A) \[{{I}_{E}}=9\,mA,\,{{I}_{B}}=-1\,mA\]
B) \[{{I}_{E}}=-1\,mA,\,{{I}_{B}}=9\,mA\]
C) \[{{I}_{E}}=1\,mA,\,{{I}_{B}}=11\,mA\]
D) \[{{I}_{E}}=11\,mA,{{I}_{B}}=1\,mA\]
Correct Answer: D
Solution :
Collector current \[{{I}_{C}}=10\,mA=0.9\,{{I}_{E}}\] (where \[{{I}_{E}}\]the emitter current) Thus \[{{I}_{E}}=\frac{10}{0.9}=11\,mA\] For \[a\text{ }n-p-n\]transistor base current is given by \[{{I}_{B}}={{I}_{E}}-{{I}_{C}}=11-10=1\,mA\]You need to login to perform this action.
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