Q. In a n-p-n transistor amplifier, the collector current is 9 mA. If 90% of the electrons from the emitter reach the collector, then
A) $\alpha = 0.9,\beta$ = 9.0
B) The base current is 10 mA
C) The emitter current is 1 mA
D) $\alpha = 9.0,\beta$ = 0.9
Explanation
(a) Let $I_{e}$ be the emitter current.
$\therefore\ \ \ \ \ \frac{90}{100} \times I_{e}$ =${\ I}_{c}$
$\Rightarrow {\ \ \ \ \ I}_{e}$ = $\frac{100{\ I}_{c}}{90} = \frac{10}{9}{\ I}_{c}$
$\therefore\alpha = \ \frac{{\ I}_{c}}{{\ I}_{e}} = \frac{9}{10} = 0.9$
And
$\beta = \frac{{\ I}_{c}}{{\ I}_{e}} = \frac{{\ I}_{c}}{{\ I}_{e} - {\ I}_{c}} = \frac{1}{\frac{{\ I}_{e}}{{\ I}_{c}} - 1} = \frac{1}{\frac{10}{9} - 1} = \frac{9}{1} = 9$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.