A) 69 mA
B) 0.69 mA
C) 6.9 mA
D) 9.6 mA
Correct Answer: C
Solution :
\[\beta =69=\frac{{{I}_{C}}}{{{I}_{B}}}\] \[\alpha =\frac{\beta }{1+\beta }\,=\frac{69}{70}=\frac{{{I}_{C}}}{{{I}_{E}}}\] \[{{I}_{C}}={{I}_{E}}\times \frac{69}{70}\] \[=\frac{69}{70}\times 7\] \[=6.9\,mA\]
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