefore \] \[\alpha =\frac{{{I}_{c}}}{{{I}_{e}}}=\frac{9}{10}=0.9\] And \[\beta =\frac{{{I}_{c}}}{{{I}_{b}}}=\frac{{{I}_{c}}}{{{I}_{e}}-{{I}_{c}}}\] \[=\frac{1}{\frac{{{I}_{e}}}{{{I}_{c}}}-1}\] \[=\frac{1}{10/9-1}\] \[=\frac{9}{1}=9\]
You need to login to perform this action.
You will be redirected in
3 sec
