For a triode,mu = 64 and g_m = 1600 , mu text | vedclass

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(C) The plate resistance $r_p$ is given by $r_p = \frac{\mu}{g_m} = \frac{64}{1600 	imes 10^{-6} \, 	ext{S}} = 4 	imes 10^4 \, \Omega = 40 \, k\Ome

Vedclass · Accepted Answer

$25.6$

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(C) The plate resistance $r_p$ is given by $r_p = \frac{\mu}{g_m} = \frac{64}{1600 	imes 10^{-6} \, 	ext{S}} = 4 	imes 10^4 \, \Omega = 40 \, k\Omega$.The voltage gain $A_v$ is given by $A_v = \frac{\mu R_L}{r_p + R_L} = \frac{64 	imes 40 	imes 10^3}{40 	imes 10^3 + 40 	imes 10^3} = \frac{64}{2} = 32$.The output signal voltage $V_0$ is $V_0 = A_v 	imes V_i = 32 	imes 1 \, V = 32 \, V 	ext{ (rms)}$.The signal power in the load $P$ is $P = \frac{V_0^2}{R_L} = \frac{(32)^2}{40 	imes 10^3} \, W = \frac{1024}{40} 	imes 10^{-3} \, W = 25.6 	imes 10^{-3} \, W = 25.6 \, mW$.

For a triode,$\mu = 64$ and $g_m = 1600 \, \mu \text{mho}$. It is used as an amplifier and an input signal of $1 \, V \text{ (rms)}$ is applied. The signal power in the load of $40 \, k\Omega$ will be......$mW$.

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