A circuit using an ideal OP-AMP is shown in the Figure. Which of the following options gives the correct value of the current I_x?

Correct Option: 3.0 mA

To find the current $I_x$ through the resistor $R_x$, we can perform a nodal analysis on the given operational amplifier (op-amp) circuit.

Step 1: Apply the Virtual Ground Property
For an ideal operational amplifier with negative feedback, the voltages at the inverting ($-$) and non-inverting ($+$) input terminals are equal. Since the non-inverting terminal is connected directly to the ground:

$$V_{+}=0\text{V}$$

Therefore, by virtual ground:

$$V_{-}=V_{+}=0\text{V}$$

Step 2: Nodal Analysis at the Inverting Node ($V_{-}$)
Let the node connecting the feedback resistor ($1\text{k}\Omega$), the resistor $R_x$ ($2\text{k}\Omega$), and the load resistor ($2\text{k}\Omega$) be denoted as $V_o\text{'}$.
Applying Kirchhoff's Current Law (KCL) at the inverting input terminal node:

$$\frac{V_i-V_{-}}{1\text{k}\Omega }+\frac{V_o\text{'}-V_{-}}{1\text{k}\Omega }=0$$

Substituting the given values $V_i=2\text{V}$ and $V_{-}=0\text{V}$:

$$\frac{2-0}{1000}+\frac{V_o\text{'}-0}{1000}=0$$

$$2+V_o\text{'}=0\Rightarrow V_o\text{'}=-2\text{V}$$

Step 3: Nodal Analysis at Node $V_o\text{'}$ to Find $I_x$
Let $I_x$ be the current flowing from the node $V_o\text{'}$ towards the op-amp output terminal through the resistor $R_x$.
Applying KCL at node $V_o\text{'}$:

$$I_{\text{feedback}}+I_{\text{load}}+I_x=0$$

Where:
- Current coming from the inverting terminal through the feedback resistor:

$$I_{\text{feedback}}=\frac{V_o\text{'}-V_{-}}{1\text{k}\Omega }=\frac{-2-0}{1000}=-2\text{mA}$$

- Current going to the ground through the $2\text{k}\Omega$ load resistor:

$$I_{\text{load}}=\frac{V_o\text{'}-0}{2\text{k}\Omega }=\frac{-2}{2000}=-1\text{mA}$$

Substituting these values back into the KCL equation:

$$-2\text{mA}-1\text{mA}+I_x=0$$

$$I_x=3\text{mA}$$

Thus, the magnitude of the current $I_x$ is indeed 3.0 mA.