1. **Problem Statement:** Calculate the output voltages $v_{O1}$ and $v_O$, and the currents $i_2$ and $i_4$ for two cascaded inverting op-amp circuits with given resistors $R_1=8\text{k}\Omega$, $R_2=90\text{k}\Omega$, $R_3=22\text{k}\Omega$, $R_4=105\text{k}\Omega$, and input voltage $v_I=0.28\text{ V}$. 2. **Formulas and Rules:** - For an inverting op-amp, the output voltage is $v_O = -\frac{R_f}{R_{in}} v_{in}$ where $R_f$ is the feedback resistor and $R_{in}$ is the input resistor. - Currents through resistors are calculated by Ohm's law: $i = \frac{v}{R}$. - The input to the second op-amp is $v_{O1}$, the output of the first. 3. **Calculate $v_{O1}$:** $$v_{O1} = -\frac{R_2}{R_1} v_I = -\frac{90\text{k}}{8\text{k}} \times 0.28 = -11.25 \times 0.28 = -3.15\text{ V}$$ 4. **Calculate $v_O$:** $$v_O = -\frac{R_4}{R_3} v_{O1} = -\frac{105\text{k}}{22\text{k}} \times (-3.15) = -4.7727 \times (-3.15) = 15.04\text{ V}$$ 5. **Calculate $i_2$ (current through $R_2$):** Voltage across $R_2$ is $v_{O1} - 0 = -3.15 - 0 = -3.15\text{ V}$ (assuming virtual ground at inverting input). $$i_2 = \frac{v_{O1} - 0}{R_2} = \frac{-3.15}{90\text{k}} = -35 \mu A$$ 6. **Calculate $i_4$ (current through $R_4$):** Voltage across $R_4$ is $v_O - 0 = 15.04 - 0 = 15.04\text{ V}$. $$i_4 = \frac{v_O - 0}{R_4} = \frac{15.04}{105\text{k}} = 143.24 \mu A$$ **Final answers:** - $v_{O1} = -3.15\text{ V}$ - $v_O = 15.04\text{ V}$ - $i_2 = -35\mu A$ - $i_4 = 143.24\mu A$