Q. In a $PNP$ transistor working as a common-base amplifier, current gain is 0.96 and emitter current is $7.2\ mA$. The base current is
A) 0.4 $mA$
B) 0.2 $mA$
C) 0.29 $mA$
D) 0.35 $mA$
Explanation
(c) $\alpha = \frac{i_{c}}{i_{e}} = 0.96$ and $i_{e} = 7.2\ mA$
$\Rightarrow i_{c} = 0.96 \times i_{e} = 0.96 \times 7.2 = 6.91\ mA$
$\therefore i_{e} = i_{c} + i_{b} \Rightarrow 7.2 = 6.91 + i_{b} \Rightarrow i_{b} = 0.29\ mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.