In the given image, we have an op-amp circuit with input voltages $v_1$ and $v_2$, and the task is to find the expression for the output voltage $v_o$ based on these inputs. The problem asks for an expression for $v_o$ in terms of $v_1$ and $v_2$, assuming the operational amplifier is ideal.

Step-by-step analysis:

• The op-amp configuration is a differential amplifier, which takes the difference between $v_1$ and $v_2$ and amplifies it by a factor set by the resistances.
• The formula for the output voltage of a differential amplifier is: $v_o = \left( \frac{R_f}{R_1} \right) v_1 - \left( \frac{R_f}{R_2} \right) v_2$ where:
  • $R_f$ is the feedback resistor (101 k$\Omega$),
  • $R_1$ is the resistor connected to $v_1$ (9.9 k$\Omega$),
  • $R_2$ is the resistor connected to $v_2$ (10.1 k$\Omega$).

Applying the values:

• From the circuit: $v_o = \left( \frac{101 \, \text{k}\Omega}{9.9 \, \text{k}\Omega} \right) v_1 - \left( \frac{101 \, \text{k}\Omega}{10.1 \, \text{k}\Omega} \right) v_2$

• Simplifying the ratios: $\frac{101}{9.9} \approx 10.202, \quad \frac{101}{10.1} \approx 10.000$

Thus, the output voltage expression becomes: $v_o = 10.202 v_1 - 10.000 v_2$

This is the desired expression for the output voltage $v_o$ in terms of $v_1$ and $v_2$.

Would you like me to go into more details or clarify anything? Here are some related questions:

1. How do ideal op-amp assumptions affect the analysis?
2. What happens if we change the feedback resistor value $R_f$?
3. Can we express $v_o$ differently if we use different resistor ratios?
4. How does the input frequency content of $v_1$ and $v_2$ affect the output?
5. What are the conditions for stability in such an op-amp circuit?

Tip: Always double-check resistor values in differential amplifier circuits, as small differences can significantly affect the output gain.