Q. For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The charge in the collector current when the base current changes by $6\ mA$ is
A) 6 $mA$
B) 4.8 $mA$
C) 24 $mA$
D) 8 $mA$
Explanation
(c) $\alpha = 0.8 \Rightarrow \beta = \frac{0.8}{(1 - 0.8)} = 4$
Also $\beta = \frac{\mathrm{\Delta}i_{c}}{\mathrm{\Delta}i_{b}} \Rightarrow \mathrm{\Delta}i_{c} = \beta \times \mathrm{\Delta}i_{b} = 4 \times 6 = 24mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.