In a triode,{g_m} = 2 times 10^{-3} , Omega^{ | vedclass

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

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$29.57$

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(B) 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 that $\mu = r_p 	imes g_m$,we can calculate the plate resistance ${r_p}$ as:${r_p} = \frac{\mu}{g_m} = \frac{42}{2 	imes 10^{-3}} = 21000 \, \Omega = 21 \, 	ext{k}\Omega$.Now,substituting the values into the voltage gain formula:${A_v} = \frac{42}{1 + \frac{21 \, 	ext{k}\Omega}{50 \, 	ext{k}\Omega}} = \frac{42}{1 + 0.42} = \frac{42}{1.42} \approx 29.57$.

In a triode,${g_m} = 2 \times 10^{-3} \, \Omega^{-1}$,$\mu = 42$,and the resistance load $R_L = 50 \, \text{k}\Omega$. The voltage amplification obtained from this triode will be:

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