A) 0.99, 90
B) 0.96, 79
C) 0.97, 99
D) 0.99, 79
Correct Answer: D
Solution :
Given that, change in emitter current, \[\Delta {{I}_{E}}=8\,mA\] and change in collector current, \[\Delta {{I}_{C}}=7.9\,mA\] We know that, \[\alpha =\frac{\Delta {{I}_{C}}}{\Delta {{I}_{E}}}\Rightarrow \alpha =\frac{7.9}{8}\Rightarrow \alpha \approx 0.99\] Also, we know that \[\beta =\frac{\alpha }{1-\alpha }\Rightarrow \beta =\frac{\frac{7.9}{8}}{-\frac{7.9}{8}}\] \[\Rightarrow \] \[\beta =\frac{7.9}{8-7.9}\Rightarrow \beta =\frac{7.9}{0.1}=79\] Hence, \[\alpha =0.99\] and \[\beta =79\]
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