A) \[9\,\,mA\]
B) \[11\,\,mA\]
C) \[1\,\,mA\]
D) \[0.1\,\,mA\]
Correct Answer: A
Solution :
In an npn-transistor emitter current \[({{i}_{e}})\] is the sum of base current \[({{i}_{b}})\] and collector current \[({{i}_{c}})\] \[{{i}_{e}}={{i}_{b}}+{{i}_{c}}\] Given, \[\frac{90}{100}{{i}_{e}}={{i}_{c}}\] ...(i) Also, \[{{i}_{c}}=10mA\] ...(ii) From Eqs. (i) and (ii), we get \[{{i}_{e}}=\frac{10\times 100}{90}=11mA\]You need to login to perform this action.
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