Q. In a common-emitter configuration of a transistor, the base current $I_{b} =$ 2$\mu$A, $\beta = 0.9$ then $I_{c}$ = ?
A) 3.0 $\mu$A
B) 2.25 $\mu$A
C) 4.9 $\mu$A
D) 1.8 $\mu$A
Explanation
(d) Current gain $\beta = \frac{I_{c}}{I_{b}}$
$\therefore$ $I_{c} = \beta \times I_{b} = 0.9 \times 2 = 1.8\ \mu$A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.