A) 8.1mA
B) 8 mA
C) 10mA
D) 9mA
Correct Answer: C
Solution :
Key Idea: Emitter current is sum of base current and collector current. Electrons are charge carriers within the n-p-n transistor, as well as in the external circuit. The small current entering the base terminal B is the base current \[{{i}_{B}},\] while the larger current entering the collector terminal C is the collector current \[{{i}_{C}}.\] Both these currents combine to leave the emitter terminal E and constitute the emitter currently. Thus, \[{{i}_{E}}={{i}_{B}}+{{i}_{C}}\] Given, \[{{i}_{C}}=9mA\] Also, \[{{i}_{C}}=\left( \frac{90}{100} \right){{i}_{E}}\] \[\Rightarrow \] \[{{i}_{C}}=0.9\,{{i}_{E}}\] \[\therefore \] \[{{i}_{E}}=\frac{{{i}_{C}}}{0.9}=\frac{9mA}{0.9}=10\,mA.\]
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