Q. A triode whose mutual conductance is $2.5\ m\ A/volt$ and anode resistance is $20\ kilo\ ohm$, is used as an amplifier whose amplification is 10. The resistance connected in plate circuit will be
A) $1\ k\Omega$
B) $5\ k\Omega$
C) $10\ k\Omega$
D) $20\ k\Omega$
Explanation
(b) $\mu = r_{p} \times g_{m} = 20 \times 2.5 = 50$
From $A = \frac{\mu R_{L}}{r_{p} + R_{L}} \Rightarrow r_{p} + R_{L} = \frac{\mu R_{L}}{A} = \frac{50R_{L}}{10} = 5R_{L}$
$\Rightarrow 4R_{L} = r_{p} \Rightarrow R_{L} = \frac{r_{p}}{4} = \frac{20}{4} = 5k\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.