To calculate the output voltage $V_0$ of the given op-amp circuit, let’s analyze it step-by-step. The circuit appears to be a difference amplifier, which outputs the amplified difference between the two input voltages, $V_1$ and $V_2$.

The output voltage $V_0$ for a difference amplifier is given by the formula:

$V_0 = \left( \frac{R_4}{R_3} \right) V_2 - \left( \frac{R_2}{R_1} \right) V_1$

Given Values:

• $V_1 = 8 \text{ V}$
• $V_2 = 9.3 \text{ V}$
• $R_1 = 1325 \ \Omega$
• $R_2 = 7425 \ \Omega$
• $R_3 = 9209 \ \Omega$
• $R_4 = 4630 \ \Omega$

Step-by-Step Solution:

1. Calculate $\frac{R_4}{R_3}$: $\frac{R_4}{R_3} = \frac{4630}{9209} \approx 0.503$

2. Calculate $\frac{R_2}{R_1}$: $\frac{R_2}{R_1} = \frac{7425}{1325} \approx 5.604$

3. Calculate each term in the formula:

   • For the $V_2$ term: $\left( \frac{R_4}{R_3} \right) V_2 = 0.503 \times 9.3 \approx 4.6779 \text{ V}$
   • For the $V_1$ term: $\left( \frac{R_2}{R_1} \right) V_1 = 5.604 \times 8 \approx 44.832 \text{ V}$

4. Calculate $V_0$: $V_0 = 4.6779 - 44.832 \approx -40.1541 \text{ V}$

Result:

$V_0 \approx -40.15 \text{ V}$

Would you like further details or have any questions about the solution?

Additional Questions

1. How does changing $R_3$ or $R_4$ affect the output $V_0$?
2. What happens to $V_0$ if $V_2$ is decreased to match $V_1$?
3. How can we adjust $R_1$ and $R_2$ to make $V_0$ closer to zero?
4. Can you explain the concept of a difference amplifier and its practical applications?
5. How would the output change if the op-amp were configured as a summing amplifier instead?

Tip

When analyzing op-amp circuits, always check if it’s inverting, non-inverting, or difference mode—each has a specific output formula.