The plate resistance of two triode valves is | vedclass

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(B) The voltage amplification $A_V$ for a triode valve is given by the formula: $A_V = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification fac

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$4/3$

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(B) The voltage amplification $A_V$ for a triode valve is given by the formula: $A_V = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,$r_p$ is the plate resistance,and $R_L$ is the load resistance.Given $\mu = 40$ and $R_L = 4 \, k\Omega$.For the first valve,$r_{p1} = 2 \, k\Omega$,so $A_1 = \frac{40 	imes 4}{2 + 4} = \frac{160}{6} = \frac{80}{3}$.For the second valve,$r_{p2} = 4 \, k\Omega$,so $A_2 = \frac{40 	imes 4}{4 + 4} = \frac{160}{8} = 20$.The ratio of voltage amplification is $\frac{A_1}{A_2} = \frac{80/3}{20} = \frac{80}{3 	imes 20} = \frac{4}{3}$.Wait,re-evaluating the ratio: $\frac{A_1}{A_2} = \frac{r_{p2} + R_L}{r_{p1} + R_L} = \frac{4 + 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}$.

The plate resistance of two triode valves is $2 \, k\Omega$ and $4 \, k\Omega$. The amplification factor of each valve is $40$. The ratio of voltage amplification,when used with a $4 \, k\Omega$ load resistance,will be:

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