1. **Problem Statement:** Find the Thévenin voltage ($V_{th}$) and Thévenin impedance ($Z_{th}$) at terminals A-B of the given circuit. Determine the maximum power delivered to a load connected across A-B for: (i) any complex load impedance, (ii) a purely resistive load. 2. **Given Data:** Voltage source: $V_s = 50 \angle 0^\circ$ V Series elements: $j100\ \Omega$ (inductive reactance), $20\ \Omega$ resistor Parallel branch: $50\ \Omega$ resistor and $-j25\ \Omega$ capacitor 3. **Step 1: Find Thévenin Voltage $V_{th}$** - $V_{th}$ is the open-circuit voltage at terminals A-B. - Calculate the voltage across the parallel branch. 4. **Calculate total series impedance before parallel branch:** $$Z_{series} = j100 + 20 = 20 + j100\ \Omega$$ 5. **Calculate parallel impedance $Z_p$ of $50\ \Omega$ resistor and $-j25\ \Omega$ capacitor:** $$\frac{1}{Z_p} = \frac{1}{50} + \frac{1}{-j25} = 0.02 - j0.04$$ $$Z_p = \frac{1}{0.02 - j0.04} = \frac{1}{0.02 - j0.04} \times \frac{0.02 + j0.04}{0.02 + j0.04} = \frac{0.02 + j0.04}{0.02^2 + 0.04^2} = \frac{0.02 + j0.04}{0.002} = 10 + j20\ \Omega$$ 6. **Total circuit impedance $Z_{total}$:** $$Z_{total} = Z_{series} + Z_p = (20 + j100) + (10 + j20) = 30 + j120\ \Omega$$ 7. **Calculate total current $I$ from source:** $$I = \frac{V_s}{Z_{total}} = \frac{50 \angle 0^\circ}{30 + j120}$$ Magnitude of denominator: $$|30 + j120| = \sqrt{30^2 + 120^2} = \sqrt{900 + 14400} = \sqrt{15300} \approx 123.69$$ Angle: $$\theta = \tan^{-1}\left(\frac{120}{30}\right) = \tan^{-1}(4) \approx 75.96^\circ$$ So, $$I = \frac{50}{123.69} \angle -75.96^\circ = 0.4045 \angle -75.96^\circ\ A$$ 8. **Voltage across parallel branch $V_{th}$:** $$V_{th} = I \times Z_p = 0.4045 \angle -75.96^\circ \times (10 + j20)$$ Calculate magnitude and angle of $Z_p$: $$|Z_p| = \sqrt{10^2 + 20^2} = \sqrt{100 + 400} = \sqrt{500} = 22.36$$ $$\phi = \tan^{-1}\left(\frac{20}{10}\right) = 63.43^\circ$$ So, $$V_{th} = 0.4045 \times 22.36 \angle (-75.96^\circ + 63.43^\circ) = 9.04 \angle -12.53^\circ\ V$$ 9. **Step 2: Find Thévenin Impedance $Z_{th}$** - Turn off independent sources: voltage source replaced by short circuit. - Calculate impedance seen from terminals A-B. 10. **Calculate $Z_{th}$:** Source shorted, so series elements $j100 + 20 = 20 + j100\ \Omega$ in series with parallel combination of $50\ \Omega$ and $-j25\ \Omega$. 11. **Parallel impedance $Z_p$ is same as before:** $$Z_p = 10 + j20\ \Omega$$ 12. **Total $Z_{th}$:** $$Z_{th} = (20 + j100) + (10 + j20) = 30 + j120\ \Omega$$ 13. **Step 3: Maximum Power Transfer** - For complex load $Z_L$, maximum power transfer occurs when: $$Z_L = Z_{th}^* = R_{th} - jX_{th}$$ where $Z_{th} = R_{th} + jX_{th}$. 14. **Calculate $R_{th}$ and $X_{th}$:** $$R_{th} = 30\ \Omega, \quad X_{th} = 120\ \Omega$$ 15. **Maximum power delivered to load:** $$P_{max} = \frac{|V_{th}|^2}{8 R_{th}}$$ Calculate $|V_{th}|$: $$|V_{th}| = 9.04\ V$$ So, $$P_{max} = \frac{9.04^2}{8 \times 30} = \frac{81.7}{240} = 0.34\ W$$ 16. **Step 4: Maximum power with purely resistive load** - Load must be real, so $Z_L = R_L$. - Maximum power occurs when $R_L = |Z_{th}|$ (magnitude of $Z_{th}$). 17. **Calculate $|Z_{th}|$:** $$|Z_{th}| = \sqrt{30^2 + 120^2} = 123.69\ \Omega$$ 18. **Calculate power delivered to $R_L = 123.69\ \Omega$:** Voltage across load: $$V_L = V_{th} \times \frac{R_L}{R_L + Z_{th}}$$ Since $Z_{th}$ is complex, power is: $$P = \frac{|V_{th}|^2 R_L}{|Z_{th} + R_L|^2}$$ Calculate denominator: $$|Z_{th} + R_L| = |(30 + j120) + 123.69| = |153.69 + j120| = \sqrt{153.69^2 + 120^2} = \sqrt{23621 + 14400} = \sqrt{38021} = 194.95$$ Power: $$P = \frac{9.04^2 \times 123.69}{194.95^2} = \frac{81.7 \times 123.69}{38021} = \frac{10105}{38021} = 0.266\ W$$ **Final answers:** - $V_{th} = 9.04 \angle -12.53^\circ$ V - $Z_{th} = 30 + j120\ \Omega$ - (i) Maximum power with complex load: $0.34$ W - (ii) Maximum power with purely resistive load: $0.266$ W