A) 0.4 mA
B) 0.2 mA
C) 0.29 mA
D) 0.35 mA
Correct Answer: C
Solution :
\[\alpha =\frac{{{i}_{c}}}{{{i}_{e}}}\]=0.96 and ie = 7.2 mA \[\Rightarrow {{i}_{c}}=0.96\times {{i}_{e}}\]= 0.96 ´ 7.2 = 6.91 mA \[\therefore {{i}_{e}}={{i}_{c}}+{{i}_{b}}\]Þ 7.2 = 6.91 + ib Þ ib = 0.29 mA.
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