To determine the output voltage $v_o$ of the operational amplifier (op-amp) circuit in Figure 3, let's break down the analysis. Given that $V_s = 2 \text{ V}$, we can apply the principles of an inverting op-amp circuit, as well as node voltage analysis, to solve for $v_o$.

Step-by-Step Solution:

1. Identify the configuration and assumptions:

   • This is an op-amp circuit with feedback, where we assume the op-amp is in an ideal state (i.e., infinite open-loop gain).
   • This assumption implies the virtual short concept: the inverting input (-) and non-inverting input (+) voltages are equal. Since the non-inverting input is grounded, the inverting input voltage is also effectively at 0 V.

2. Set up node voltage at the inverting input:

   • Let’s denote the current through each resistor:
     • Current through $4 \, k\Omega$ resistor connected to $9 \text{ V}$: $I_1 = \frac{9 \, \text{V} - 0 \, \text{V}}{4 \, k\Omega}$.
     • Current through $4 \, k\Omega$ resistor connected to $V_s = 2 \text{ V}$: $I_2 = \frac{2 \, \text{V} - 0 \, \text{V}}{4 \, k\Omega}$.
     • Current through $8 \, k\Omega$ feedback resistor: $I_3 = \frac{0 \, \text{V} - v_o}{8 \, k\Omega}$.

3. Apply Kirchhoff's Current Law (KCL) at the inverting input:

   • At the inverting input, the sum of currents entering and leaving must equal zero: $I_1 + I_2 = I_3$
   • Substituting the currents: $\frac{9 \, \text{V}}{4 \, k\Omega} + \frac{2 \, \text{V}}{4 \, k\Omega} = \frac{-v_o}{8 \, k\Omega}$

4. Solve for $v_o$:

   • Calculate $I_1$ and $I_2$: $I_1 = \frac{9}{4} = 2.25 \, \text{mA}$ $I_2 = \frac{2}{4} = 0.5 \, \text{mA}$
   • Substitute into the equation: $2.25 + 0.5 = \frac{-v_o}{8}$
   • Simplify: $2.75 = \frac{-v_o}{8}$ $v_o = -2.75 \times 8 = -22 \, \text{V}$

Final Answer:

The output voltage $v_o$ of the op-amp circuit is $-22 \text{ V}$, which differs from the expected answer. Double-checking values might reveal assumptions or additional components affecting the circuit's behavior.

Would you like me to clarify any steps or assumptions in this solution?

Related Questions

1. How would the output change if $V_s$ were increased to 3 V?
2. What if the op-amp had a finite gain instead of being ideal?
3. How does feedback resistance affect the gain of this op-amp configuration?
4. Could we calculate $v_o$ if the op-amp were in a non-inverting configuration?
5. What role does the $2 \, k\Omega$ resistor play in this circuit?

Tip

In ideal op-amp analysis, always check for virtual short assumptions between inputs in feedback configurations to simplify node analysis.