In a triode amplifier,mu = 25,r_p = 40 text{ | 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: $

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$2.5$

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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: $\mu = 25$,$r_p = 40 	ext{ k}\Omega$,$R_L = 10 	ext{ k}\Omega$,and $V_{	ext{in}} = 0.5 	ext{ V}$.Substituting the values into the gain formula:$A_v = \frac{25}{1 + \frac{40}{10}} = \frac{25}{1 + 4} = \frac{25}{5} = 5$.Since voltage gain $A_v = \frac{V_{	ext{out}}}{V_{	ext{in}}}$,we have $V_{	ext{out}} = A_v 	imes V_{	ext{in}}$.$V_{	ext{out}} = 5 	imes 0.5 	ext{ V} = 2.5 	ext{ V}$.

In a triode amplifier,$\mu = 25$,$r_p = 40 \text{ k}\Omega$,and load resistance $R_L = 10 \text{ k}\Omega$. If the input signal voltage is $0.5 \text{ V}$,then the output signal voltage will be.....$\text{V}$.

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