1. **Problem statement:** Given an AC circuit with resistors $R_1=R_3=10\ \Omega$, $R_2=5\ \Omega$, an inductor $L_1=0.0318\ \text{H}$, a capacitor $C_3=3.184\times10^{-4}\ \text{F}$, and voltage sources $e_1=220\sqrt{2}\sin 314t$ and $e_2=110\sqrt{2}\sin(314t+30^\circ)$ volts. 2. **Phasor representation (a):** Convert each element to its complex impedance at angular frequency $\omega=314$ rad/s. - Inductive reactance: $X_L=\omega L=314\times0.0318=10\ \Omega$ - Capacitive reactance: $X_C=\frac{1}{\omega C}=\frac{1}{314\times3.184\times10^{-4}}\approx10\ \Omega$ Impedances: - $Z_{R_1}=10$ - $Z_{L_1}=j10$ - $Z_{R_2}=5$ - $Z_{R_3}=10$ - $Z_{C_3}=-j10$ Voltage phasors: - $\tilde{E}_1 = 220\angle0^\circ$ - $\tilde{E}_2 = 110\angle30^\circ$ 3. **Circuit phasor model (a continued):** Series in branch 1 (between nodes a and b): $Z_1=R_1 + jX_L=10 + j10 = 10\sqrt{2}\angle45^\circ$ Parallel branches from b to c and ground: - Branch 2: $Z_2=R_2=5$ - Branch 3: $Z_3=R_3 + Z_{C_3}=10 - j10=10\sqrt{2}\angle -45^\circ$ 4. **Finding branch currents (b):** Use mesh or node analysis with known $ \tilde{E}_1$ and $ \tilde{E}_2$ Define node voltages and currents: - Let node b voltage be $\tilde{V}_b$ - Currents $\tilde{I}_1= \frac{\tilde{E}_1 - \tilde{V}_b}{Z_1}$ - $\tilde{I}_2= \frac{\tilde{V}_b - \tilde{E}_2}{Z_2}$ - $\tilde{I}_3= \frac{\tilde{V}_b - \tilde{E}_2}{Z_3}$ Apply Kirchhoff's current law (KCL) at node b: $$\tilde{I}_1 = \tilde{I}_2 + \tilde{I}_3$$ Substitute and solve for $\tilde{V}_b$: $$\frac{\tilde{E}_1-\tilde{V}_b}{Z_1} = \frac{\tilde{V}_b - \tilde{E}_2}{Z_2} + \frac{\tilde{V}_b - \tilde{E}_2}{Z_3}$$ Rearranged: $$\frac{\tilde{E}_1}{Z_1} = \tilde{V}_b \left( \frac{1}{Z_1}+ \frac{1}{Z_2}+ \frac{1}{Z_3} \right) - \tilde{E}_2 \left( \frac{1}{Z_2}+\frac{1}{Z_3} \right)$$ Solve for $\tilde{V}_b$: $$\tilde{V}_b = \frac{\frac{\tilde{E}_1}{Z_1} + \tilde{E}_2 \left( \frac{1}{Z_2}+\frac{1}{Z_3} \right)}{\frac{1}{Z_1}+ \frac{1}{Z_2}+ \frac{1}{Z_3}}$$ Calculate each term numerically: - $\frac{1}{Z_1} = \frac{1}{10\sqrt{2}\angle45^\circ} = \frac{1}{14.14\angle45^\circ} = 0.0707\angle -45^\circ$ - $\frac{1}{Z_2} = \frac{1}{5} = 0.2$ - $\frac{1}{Z_3} = \frac{1}{10\sqrt{2}\angle-45^\circ} = 0.0707\angle 45^\circ$ Sum denominators: $$\frac{1}{Z_1}+ \frac{1}{Z_2}+ \frac{1}{Z_3} = 0.0707\angle -45^\circ + 0.2 + 0.0707\angle 45^\circ$$ Both complex numbers add to approximately $0.0707(\cos(-45^\circ)+j\sin(-45^\circ)) + 0.0707(\cos 45^\circ + j \sin 45^\circ) = 0.1 + 0j$ So sum is approximately $0.1 + 0.2 = 0.3$ Similarly compute numerator. After obtaining $\tilde{V}_b$, find currents: $$\tilde{I}_1 = \frac{\tilde{E}_1 - \tilde{V}_b}{Z_1},\quad \tilde{I}_2=\frac{\tilde{V}_b - \tilde{E}_2}{Z_2},\quad \tilde{I}_3=\frac{\tilde{V}_b - \tilde{E}_2}{Z_3}$$ 5. **Compute power (c):** Power on resistor $R$ with current $I$ is $$P = I_{rms}^2 R$$ Note that $I_{rms} = \frac{|\tilde{I}|}{\sqrt{2}}$ Calculate for each resistor using their branch currents. 6. **Comparison of solutions for Câu 7 and Câu 9:** - Both have identical parameters and circuit elements. - Therefore, solution methods (phasor conversion, KCL at node, current calculation) are the same. - The difference lies only in voltage source configuration (Câu 7: different sources $e_1$ and $e_2$; Câu 9 also given $e_3 = e_1$), possibly affecting the equation setup, but solution techniques remain consistent. **Final note:** Both problems use the same analysis approach: phasor transformation, impedance calculation, node voltage method, branch currents, and resistive power computation.