If the op - amp in the circuit shown below is ideal, then the output voltage $V_{0}$ will be

-8V

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-12 V

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6 V

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Solution

The correct option is B -12 V

let I be I(mA)

$V_{x} - V_{0} = 4 I$ ...(i)
$V_{y} + 12 V - V_{x} = 5 I$ ...(ii)
$V_{x} = V_{y}$ (virtual short)
So, from equation (ii),
$5 I = 12 V$
$V_{y} = - I = V_{x}$
So, from equation(i) ,
$V_{y} - V_{0} = 4 I$
$V_{0} = V_{y} - 4 I = - 5 I$
$V_{0} = - 12 V$

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