A) 1.57 mA, 49
B) 1.92 mA, 70
C) 2 mA, 25
D) 2.25 mA, 100
Correct Answer: A
Solution :
\[{{I}_{e}}={{10}^{10}}\times 1.6\times {{10}^{-19}}\times \frac{1}{{{10}^{-6}}}=1.6\,mA\] \[\left( \because I=\frac{Q}{t} \right)\] Since 2% electrons are absorbed by base, hence 98% electrons reaches the collector i.e. a = 0.98 Þ \[{{I}_{c}}=\alpha {{I}_{e}}=0.98\times 1.6=1.568\,mA\approx 1.57\,mA\] Also current amplification factor\[\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{0.02}=49\]
You need to login to perform this action.
You will be redirected in
3 sec
