A triode has a mutual conductance of 2 times | vedclass

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Question

(B) Given:Mutual conductance $(g_m)$ = $2 	imes 10^{-3} \, 	ext{mho}$Amplification factor $(\mu)$ = $50$Load resistance $(R_L)$ = $25 	imes 10^3 \,

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$25$

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(B) Given:Mutual conductance $(g_m)$ = $2 	imes 10^{-3} \, 	ext{mho}$Amplification factor $(\mu)$ = $50$Load resistance $(R_L)$ = $25 	imes 10^3 \, \Omega$First,calculate the plate resistance $(r_p)$ using the relation $\mu = r_p 	imes g_m$:$r_p = \frac{\mu}{g_m} = \frac{50}{2 	imes 10^{-3}} = 25 	imes 10^3 \, \Omega$The voltage gain $(A_V)$ of a triode amplifier is given by the formula:$A_V = \frac{\mu 	imes R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$Substituting the values:$A_V = \frac{50}{1 + \frac{25 	imes 10^3}{25 	imes 10^3}} = \frac{50}{1 + 1} = \frac{50}{2} = 25$Therefore,the voltage gain is $25$.

$A$ triode has a mutual conductance of $2 \times 10^{-3} \text{ mho}$ and an amplification factor of $50$. The anode is connected through a resistance of $25 \times 10^3 \, \Omega$ to a $250 \, \text{V}$ supply. The voltage gain of this amplifier is

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