To analyze these circuits and determine $V_{\text{out}}$, I'll go through each circuit step-by-step, applying principles for operational amplifiers (op-amps), particularly inverting and non-inverting configurations, assuming ideal op-amp conditions (input currents $I_{\text{in}} = 0$ and $V_{+} = V_{-}$).

Circuit-a Analysis

1. Identify Known Voltages and Resistances:

   • $V_{\text{s}} = \pm 15 \text{V}$
   • Input voltage $V_{\text{in}} = 10 \text{V}$
   • Resistors: $R_1 = 6 \text{k}\Omega$ and $R_2 = 4 \text{k}\Omega$

2. Determine the Voltage at the Inverting Input ($V_{-}$): Since this is an inverting amplifier configuration, the virtual ground concept applies, meaning $V_{+} = V_{-}$.

   • With $V_{+}$ connected to ground, $V_{-}$ is also 0 V.

3. Apply Kirchhoff’s Current Law (KCL) at the Inverting Input: Using the inverting amplifier formula: $V_{\text{out}} = - \left( \frac{R_2}{R_1} \right) V_{\text{in}}$ Substituting values: $V_{\text{out}} = - \left( \frac{4 \text{k}\Omega}{6 \text{k}\Omega} \right) \times 10 \text{ V}$ $V_{\text{out}} = - \left( \frac{2}{3} \right) \times 10 \text{ V} = -6.67 \text{ V}$

Final Answer for Circuit-a: $V_{\text{out}} = -6.67 \text{ V}$

Circuit-b Analysis

1. Identify Known Voltages and Resistances:

   • $V_{\text{s}} = \pm 15 \text{V}$
   • The voltage divider with resistors $R_1 = 12 \text{k}\Omega$, $R_2 = 13 \text{k}\Omega$, and $R_3 = 5 \text{k}\Omega$ divides the input voltage (15 V).

2. Calculate the Voltage at the Non-Inverting Input ($V_{+}$): The non-inverting input voltage $V_{+}$ is derived from the voltage divider: $V_{+} = V_{\text{in}} \times \frac{R_3}{R_1 + R_2 + R_3}$ Substituting values: $V_{+} = 15 \text{ V} \times \frac{5 \text{k}\Omega}{12 \text{k}\Omega + 13 \text{k}\Omega + 5 \text{k}\Omega}$ $V_{+} = 15 \text{ V} \times \frac{5}{30} = 15 \text{ V} \times \frac{1}{6} = 2.5 \text{ V}$

3. Calculate the Output Voltage $V_{\text{out}}$: Since this is a non-inverting amplifier: $V_{\text{out}} = \left( 1 + \frac{R_f}{R_{in}} \right) V_{+}$ Here, no feedback resistors are given directly in the diagram, which could imply $V_{\text{out}} = V_{+}$.

Thus: $V_{\text{out}} = 2.5 \text{ V}$

Final Answer for Circuit-b: $V_{\text{out}} = 2.5 \text{ V}$

Summary of Results

• Circuit-a: $V_{\text{out}} = -6.67 \text{ V}$
• Circuit-b: $V_{\text{out}} = 2.5 \text{ V}$

Would you like further details on any step?

Additional Questions for Deeper Understanding:

1. How would the output change in Circuit-a if $R_1$ or $R_2$ were adjusted?
2. How does the virtual ground principle work in inverting op-amp configurations?
3. What would happen if the supply voltages $+V_s$ and $-V_s$ were different from $\pm 15 \text{V}$?
4. Can you analyze the circuit if feedback resistors were included in Circuit-b?
5. What effect does an op-amp's ideal properties have on simplifying circuit analysis?

Tip: Always double-check resistor values and configurations, as small changes can significantly impact op-amp circuit outputs.