1. **Problem Statement:** Find the output voltage $v_o$ and output current $i_o$ in the op amp circuit given $v_s = 12 \cos 5000t$ V. 2. **Circuit Analysis Setup:** The circuit has resistors and capacitors connected to an ideal op amp. We assume the op amp input currents are zero and the voltage difference between the inputs is zero (virtual short). 3. **Impedances:** - Resistor $R_1 = 10\text{k}\Omega = 10,000\ \Omega$ - Capacitor $C_1 = 10\text{nF} = 10 \times 10^{-9} F$ - Resistor $R_2 = 20\text{k}\Omega = 20,000\ \Omega$ - Capacitor $C_2 = 20\text{nF} = 20 \times 10^{-9} F$ Angular frequency $\omega = 5000$ rad/s. Capacitive reactances: $$X_{C1} = \frac{1}{\omega C_1} = \frac{1}{5000 \times 10 \times 10^{-9}} = 20,000\ \Omega$$ $$X_{C2} = \frac{1}{\omega C_2} = \frac{1}{5000 \times 20 \times 10^{-9}} = 10,000\ \Omega$$ 4. **Node Voltage at Non-inverting Input (+):** The node between $R_1$, $C_1$, and $R_2$ is connected to the non-inverting input. Calculate the equivalent impedance from $v_s$ to the node: - $R_1$ in series with $C_1$ impedance $Z_{C1} = -jX_{C1} = -j20,000\ \Omega$ Calculate the voltage divider between $R_1 + C_1$ and $R_2$: Impedance of $R_1$ and $C_1$ in series: $$Z_1 = R_1 - jX_{C1} = 10,000 - j20,000\ \Omega$$ Voltage at node $v_+$: $$v_+ = v_s \times \frac{R_2}{R_2 + Z_1} = 12 \cos 5000t \times \frac{20,000}{20,000 + 10,000 - j20,000}$$ Calculate denominator magnitude and phase: $$20,000 + 10,000 - j20,000 = 30,000 - j20,000$$ Magnitude: $$|30,000 - j20,000| = \sqrt{30,000^2 + 20,000^2} = \sqrt{900,000,000 + 400,000,000} = \sqrt{1,300,000,000} \approx 36,056. \Omega$$ Phase: $$\theta = -\tan^{-1}\left(\frac{20,000}{30,000}\right) = -33.69^\circ$$ Voltage magnitude at node: $$|v_+| = 12 \times \frac{20,000}{36,056} = 6.65 V$$ Phase at node: $$\phi = 0 - (-33.69^\circ) = 33.69^\circ$$ So, $$v_+ = 6.65 \cos(5000t + 33.69^\circ)$$ 5. **Voltage at Inverting Input (-):** Because of the virtual short, $$v_- = v_+ = 6.65 \cos(5000t + 33.69^\circ)$$ 6. **Output Voltage $v_o$:** The output is connected through $C_2$ to ground and the output current $i_o$ flows out of the op amp. The output voltage $v_o$ is related to $v_-$ by the feedback network. Since the inverting input is connected to output through $C_2$, and the input current to the op amp is zero, the current through $C_2$ is: $$i_o = C_2 \frac{d}{dt}(v_o - 0) = C_2 \frac{dv_o}{dt}$$ But also, the current through $R_2$ and $C_2$ must satisfy the node equations. Using the virtual short and KCL, the output voltage is: Since the inverting input voltage equals $v_+$, and the output voltage is across $C_2$, the output voltage is: $$v_o = v_- = 6.65 \cos(5000t + 33.69^\circ)$$ 7. **Output Current $i_o$:** Calculate $i_o$ through $C_2$: $$i_o = C_2 \frac{dv_o}{dt} = 20 \times 10^{-9} \times \frac{d}{dt} \left[6.65 \cos(5000t + 33.69^\circ)\right]$$ Derivative: $$\frac{d}{dt} \cos(5000t + 33.69^\circ) = -5000 \sin(5000t + 33.69^\circ)$$ So, $$i_o = 20 \times 10^{-9} \times 6.65 \times (-5000) \sin(5000t + 33.69^\circ) = -0.000665 \sin(5000t + 33.69^\circ)$$ Or, $$i_o = 0.665 \times 10^{-3} \sin(5000t + 33.69^\circ + 180^\circ) = 0.665 \text{mA} \sin(5000t + 213.69^\circ)$$ **Final answers:** $$v_o = 6.65 \cos(5000t + 33.69^\circ) \text{ V}$$ $$i_o = 0.665 \text{ mA} \sin(5000t + 213.69^\circ)$$