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In the circuit shown above, an input of $1 \mathrm{~V}$ is fed into the inverting input of an ideal Op-amp A. The output signal $\mathrm{V}_{\text {out }}$ will be
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$-10 \mathrm{~V}$
For the Op-amp shown above, we have
$\frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{V}_{\mathrm{I}}}=-\frac{\mathrm{R}_{\mathrm{f}}}{\mathrm{R}_{\mathrm{I}}}$
Comparing this circuit with the given one, We get $\mathrm{V}_{\mathrm{I}}=1 \mathrm{~V}, \mathrm{R}_{\mathrm{f}}=10 \mathrm{k} \Omega=10 \times 10^{3} \Omega$ $\mathrm{R}_{\mathrm{I}}=1 \mathrm{k} \Omega=1 \times 10^{3} \Omega$
$\therefore \frac{\mathrm{V}_{\mathrm{o}}}{1}=-\frac{10 \times 10^{3}}{1 \times 10^{3}} \Rightarrow \mathrm{V}_{\mathrm{o}}=-10 \mathrm{~V}$
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