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(D) The plate resistance $r_p$ is defined as $r_p = \left( \frac{\Delta V_p}{\Delta I_p} ight)_{V_g = 	ext{constant}}$.Taking $V_g = 0$ $V$, we hav

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$6000$

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(D) The plate resistance $r_p$ is defined as $r_p = \left( \frac{\Delta V_p}{\Delta I_p} ight)_{V_g = 	ext{constant}}$.Taking $V_g = 0$ $V$, we have $\Delta V_p = 150 - 120 = 30$ $V$ and $\Delta I_p = (15 - 10) 	ext{ mA} = 5 	imes 10^{-3}$ $A$.Thus, $r_p = \frac{30}{5 	imes 10^{-3}} = 6 	imes 10^3 \Omega = 6000 \Omega$.Alternatively, using $\mu = \left( \frac{\Delta V_p}{\Delta V_g} ight)_{I_p = 	ext{constant}}$ and $g_m = \left( \frac{\Delta I_p}{\Delta V_g} ight)_{V_p = 	ext{constant}}$:For $I_p = 10$ mA, $\mu = \frac{150 - 120}{0 - (-4)} = \frac{30}{4} = 7.5$.For $V_p = 150$ $V$, $g_m = \frac{(15 - 10) 	imes 10^{-3}}{0 - (-4)} = \frac{5 	imes 10^{-3}}{4} = 1.25 	imes 10^{-3} 	ext{ S}$.Then $r_p = \frac{\mu}{g_m} = \frac{7.5}{1.25 	imes 10^{-3}} = 6000 \Omega$.

$V_g$ (volt)	$0$	$-2$	$-4$	$-6$
$I_p$ (mA) for $V_p = 150$ $V$	$15$	$12.5$	$10$	$7.5$
$I_p$ (mA) for $V_p = 120$ $V$	$10$	$7.5$	$5$	$2.5$

The linear portions of the characteristic curves of a triode valve give the following readings. The plate resistance is ....... $\Omega$.
$V_g$ (volt) $0$ $-2$ $-4$ $-6$
$I_p$ (mA) for $V_p = 150$ $V$ $15$ $12.5$ $10$ $7.5$
$I_p$ (mA) for $V_p = 120$ $V$ $10$ $7.5$ $5$ $2.5$

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The linear portions of the characteristic curves of a triode valve give the following readings. The plate resistance is ....... $\Omega$.$V_g$ (volt)$0$$-2$$-4$$-6$$I_p$ (mA) for $V_p = 150$ $V$$15$$12.5$$10$$7.5$$I_p$ (mA) for $V_p = 120$ $V$$10$$7.5$$5$$2.5$