In a triode, g_m=2 times 10^{-3} ohm ^{-1} ; mu=42, resistance load, R=50 kilo ohm. The voltage amplification obtained from this triode will be

Q: In a triode, $g_m=2 \times 10^{-3} ohm ^{-1} ; \mu=42$, resistance load, $R=50$ kilo ohm. The voltage amplification obtained from this triode will be

$ \text { Voltage gain } A_v=\frac{\mu}{1+\frac{r_p}{R_L}} \text { and } \mu=r_p \times g_m$ $ \Rightarrow r_p=\frac{42}{2 \times 10^{-3}}=21000 \Omega$ $ \Rightarrow A_v=\frac{42}{1+\frac{21000}{50 \times 10^3}}=29.57$

MCQ

CBSE - English Medium

In a triode, $g_m=2 \times 10^{-3} ohm ^{-1} ; \mu=42$, resistance load, $R=50$ kilo ohm. The voltage amplification obtained from this triode will be

A$30.42$
B$29.57$
C$28.18$
D$ 27.15$

[MNR 1999]

STD 12 Science
JEE physics
Electronics
NCERT

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