For a triode, {r_p} = 10 text{ k}Omega and {g | vedclass

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(B) The voltage gain ${A_V}$ of a triode is given by the formula: ${A_V} = \frac{\mu}{1 + \frac{r_p}{R_L}}.$
First, we calculate the amplification fa

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$20$

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(B) The voltage gain ${A_V}$ of a triode is given by the formula: ${A_V} = \frac{\mu}{1 + \frac{r_p}{R_L}}.$
First, we calculate the amplification factor $\mu$ using the relation $\mu = {r_p} 	imes {g_m}.$
Given ${r_p} = 10 	imes 10^3 \ \Omega$ and ${g_m} = 3 	imes 10^{-3} \ \mho$, we get $\mu = (10 	imes 10^3) 	imes (3 	imes 10^{-3}) = 30.$
It is given that the load resistance ${R_L} = 2{r_p}.$
Substituting these values into the voltage gain formula:
${A_V} = \frac{30}{1 + \frac{r_p}{2r_p}} = \frac{30}{1 + 0.5} = \frac{30}{1.5} = 20.$
Therefore, the voltage gain is $20$.

For a triode, ${r_p} = 10 \text{ k}\Omega$ and ${g_m} = 3 \text{ m}\mho$. If the load resistance is double the plate resistance, then the value of the voltage gain will be:

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