A) 89
B) 96
C) 79
D) 99
Correct Answer: C
Solution :
As,\[{{I}_{E}}={{I}_{B}}+{{I}_{C}}\]\[\Rightarrow \]\[\Delta {{I}_{E}}=\Delta {{I}_{B}}+\Delta {{I}_{C}}\] As per question, \[\Delta {{I}_{E}}=8.0mA\] and \[\Delta {{I}_{C}}=7.9mA\] Thus, \[\Delta {{I}_{B}}=0.8-7.9\] \[=0.1mA\] Now, \[\beta =\frac{{{I}_{C}}}{{{I}_{B}}}=\frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}}=\frac{7.9}{0.1}=79\]
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