In the given op-amp circuit, the non-inverting terminal is grounded. The input voltage is 2 V applied through 1 k$\Omega$. The feedback resistor is 1 k$\Omega$. The output is connected to a 2 k$\Omega$ load to ground and also through a 2 k$\Omega$ resistor to the op-amp output. Find the output voltage $V_0$ and currents $I_1$, $I_0$, and $I_x$.
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In an inverting amplifier, first find $V_0$ using gain formula.
Then apply KCL at the output node to compute additional load currents.
Step 1: Identify configuration.
Since the non-inverting terminal is grounded, the circuit is an inverting amplifier.
Using virtual ground concept:
\[
V_- = V_+ = 0.
\] Step 2: Calculate input current.
Input resistor = 1 k$\Omega$, input voltage = 2 V.
Voltage at inverting node = 0 V.
Thus,
\[
I_{in} = \frac{2 - 0}{1k} = 2\,\text{mA}.
\] Step 3: Apply KCL at inverting node.
Since op-amp input current is zero, entire 2 mA flows through feedback resistor (1 k$\Omega$).
Thus,
\[
I_1 = 2\,\text{mA}.
\] Step 4: Find output voltage.
Voltage drop across feedback resistor:
\[
V = (2\,\text{mA})(1k) = 2\,\text{V}.
\]
Since current flows from virtual ground toward output, output must be negative:
\[
V_0 = -2\,\text{V}.
\] Step 5: Current through load resistor (2 k$\Omega$ to ground).
\[
I_0 = \frac{V_0}{2k}
= \frac{-2}{2k}
= -1\,\text{mA}.
\]
Magnitude is 1 mA (flowing upward toward node). Step 6: Current through 2 k$\Omega$ series resistor from op-amp output.
Voltage across this resistor is difference between internal op-amp output and $V_0$.
To maintain $V_0 = -2V$, internal node must supply both feedback current (2 mA) and load current (1 mA).
Thus total current:
\[
I_x = 2\,\text{mA} + 1\,\text{mA} = 3\,\text{mA}.
\] Final Results:
\[
V_0 = -2\,\text{V}
\]
\[
I_1 = 2\,\text{mA}
\]
\[
I_0 = 1\,\text{mA}
\]
\[
I_x = 3\,\text{mA}
\]
