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(B) The amplification factor $\mu$ of a triode is given by the product of mutual conductance $(g_m)$ and anode resistance $(r_p)$: $\mu = g_m 	imes r

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$5$

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(B) The amplification factor $\mu$ of a triode is given by the product of mutual conductance $(g_m)$ and anode resistance $(r_p)$: $\mu = g_m 	imes r_p = 2.5 	imes 10^{-3} 	ext{ A/V} 	imes 20 	imes 10^3 	ext{ }\Omega = 50$. For a triode amplifier,the voltage gain $(A)$ is given by the formula: $A = \frac{\mu R_L}{r_p + R_L}$,where $R_L$ is the load resistance. Substituting the given values: $10 = \frac{50 	imes R_L}{20 + R_L}$. Multiplying both sides by $(20 + R_L)$: $10(20 + R_L) = 50 R_L$. $200 + 10 R_L = 50 R_L$. $200 = 40 R_L$. $R_L = \frac{200}{40} = 5 	ext{ k}\Omega$.

$A$ triode whose mutual conductance is $2.5 \text{ mA/V}$ and anode resistance is $20 \text{ k}\Omega$,is used as an amplifier whose amplification factor is $10$. The load resistance connected in the plate circuit will be.....$\text{k}\Omega$.

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