Q. For a given triode $\mu = 20$. The load resistance is 1.5 times the anode resistance. The maximum gain will be
A) 16
B) 12
C) 10
D) None of the above
Explanation
(b) Voltage gain $A_{v} = \frac{\mu}{1 + \frac{r_{p}}{R_{L}}}$
$\because R_{L} = 1.5r_{p} \Rightarrow A_{v} = \frac{\mu}{1 + \frac{r_{p}}{1.5r_{p}}} = \frac{3}{5}\mu = \frac{3}{5} \times 20 = 12$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.