1. **Problem statement:** Given an electrical circuit with resistors $R_1 = R_3 = 10\,\Omega$, $R_2 = 5\,\Omega$, inductor $L_1 = 0.0318\,H$, and capacitor $C_3 = 3.184 \times 10^{-4}\,F$. The voltage sources are $e_1 = 220\sqrt{2} \sin 314t$ (V) and $e_2 = 110\sqrt{2} \sin (314 t + 30^\circ)$ (V). We need to: a. Convert the circuit to phasor (complex) form. b. Find the branch currents. c. Calculate the power on the resistors. 2. **Phasor conversion (a):** - Angular frequency $\omega = 314\ rad/s$ because $\omega = 2\pi f$. - Compute impedances: - For the inductor: $$Z_L = j \omega L = j \times 314 \times 0.0318 = j10\ \Omega$$ - For the capacitor: $$Z_C = \frac{1}{j \omega C} = \frac{1}{j \times 314 \times 3.184 \times 10^{-4}} = -j10\ \Omega$$ - Resistances remain real: $Z_{R1} = 10\ \Omega$, $Z_{R2} = 5\ \Omega$, $Z_{R3} = 10\ \Omega$ - Voltage source phasors: - $\tilde{E}_1 = 220\sqrt{2} \angle 0^\circ = 311\angle 0^\circ$ V - $\tilde{E}_2 = 110\sqrt{2} \angle 30^\circ = 155.56 \angle 30^\circ$ V 3. **Circuit simplification:** - Series combination from source $e_1$: resistor $R_1$ and inductor $L_1$ in series $$Z_{1} = R_1 + Z_L = 10 + j10 = 10 + j10\ \Omega$$ - At node b, parallel branches: - Branch 1: nothing further after $Z_1$ - Branch 2: resistor $R_2 = 5\ \Omega$ - Branch 3: resistor $R_3$ in series with capacitor $C_3$ $$Z_{3} = R_3 + Z_C = 10 - j10\ \Omega$$ 4. **Node voltage method:** - Assign node voltages: let node $a$ be reference (ground) for $e_1$ phasor, voltage at node $b$ is $V_b$, and node $c$ voltage $V_c$ - Voltage source $e_1$ connected at node $a$, voltage source $e_2$ connected at node $c$ - Using KCL at node $b$: $$\frac{V_b - \tilde{E}_1}{Z_1} + \frac{V_b}{R_2} + \frac{V_b - V_c}{Z_3} = 0$$ - At node $c$ (considering $e_2$ connected here): $$V_c = \tilde{E}_2$$ 5. **Calculate admittances:** $$Y_1 = \frac{1}{Z_1} = \frac{1}{10 + j10} = \frac{10 - j10}{(10)^2 + (10)^2} = \frac{10 - j10}{200} = 0.05 - j0.05$$ $$Y_2 = \frac{1}{R_2} = \frac{1}{5} = 0.2$$ $$Y_3 = \frac{1}{Z_3} = \frac{1}{10 - j10} = \frac{10 + j10}{200} = 0.05 + j0.05$$ 6. **Substitute values into KCL at node $b$:** $$ (V_b - 311) \times (0.05 - j0.05) + V_b \times 0.2 + (V_b - V_c) \times (0.05 + j0.05) = 0$$ - Recall $V_c = 155.56 \angle 30^\circ = 134.55 + j77.78$ V 7. **Simplify and collect terms:** $$ V_b (0.05 - j0.05 + 0.2 + 0.05 + j0.05) - 311(0.05 - j0.05) - (0.05 + j0.05)V_c = 0$$ $$ V_b (0.3) = 311(0.05 - j0.05) + (0.05 + j0.05) V_c $$ 8. **Calculate the right side:** - $311 \times 0.05 = 15.55$ - $311 \times (-j0.05) = -j15.55$ - Left term: $15.55 - j15.55$ - $(0.05 + j0.05)(134.55 + j77.78) = 0.05 \times 134.55 + 0.05 \times 77.78 j + j0.05 \times 134.55 + j0.05 \times j77.78$ - Simplify: $$= 6.7275 + 3.889 j + 6.7275 j + j^2 3.889 = 6.7275 + (3.889+6.7275)j - 3.889$$ $$= (6.7275 - 3.889) + 10.6165j = 2.8385 + 10.6165j$$ 9. **Sum right side:** $$ (15.55 - j15.55) + (2.8385 + j10.6165) = (15.55 + 2.8385) + (-15.55 + 10.6165)j = 18.3885 - 4.9335j$$ 10. **Solve for $V_b$:** $$V_b = \frac{18.3885 - 4.9335 j}{0.3} = 61.295 - 16.445 j$$ - Calculate magnitude and angle for $V_b$: $$|V_b| = \sqrt{61.295^2 + (-16.445)^2} = 63.4\ V$$ $$\theta = \arctan \left( \frac{-16.445}{61.295} \right) = -15.03^\circ$$ 11. **Calculate branch currents:** - $\tilde{I}_1 = \frac{V_b - \tilde{E}_1}{Z_1} = \frac{(61.295 - 16.445 j) - 311}{10 + j10}$ - Numerator: $61.295 - 16.445 j - 311 = -249.705 - 16.445 j$ - Multiply numerator and denominator conjugate: $$\tilde{I}_1 = \frac{(-249.705 - 16.445 j)(10 - j10)}{(10 + j10)(10 - j10)} = \frac{-2497.05 + 2497.05 j - 164.45 j - 164.45 j^2}{200}$$ - Since $j^2 = -1$, $$= \frac{-2497.05 + 2332.6 j + 164.45}{200} = \frac{-2332.6 + 2332.6 j}{200} = -11.66 + 11.66 j$$ - Magnitude and angle: $$|I_1|=16.49 A, \angle 135^\circ$$ - $\tilde{I}_2 = \frac{V_b}{R_2} = \frac{61.295 - 16.445 j}{5} = 12.259 - 3.289 j$\ A - $\tilde{I}_3 = \frac{V_b - V_c}{Z_3} = \frac{(61.295 - 16.445 j) - (134.55 + 77.78 j)}{10 - j10} = \frac{-73.255 - 94.225 j}{10 - j10}$ - Multiply numerator and denominator conjugate: $$= \frac{(-73.255 - 94.225 j)(10 + j10)}{(10 - j10)(10 + j10)} = \frac{-732.55 - 732.55 j - 942.25 j - 942.25 j^2}{200}$$ - Since $j^2 = -1$, $$= \frac{-732.55 - 1674.8 j + 942.25}{200} = \frac{209.7 - 1674.8 j}{200} = 1.0485 - 8.374 j$$ - Magnitude and angle: $$|I_3| = 8.45 A, \angle -82.86^\circ$$ 12. **Calculate power on resistors (c):** - Power dissipated on resistor $R_k$ is \ $P_k = R_k |I_k|^2$ (only the resistive part consumes power) - $P_1 = 10 \times (16.49)^2 = 10 \times 271.9 = 2719$ W - $P_2 = 5 \times \left((12.259)^2 + (-3.289)^2\right) = 5 \times (150.3 + 10.81) = 5 \times 161.11 = 805.55$ W - $P_3 = 10 \times (8.45)^2 = 10 \times 71.4 = 714$ W **Final answers:** - a. Phasor impedances: $Z_1 = 10 + j10\,\Omega$, $Z_3 = 10 - j10\,\Omega$ - b. Branch currents magnitudes: $|I_1|=16.49 A$, $|I_2|=12.62 A$, $|I_3|=8.45 A$ - c. Power on resistors: $P_1=2719 W$, $P_2=805.55 W$, $P_3=714 W$