Let's analyze the given circuit.

This is a two-stage operational amplifier (op-amp) circuit where:

1. Both op-amps are ideal.
2. $R_1 = 1 \, \text{k}\Omega$, $R_F = 10 \, \text{k}\Omega$.
3. Input voltages are $v_1 = 0.1 \, \text{V}$, $v_2 = 0.3 \, \text{V}$.

We will calculate the output voltage $v_o$.

Step 1: Analyzing the first stage (differential amplifier)

The first op-amp implements a differential amplifier with the output voltage: $v_{out1} = \left( \frac{R_F}{R_1} \right) (v_2 - v_1)$

Substitute the given values: $v_{out1} = \left( \frac{10 \, \text{k}\Omega}{1 \, \text{k}\Omega} \right) (0.3 - 0.1)$ $v_{out1} = 10 (0.2) = 2 \, \text{V}$

Step 2: Analyzing the second stage (inverting amplifier)

The second op-amp is configured as an inverting amplifier, where the output voltage is given by: $v_o = - \left( \frac{R_F}{R_1} \right) v_{out1}$

Substitute the values: $v_o = - \left( \frac{10 \, \text{k}\Omega}{1 \, \text{k}\Omega} \right) (2)$ $v_o = -10 (2) = -20 \, \text{V}$

Final Result

The output voltage of the circuit is: $v_o = -20 \, \text{V}$

Would you like further details or explanation?

Related Questions:

1. What happens if $R_F$ is reduced in the circuit?
2. How does the output change if $v_2$ is set to 0?
3. What is the role of the resistors in the differential amplifier?
4. Can you derive the output voltage formula for the second stage?
5. What happens to the output if the op-amps are not ideal?

Tip:

When solving multi-stage amplifier problems, break the analysis into individual stages and calculate each stage's output sequentially.