1. **Problem statement:** Calculate the output voltage $V_o$ of the given ideal op-amp circuit. 2. **Step 1: Analyze Op-Amp 1** - The non-inverting input of Op-Amp 1 is at $-1\text{ V}$. - For an ideal op-amp, the voltage at the inverting input equals the non-inverting input due to the virtual short concept. - Therefore, the inverting input of Op-Amp 1 is at $-1\text{ V}$. 3. **Step 2: Calculate current through 2$\Omega$ resistor connected to 11 V source** - Voltage across the 2$\Omega$ resistor is $11 - (-1) = 12\text{ V}$. - Current $I = \frac{12}{2} = 6\text{ A}$ flowing from 11 V source towards the node. 4. **Step 3: Apply KCL at the node connecting 11 V source, 2$\Omega$ resistor, 1 A current source, and Op-Amp 1 input** - Incoming currents: 6 A (from 11 V source), 1 A (current source). - Outgoing current: through 10$\Omega$ resistor and 1 V source in feedback path. - Total current leaving node: $6 + 1 = 7\text{ A}$. 5. **Step 4: Calculate voltage at output of Op-Amp 1 ($V_1$)** - Voltage drop across 10$\Omega$ resistor with current 7 A is $V = IR = 7 \times 10 = 70\text{ V}$. - The 1 V source is in series, so total voltage at output of Op-Amp 1 is $V_1 = -1 + 1 + 70 = 70\text{ V}$. 6. **Step 5: Analyze Op-Amp 2 feedback network** - Op-Amp 2 has a feedback resistor of 4$\Omega$ and a resistor to ground of 2$\Omega$. - The voltage at the inverting input of Op-Amp 2 equals the non-inverting input voltage (virtual short). - Non-inverting input of Op-Amp 2 is connected to output of Op-Amp 1, so $V_+ = 70\text{ V}$. - Let $V_-$ be voltage at inverting input of Op-Amp 2. - By virtual short, $V_- = 70\text{ V}$. 7. **Step 6: Calculate output voltage $V_o$ of Op-Amp 2** - Using voltage divider rule: $$V_- = V_o \times \frac{2}{4 + 2} = V_o \times \frac{2}{6} = \frac{V_o}{3}$$ - Since $V_- = 70$, we have: $$\frac{V_o}{3} = 70 \implies V_o = 210\text{ V}$$ 8. **Step 7: Potential issues with real op-amps** - The calculated output voltage $V_o = 210\text{ V}$ is very high and likely beyond the supply voltage of real op-amps. - Real op-amps saturate at voltages close to their supply rails, so $V_o$ would saturate and not reach 210 V. - The ideal assumptions ignore input bias currents, offset voltages, and finite gain, which affect real circuit behavior. **Final answer:** $$V_o = 210\text{ V}$$