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