Q. For the transistor circuit shown below, if $\beta = 100$, voltage drop between emitter and base is $0.7\ V$ then value of $V_{CE}$ will be
A) 10 $V$
B) 5 $V$
C) 13 $V$
D) 0 $V$
Explanation
(c) $i_{b} = \frac{5 - 0.7}{8.6} = 0.5\ mA \Rightarrow I_{c} = \beta\ I_{b} = 100 \times 0.5\ mA$
By using $V_{CE} = V_{CC} - I_{c}R_{L} = 18 - 50 \times 10^{- 3} \times 100 = 13V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.