Q. Consider an $NPN$ transistor amplifier in common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by $1\ mA$, what will be the change in emitter current
A) 1.1 $mA$
B) 1.01 $mA$
C) 0.01 $mA$
D) 10 $mA$
Explanation
(b) Current gain $\beta = \frac{\Delta i_{c}}{\Delta i_{b}} \Rightarrow \Delta_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5}A = 0.01\ mA$
By using $\mathrm{\Delta}i_{e} = \mathrm{\Delta}i_{b} + \mathrm{\Delta}i_{c} \Rightarrow \mathrm{\Delta}i_{e} = 0.01 + 1 = 1.01mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.