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In the circuit shown below, an input of 1 \mathrm{~V} is fed into the inverting input of an ideal OPamplifier. The output signal V_{\text {out }} will be

Option: 1

+10 \mathrm{~V}


Option: 2

-10 \mathrm{~V}


Option: 3

0 \mathrm{~V}


Option: 4

Infinity


Answers (1)

best_answer

This is operational or OP-inverting amplifier

\mathrm{\begin{aligned} & A=\frac{V_0}{V_i}=-\frac{R_f}{R_i} \\ & \text { Given } V_i=1 V, R_f=10 \mathrm{k} \Omega, R_i=1 \mathrm{k} \Omega \\ & \therefore V_0=-V_i \frac{R_f}{R_i}=-1 \times \frac{10}{1} \Rightarrow V_0=-10 \mathrm{~V} \end{aligned}}

\mathrm{V_0} is negative because \mathrm{V_{\text {input }} is +1 V (positive)}

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Nehul

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