Q. In a $n - p - n$ transistor $10^{10}$ electrons enter the emitter in $10^{- 6}$ s. 4% of the electrons are lost in base. The current transfer ratio will be
A) 0.98
B) 0.97
C) 0.96
D) 0.94
Explanation
(c) No. of electrons reaching the collector,
$n_{c} = \frac{96}{100} \times 10^{10} = 0.96 \times 10^{10}$
Emitter current, $I_{e} = \frac{n_{e} \times e}{t}$
Collector current, $I_{c} = \frac{n_{c} \times e}{t}$
Current transfer ratio, $\alpha = \frac{I_{c}}{I_{e}} = \frac{n_{c}}{n_{e}}$
$= \frac{0.96 \times 10^{10}}{10^{10}} = 0.96$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.