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This circuit represents an operational amplifier (op-amp) in a non-inverting configuration with a voltage divider at the input.

### Step 1: Voltage at Non-inverting Terminal (Input Voltage)
The circuit shows a voltage \( v_1 = 2 \, \text{V} \) applied to a resistive voltage divider, which is made up of two \( 2 \, \text{k}\Omega \) resistors in series. The voltage at the non-inverting input of the op-amp can be calculated as the voltage at the midpoint of the voltage divider:

\[
v_+ = \frac{2 \, \text{k}\Omega}{2 \, \text{k}\Omega + 2 \, \text{k}\Omega} \times 2 \, \text{V} = \frac{1}{2} \times 2 \, \text{V} = 1 \, \text{V}
\]

So, the voltage at the non-inverting terminal \( v_+ = 1 \, \text{V} \).

### Step 2: Gain of the Non-Inverting Amplifier
For a non-inverting op-amp configuration, the output voltage \( v_o \) is related to the input voltage \( v_+ \) by the following formula:

\[
v_o = v_+ \left( 1 + \frac{R_f}{R_{in}} \right)
\]

Where:
- \( v_+ \) is the voltage at the non-inverting terminal.
- \( R_f = 10 \, \text{k}\Omega \) is the feedback resistor.
- \( R_{in} = 1 \, \text{k}\Omega \) is the resistor connected between the inverting terminal and ground.

Substituting the values:

\[
v_o = 1 \, \text{V} \left( 1 + \frac{10 \, \text{k}\Omega}{1 \, \text{k}\Omega} \right) = 1 \, \text{V} \left( 1 + 10 \right) = 1 \, \text{V} \times 11 = 11 \, \text{V}
\]

Thus, the output voltage \( v_o = 11 \, \text{V} \).

### Final Answer:
The output voltage \( v_o \) is \( 11 \, \text{V} \).

Let me know if you would like more details or have any questions!

Here are five related questions to deepen your understanding:
1. How does the voltage divider affect the input signal to the op-amp?
2. How would the output change if the feedback resistor \( R_f \) were reduced to \( 5 \, \text{k}\Omega \)?
3. What would happen to the output voltage if the input voltage \( v_1 \) was increased to \( 5 \, \text{V} \)?
4. Why is the inverting terminal connected to ground in this configuration?
5. How would the circuit behave if the op-amp was not ideal (e.g., considering offset voltage)?

**Tip:** In non-inverting amplifiers, the gain depends on the ratio of the feedback resistor to the input resistor. Changing either can significantly alter the output.

The circuit consists of an ideal operational amplifier. The output voltage v₀ is determined based on the circuit parameters.

This problem involves calculating the output voltage of a non-inverting operational amplifier circuit with a voltage divider. Key concepts include op-amp gain, voltage dividers, and feedback networks.

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