A) the emitter current will be nearly 9 mA and the base current will be nearly 1 mA
B) the emitter current will be nearly 11 mA and the base current will be nearly 9 mA
C) the emitter current will be nearly 11 mA and the base current will be nearly 1 mA
D) the emitter and base currents will be 10 mA and 1 mA respectively
Correct Answer: C
Solution :
Given that \[{{i}_{c}}=10\,mA\] According to question \[{{i}_{E}}\times \frac{90}{100}=10\] \[{{i}_{E}}\approx 11\,mA\] Now from \[{{i}_{E}}={{i}_{b}}+{{i}_{c}}\] \[{{i}_{b}}=1\,mA\]
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