The amplification factor of a triode is 18 an | vedclass

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(C) The voltage gain $A_V$ of a triode amplifier is given by the formula:$A_V = \frac{\mu R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$Given:Ampl

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$10$

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(C) The voltage gain $A_V$ of a triode amplifier is given by the formula:$A_V = \frac{\mu R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$Given:Amplification factor $\mu = 18$Plate resistance $r_p = 8 	imes 10^{3} \ \Omega$Load resistance $R_L = 10^{4} \ \Omega = 10 	imes 10^{3} \ \Omega$Substituting the values:$A_V = \frac{18}{1 + \frac{8 	imes 10^{3}}{10 	imes 10^{3}}} = \frac{18}{1 + 0.8} = \frac{18}{1.8} = 10$Therefore,the voltage gain is $10$.

The amplification factor of a triode is $18$ and its plate resistance is $8 \times 10^{3} \ \Omega$. $A$ load resistance of $10^{4} \ \Omega$ is connected in the plate circuit. The voltage gain will be

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