Q. Mutual conductance of triode is $2\ m\Omega^{- 1}$ and the amplification factor is 50. Its anode is connected with a source of 250 V and a resistance of $25\ k\Omega$. The voltage gain of this amplifier is
A) 12.5
B) 10
C) 25
D) 50
Explanation
(c) $R_{a} = \frac{\mu}{g_{m}} = \frac{50}{2 \times 10^{- 3}} = 25 \times 10^{3}\Omega;$
$A_{v} = \frac{\mu R_{L}}{R_{a} + R_{L}} = \frac{50 \times 25 \times 10^{3}}{25 \times 10^{3} + 25 \times 10^{3}} = 25$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.