The levels of $V_{D S}$ and $I_{D}$ are specified for the network of figure. The level of $R_{D}$ is ___k $Ω$ .

$V_{D S} = \frac{1}{2} V_{D D} a n d I_{D} = I_{D (o n)}$

1.5

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Solution

The correct option is A 1.5
Given, $I_{D} = I_{D (o n)} = 4 m A$

and $V_{G S} = V_{G S (o n)} = 6 V$

For this configuration

$V_{D S} = V_{G S} = \frac{1}{2} V_{D D}$

and $6 V = \frac{1}{2} V_{D D}$
$V_{D D} = 12 V$

Now, $R_{D} = \frac{V_{D D} - V_{D S}}{I_{D} (o n)} = \frac{V_{D D} - \frac{1}{2} V_{D D}}{I_{D} (o n)} = \frac{\frac{1}{2} V_{D D}}{I_{D} (o n)}$

$= \frac{6 V}{4 m A} = 1.5 k Ω$

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