GATE Electrical Engineering (EE) 2019 Solved Paper

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Question : 51 of 65

Marks: +1, -0

In the circuit below, the operational amplifier is ideal. If

{V}_{1}=10 {mV}

and

{V}_{2}=50 {mV}

, the output voltage

( {V}_{\;\text "out"\;})

is

$500 {mV}$
$600 {mV}$
$400 {mV}$
$100 {mV}$

Solution:

KCL at (a)

{I}_{1}= {I}_{2}

\;{V_{1}-V_{a}}/{10 {K}}=\;{V_{a}-V_{0}}/{100 {K}}

10 {V}_{1}-10 {V}_{ {a}}= {V}_{ {a}}- {V}_{0}

{V}_{0}=11 {V}_{ {a}}-10 {V}_{1}

=11(\;{10}/{11}) V_{2}-10 V_{1}

=10( {V}_{2}- {V}_{1})

=10(40 × 10^{-3})

=400 {mV}

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