The amplification factor of a triode is 10. W | vedclass

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(A) The plate current $I_p$ in a triode is given by the Child-Langmuir law: $I_p = k(V_p + \mu V_g)^{3/2}$,where $V_p$ is the plate potential,$V_g$ is

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$1.69$

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(A) The plate current $I_p$ in a triode is given by the Child-Langmuir law: $I_p = k(V_p + \mu V_g)^{3/2}$,where $V_p$ is the plate potential,$V_g$ is the grid potential,and $\mu$ is the amplification factor.Given $\mu = 10$. For the first case: $4 = k(200 + 10 	imes (-4))^{3/2} = k(200 - 40)^{3/2} = k(160)^{3/2} \dots (i)$.For the second case: $I_p' = k(160 + 10 	imes (-7))^{3/2} = k(160 - 70)^{3/2} = k(90)^{3/2} \dots (ii)$.Dividing equation $(ii)$ by equation $(i)$:$\frac{I_p'}{4} = \frac{k(90)^{3/2}}{k(160)^{3/2}} = \left(\frac{90}{160}ight)^{3/2} = \left(\frac{9}{16}ight)^{3/2}$.$I_p' = 4 	imes \left(\frac{3}{4}ight)^3 = 4 	imes \frac{27}{64} = \frac{27}{16} = 1.6875 \, mA \approx 1.69 \, mA$.

The amplification factor of a triode is $10$. When the plate potential is $200 \, V$ and the grid potential is $-4 \, V$,a plate current of $4 \, mA$ is observed. If the plate potential is changed to $160 \, V$ and the grid potential is kept at $-7 \, V$,then the plate current will be.....$mA$.

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