In the FET amplifier circuit shown in figure ,the signal outputs $V_{1}$ and $V_{2}$ are related as

V_{2} = \frac{R_{d}}{R_{s}} . V_{1}

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V_{2} = \frac{- R_{s}}{R_{d}} . V_{1}

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V_{2} = \frac{- R_{p}}{R_{d}} . V_{1}

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V_{2} = \frac{- R_{d}}{R_{s}} . V_{1}

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Solution

The correct option is D $V_{2} = \frac{- R_{d}}{R_{s}} . V_{1}$
$I_{D}$ is the current flowing through $R_{d}$

$V_{D D} = 0 V$ for AC analysis

From the given amplifier circuit,

$V_{2} = - I_{D} R_{d}$ .... (i)

and $V_{1} = I_{D} R_{S}$ .... (ii)

From equation (i) and (ii) , we get

$∴ \frac{V_{2}}{V_{1}} = \frac{- R_{d}}{R_{s}}$
$V_{2} = \frac{- R_{d}}{R_{s}} V_{1}$

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In the FET amplifier circuit shown in figure ,the signal outputs V1 and V2 are related as

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