Q. $S$ certain triode shows the following readings $V_{p}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I_{p}$ 150V $-$2V, 5 mA 150V $-$3.5V, 3.2 mA 195V $-$3.5V, 5 mA The amplification factor of the triode is

A) 22.5
B) 45
C) 30
D) 60
Explanation

(c) $\mu = \left( \frac{\mathrm{\Delta}V_{p}}{\mathrm{\Delta}V_{g}} \right)_{{\mathrm{\Delta}I}_{p}\ is\ constant} = \frac{195 - 150}{- 2 - ( - 3.5)} = \frac{45}{1.5} = 30$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.