For a given triode,mu = 20. The load resistan | vedclass

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(B) The voltage gain $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu}{1 + \frac{r_p}{R_L}}$.Given that the load resistance $R_L

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$12$

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(B) The voltage gain $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu}{1 + \frac{r_p}{R_L}}$.Given that the load resistance $R_L = 1.5 \, r_p$,where $r_p$ is the anode resistance.Substituting the value of $R_L$ into the formula:$A_v = \frac{\mu}{1 + \frac{r_p}{1.5 \, r_p}} = \frac{\mu}{1 + \frac{1}{1.5}} = \frac{\mu}{1 + \frac{2}{3}} = \frac{\mu}{\frac{5}{3}} = \frac{3}{5} \mu$.Given $\mu = 20$,we calculate:$A_v = \frac{3}{5} 	imes 20 = 3 	imes 4 = 12$.

For a given triode,$\mu = 20$. The load resistance is $1.5$ times the anode resistance. The voltage gain will be:

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