Q. In a common emitter amplifier, using output resistance of $5000\ \Omega$ and input resistance of $2000\ \Omega$, if the peak value of input signal voltage is 10 mV and $\beta = 50$, then peak value of output voltage is
A) $5 \times 10^{- 6}\ V$
B) $12.50 \times 10^{- 6}\ V$
C) 1.25 V
D) 125.0 V
Explanation
(c) $A_{v} = \frac{\Delta V_{0}}{\Delta V_{i}} = \beta\frac{R_{0}}{R_{i}};\ \ So,\ \mathrm{\Delta}V_{0} = \mathrm{\Delta}V_{i} \times \beta\frac{R_{0}}{R_{i}}$
$\therefore\ \mathrm{\Delta}V_{0} = 10 \times 50 \times \frac{500}{2000} = 1250\ mV = 1.25\ V.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.