1. **Problem statement:** Given an AC circuit with voltage sources and components as follows: - $e_1 = e_3 = 220\sqrt{2} \sin 314t\ (V)$ - $e_2 = 110\sqrt{2} \sin(314t + 30^\circ)\ (V)$ - Resistances: $R_1 = R_3 = 10\ \Omega$, $R_2 = 5\ \Omega$ - Inductance: $L_1 = 0.0318\ H$ - Capacitance: $C_3 = 3.184 \times 10^{-4} F$ Tasks: (a) Find the phasor (complex) form of the circuit. (b) Calculate branch currents $i_1$, $i_2$, and $i_3$. (c) Calculate the power dissipated on each resistor. 2. **Phasor conversion of voltages:** Angular frequency: $\omega = 314\ rad/s$ - $\tilde{E}_1 = 220\sqrt{2} \angle 0^\circ$ - $\tilde{E}_2 = 110\sqrt{2} \angle 30^\circ$ - $\tilde{E}_3 = 220\sqrt{2} \angle 0^\circ$ 3. **Impedances:** - Inductor impedance: $Z_L = j\omega L_1 = j(314)(0.0318) = j9.9852\ \Omega$ - Capacitor impedance: $$Z_C = \frac{1}{j\omega C_3} = \frac{1}{j \times 314 \times 3.184 \times 10^{-4}} = \frac{1}{j0.1} = -j10\ \Omega$$ - Resistors: $Z_{R_1} = Z_{R_3} = 10\ \Omega$, $Z_{R_2} = 5\ \Omega$ 4. **Branch impedances:** - Branch 1 (with $R_1$ and $L_1$ in series): $$Z_1 = R_1 + Z_L = 10 + j9.9852\ \Omega$$ - Branch 2 (only $R_2$): $$Z_2 = 5\ \Omega$$ - Branch 3 ($R_3$ in series with $C_3$): $$Z_3 = R_3 + Z_C = 10 - j10\ \Omega$$ 5. **Set up node equations:** Define voltages and currents with respect to a reference node. Using mesh or node analysis is appropriate here but given the branches and voltages, we use mesh analysis on three loops. 6. **Apply mesh current method:** Define mesh currents $I_1$, $I_2$, and $I_3$ corresponding to branches with impedances $Z_1$, $Z_2$, and $Z_3$ respectively. 7. **Write mesh voltage equations:** - Mesh 1 (with $E_1$): $$E_1 = I_1 Z_1 + (I_1 - I_2) Z_{shared12} + (I_1 - I_3) Z_{shared13}$$ Since branches are separated, conflicts are absent, so simplified as: $$E_1 = I_1 Z_1$$ - Mesh 2 (with $E_2$): $$E_2 = I_2 Z_2$$ - Mesh 3 (with $E_3$): $$E_3 = I_3 Z_3$$ 8. **Calculate branch currents:** - $$I_1 = \frac{E_1}{Z_1} = \frac{220\sqrt{2} \angle 0^\circ}{10 + j9.9852}$$ Calculate magnitude of $Z_1$: $$|Z_1| = \sqrt{10^2 + 9.9852^2} \approx 14.14$$ Phase angle: $$\theta_1 = \arctan \frac{9.9852}{10} \approx 45^\circ$$ Therefore: $$I_1 = \frac{220\sqrt{2}}{14.14} \angle (0^\circ - 45^\circ) \approx 22 \angle -45^\circ \ A$$ - $$I_2 = \frac{E_2}{Z_2} = \frac{110\sqrt{2} \angle 30^\circ}{5} = \frac{155.56 \angle 30^\circ}{5} = 31.11 \angle 30^\circ\ A$$ - $$I_3 = \frac{E_3}{Z_3} = \frac{220\sqrt{2} \angle 0^\circ}{10 - j10}$$ Magnitude of $Z_3$: $$|Z_3| = \sqrt{10^2 + (-10)^2} = \sqrt{200} = 14.14$$ Phase: $$\theta_3 = \arctan \frac{-10}{10} = -45^\circ$$ So: $$I_3 = \frac{220\sqrt{2}}{14.14} \angle (0^\circ - (-45^\circ)) = 22 \angle 45^\circ \ A$$ 9. **Calculate power on resistors:** Power on resistor $R$ is $P = I_{rms}^2 R$ Since current magnitude above is peak ($I_{peak}$), convert to RMS: $$I_{rms} = \frac{I_{peak}}{\sqrt{2}}$$ - For $R_1$: $$P_{R_1} = I_{rms}^2 R_1 = \left(\frac{22}{\sqrt{2}}\right)^2 \times 10 = 11^2 \times 10 = 1210\ W$$ - For $R_2$: $$P_{R_2} = \left(\frac{31.11}{\sqrt{2}}\right)^2 \times 5 = (22)^2 \times 5 = 484 \times 5 = 2420\ W$$ - For $R_3$: $$P_{R_3} = \left(\frac{22}{\sqrt{2}}\right)^2 \times 10 = 11^2 \times 10 = 1210\ W$$ **Final answers:** - Phasor voltages and impedances as above. - Branch currents: $$I_1 = 22 \angle -45^\circ A,$$ $$I_2 = 31.11 \angle 30^\circ A,$$ $$I_3 = 22 \angle 45^\circ A$$ - Powers dissipated: $$P_{R_1} = 1210 W,$$ $$P_{R_2} = 2420 W,$$ $$P_{R_3} = 1210 W$$