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