1. **Problem Statement:** Find the Thévenin voltage ($V_{th}$) and Thévenin impedance ($Z_{th}$) seen from terminals A-B of the given circuit. Then determine the maximum power delivered to a load connected across A-B for (i) any complex load and (ii) a purely resistive load. 2. **Given:** Voltage source $V_s = 50 \angle 0^\circ$ V Impedances: $j100 \Omega$ (inductor), $20 \Omega$ (resistor), $-j25 \Omega$ (capacitor), and $50 \Omega$ resistor. 3. **Step 1: Find $V_{th}$ (open-circuit voltage at A-B):** - Terminal B is at the bottom node of the $50 \Omega$ resistor. - The series path is $V_s$ in series with $j100 \Omega$, $20 \Omega$, and $-j25 \Omega$. - The $50 \Omega$ resistor is connected in parallel with the series combination of $j100 \Omega$ and $20 \Omega$. Calculate the parallel combination of $j100 \Omega$ and $50 \Omega$: $$Z_p = \frac{j100 \times 50}{j100 + 50} = \frac{j5000}{50 + j100}$$ Multiply numerator and denominator by conjugate $50 - j100$: $$Z_p = \frac{j5000(50 - j100)}{(50 + j100)(50 - j100)} = \frac{j5000 \times 50 - j5000 \times j100}{50^2 + 100^2} = \frac{j250000 + 500000}{2500 + 10000} = \frac{500000 + j250000}{12500} = 40 + j20 \Omega$$ Now, $Z_p$ is in series with $20 \Omega$ resistor: $$Z_{series} = 20 + Z_p = 20 + 40 + j20 = 60 + j20 \Omega$$ The capacitor $-j25 \Omega$ is connected in series after this, so total impedance from source to terminal A is: $$Z_{total} = j100 + Z_{series} - j25 = j100 + 60 + j20 - j25 = 60 + j(100 + 20 - 25) = 60 + j95 \Omega$$ Calculate current from source: $$I = \frac{V_s}{Z_{total}} = \frac{50 \angle 0^\circ}{60 + j95}$$ Magnitude of denominator: $$|Z_{total}| = \sqrt{60^2 + 95^2} = \sqrt{3600 + 9025} = \sqrt{12625} \approx 112.36$$ Angle of denominator: $$\theta = \tan^{-1}\left(\frac{95}{60}\right) \approx 57.5^\circ$$ So, $$I = \frac{50}{112.36} \angle -57.5^\circ = 0.445 \angle -57.5^\circ \text{ A}$$ Voltage at terminal A (across $-j25 \Omega$ capacitor): $$V_A = I \times (-j25) = 0.445 \angle -57.5^\circ \times 25 \angle -90^\circ = 11.125 \angle (-57.5 - 90)^\circ = 11.125 \angle -147.5^\circ \text{ V}$$ Terminal B is at the bottom node of the $50 \Omega$ resistor, which is reference (0 V), so: $$V_{th} = V_A - V_B = 11.125 \angle -147.5^\circ - 0 = 11.125 \angle -147.5^\circ \text{ V}$$ 4. **Step 2: Find $Z_{th}$ (Thévenin impedance):** - Turn off independent sources: replace $V_s$ by short circuit. - Calculate impedance seen from terminals A-B. From terminal A looking back: - Capacitor $-j25 \Omega$ in series with the parallel combination of $j100 \Omega$ and $50 \Omega$ plus $20 \Omega$ resistor. Calculate parallel of $j100$ and $50$ again: $$Z_p = 40 + j20 \Omega$$ Add $20 \Omega$ resistor: $$Z_{series} = 20 + Z_p = 60 + j20 \Omega$$ Add capacitor $-j25 \Omega$ in series: $$Z_{th} = Z_{series} - j25 = 60 + j(20 - 25) = 60 - j5 \Omega$$ 5. **Step 3: Maximum power delivered to load:** (i) For any complex load $Z_L$: - Maximum power transfer occurs when $Z_L = Z_{th}^*$ (complex conjugate of $Z_{th}$). - So, $$Z_L = 60 + j5 \Omega$$ - Maximum power: $$P_{max} = \frac{|V_{th}|^2}{8 R_{th}}$$ where $R_{th} = 60 \Omega$ and $$|V_{th}| = 11.125 V$$ Calculate: $$P_{max} = \frac{11.125^2}{8 \times 60} = \frac{123.77}{480} \approx 0.2579 \text{ W}$$ (ii) For purely resistive load $R_L$: - Maximum power transfer when $R_L = R_{th} = 60 \Omega$ (since load must be real). - Power delivered: $$P = \frac{|V_{th}|^2}{(R_{th} + R_L)^2} R_L = \frac{123.77}{(60 + 60)^2} \times 60 = \frac{123.77}{14400} \times 60 = 0.515 \text{ W}$$ **Final answers:** - $V_{th} = 11.125 \angle -147.5^\circ$ V - $Z_{th} = 60 - j5 \Omega$ - Maximum power (complex load): $0.258$ W - Maximum power (pure resistance load): $0.515$ W