Q. In $NPN$ transistor, $10^{10}$ electrons enter in emitter region in $10^{- 6}s.$ If 2% electrons are lost in base region then collector current and current amplification factor $(\beta)$ respectively are
A) $1.57\ mA,\ 49$
B) $1.92\ mA,\ 70$
C) $2\ mA,\ 25$
D) $2.25\ mA,\ 100$
Explanation
(a) $I_{e} = 10^{10} \times 1.6 \times 10^{- 19} \times \frac{1}{10^{- 6}} = 1.6\ mA\ \ \left\lbrack \because I = \frac{Q}{t} \right\rbrack$
Since 2% electrons are absorbed by base, hence 98% electrons reach the collector, $i.e.,\ \alpha = 0.98$
$\Rightarrow I_{c} = \omega I_{e} = 0.98 \times 1.6 = 1.568\ mA \approx 1.57\ mA$
Also current amplification factor $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.98}{0.02} = 49$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.