Q.
In the following common emitter circuit if $\beta = 100,\ V_{CE} = 7V,\ V_{BE} =$ negligible, $R_{C} = 2\ k\Omega$ then $I_{B} = ?$
A) 0.01 $mA$
B) 0.04 $mA$
C) 0.02 $mA$
D) 0.03 $mA$
Explanation
(b) $V = V_{CE} + I_{C}R_{L}$
$\Rightarrow 15 = 7 + I_{C} \times 2 \times 10^{3} \Rightarrow i_{C} = 4\ mA$
$\because\ \beta = \frac{i_{C}}{i_{B}} \Rightarrow i_{B} = \frac{4}{100} = 0.04\ mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.