Q. In a NPN transistor, 108 electrons enter the emitter in $10^{- 8}s$. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively
A) 0.8 and 49
B) 0.9 and 90
C) 0.7 and 50
D) 0.99 and 99
Explanation
(d) 108 electrons enter the emitter in $10^{- 8}\ s$
$i.e.,\ i_{E} = \frac{108 \times 1.6 \times 10^{- 19}}{10^{- 8}}A = 172.8 \times 10^{- 11}A$
$\therefore 1\%$ of $i_{E}$ is lost in base $i.e.,\ i_{B} = \frac{i_{E}}{100}$
$\Rightarrow 99\%\ i_{E}\ i.e.,\frac{99}{100}i_{E}$ enters the collector
$\Rightarrow I_{C} = 0.99\ i_{E}$
Current amplification factor
$\beta = \frac{i_{C}}{i_{B}} = \frac{0.99i_{E}}{0.01i_{E}} = 99$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.