1. **Problem Statement:** Derive the output voltage $V_o$, output current $i_o$, and voltage gain for the first circuit. 2. **Given:** - Voltage source: 3.3V - Resistors: $R_1 = 1k\Omega$, $R_2 = 100k\Omega$, $R_3 = 2.2k\Omega$ - Additional voltage source at node between $R_1$ and $R_2$: +2V 3. **Assumptions and Formula:** - The op-amp is ideal, so input current is zero and voltage at inverting and non-inverting inputs are equal. - Output voltage $V_o$ is across $R_2$ and $R_3$. - Output current $i_o$ flows through $R_3$. 4. **Step 1: Find voltage at node between $R_1$ and $R_2$ (call it $V_x$).** Since the node is connected to +2V source, $V_x = 2V$. 5. **Step 2: Calculate output voltage $V_o$.** The output voltage is at the node after $R_2$ resistor. Using voltage division and considering the op-amp output drives the node: $$V_o = V_x + i_o \times R_2$$ But $i_o$ is the current through $R_3$, which is: $$i_o = \frac{V_o - 0}{R_3} = \frac{V_o}{2.2k}$$ Substitute $i_o$ into the first equation: $$V_o = 2V + \frac{V_o}{2.2k} \times 100k$$ Simplify: $$V_o = 2 + V_o \times \frac{100k}{2.2k} = 2 + V_o \times 45.4545$$ Rearranged: $$V_o - 45.4545 V_o = 2$$ $$-44.4545 V_o = 2$$ $$V_o = -\frac{2}{44.4545} \approx -0.045 V$$ 6. **Step 3: Calculate output current $i_o$.** $$i_o = \frac{V_o}{2.2k} = \frac{-0.045}{2200} \approx -2.05 \times 10^{-5} A = -20.5 \mu A$$ 7. **Step 4: Calculate voltage gain $A_v$.** Voltage gain is output voltage over input voltage difference. Input difference is $3.3V - 2V = 1.3V$. $$A_v = \frac{V_o}{1.3} = \frac{-0.045}{1.3} \approx -0.0346$$ **Final answers:** - Output voltage $V_o \approx -0.045 V$ - Output current $i_o \approx -20.5 \mu A$ - Voltage gain $A_v \approx -0.0346$