question_answer

In a npn-transistor, the collector current is\[10\text{ }mA\]. If \[90%\] of the electrons emitted reach the collector, then the emitter current will be

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 \[{{i}_{e}}=\frac{10\times 100}{90}=11mA\]