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The circuit shown in the figure employs an operational amplifier that can be considered ideal. Determine the value of the current $I_{0}$ delivered to the load resistance $R_{L}$.
Data: $R_{1}=R_{2}=2[k \Omega], R_{L}=1[k \Omega], V_{\mathbb{N}}=-5[V]$.
(Glossary: $\mathrm{V}=$ volt, $\mathrm{m}=$ meter, $\mathrm{A}=$ ampere, $\mathrm{W}=$ watt, $\Omega=$ ohm, $\mathrm{F}=$ farad, $\mathrm{Hz}=$ hertz, $\mathrm{s}=$ second)
Select one:
a. $I_{0}=-5[\mathrm{~mA}]$
b. $I_{0}=5[\mathrm{~mA}]$
c. $I_{0}=1[\mathrm{~mA}]$
d. $\mathrm{I}_{0}=-1[\mathrm{~mA}]$
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Answer from Sia

Posted about 2 months ago

D

Key Concept

Inverting Operational Amplifier

Explanation

In an inverting operational amplifier, the output voltage is given by $V_o = -\left(\frac{R_2}{R_1}\right) V_{in}$. Given $R_1 = R_2 = 2 \, k\Omega$ and $V_{in} = -5 \, V$, the output voltage $V_o = -\left(\frac{2}{2}\right) (-5) = 5 \, V$. The current through the load resistor $R_L = 1 \, k\Omega$ is $I_0 = \frac{V_o}{R_L} = \frac{5 \, V}{1 \, k\Omega} = 5 \, mA$. Since the current direction is opposite to the assumed direction, $I_0 = -5 \, mA$.

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