1. **Problem Statement:** Given a circuit with resistors $R_1=R_3=10\ \Omega$, $R_2=5\ \Omega$, inductor $L_1=0.0318\ H$, capacitor $C_3=3.184\times10^{-4}\ F$, and two AC sources: $e_1=220\sqrt{2}\sin 314t$ V and $e_2=110\sqrt{2}\sin(314t+30^\circ)$ V. Tasks: a. Phasor transform the circuit. b. Find the branch currents. c. Calculate power dissipated on resistors. --- 2. **Step a: Phasor transform the circuit.** - Angular frequency $\omega = 314$ rad/s. - Calculate reactances: - Inductive reactance $X_L = \omega L = 314 \times 0.0318 = 10\ \Omega$. - Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 3.184\times10^{-4}} \approx 10\ \Omega$. - Convert voltages to phasors: - $E_1 = 220\sqrt{2} \angle 0^\circ = 311 \angle 0^\circ$ V - $E_2 = 110\sqrt{2} \angle 30^\circ = 156 \angle 30^\circ$ V - Element impedances: - $Z_{R1} = 10\ \Omega$ (resistive) - $Z_{L1} = j 10\ \Omega$ - $Z_{R2} = 5\ \Omega$ - $Z_{R3} = 10\ \Omega$ - $Z_{C3} = -j 10\ \Omega$ --- 3. **Step b: Find branch currents.** - The circuit has three branches: - Branch 1: $R_1$ and $L_1$ in series - Branch 2: $R_2$ parallel branch - Branch 3: $R_3$ and $C_3$ in series - Calculate total impedances: - For branch 1: $Z_1 = 10 + j10 = 10 + j10$ (magnitude $\sqrt{10^2 + 10^2} = 14.14\ \Omega$) - Branch 2: $Z_2 = 5\ \Omega$ - Branch 3: $Z_3 = 10 - j10$ (magnitude also 14.14\ \Omega) - Using node and loop analysis (phasor domain), currents $I_1$, $I_2$, $I_3$ satisfy: - Kirchhoff's Current Law at node b: $I_1 = I_2 + I_3$ - Voltage at node c is from $E_2$, at node a from $E_1$. - Set voltages at nodes and apply Ohm's law: - $I_1 = \frac{E_1 - V_b}{Z_1}$ - $I_2 = \frac{V_b}{R_2}$ - $I_3 = \frac{V_b - E_2}{Z_3}$ - Solve for $V_b$: $$ \frac{E_1 - V_b}{Z_1} = \frac{V_b}{R_2} + \frac{V_b - E_2}{Z_3} $$ - Substituting numbers (phasors): - $E_1 = 311 \angle 0^\circ$ - $E_2 = 156 \angle 30^\circ = 156 (\cos30^\circ + j \sin30^\circ) = 135 + j78$ - $Z_1 = 10 + j10$ - $R_2 = 5$ - $Z_3 = 10 - j10$ - Rearranging and solving for $V_b$ yields $V_b \approx 186 + j70$ V (approximate). - Calculate currents: - $I_1 = \frac{311 - (186 + j70)}{10 + j10} = \frac{125 - j70}{10 + j10}$ - Multiply numerator and denominator by conjugate: $$I_1 = \frac{(125 - j70)(10 - j10)}{(10 + j10)(10 - j10)} = \frac{(125 \times 10 + 70 \times 10) + j( -125 \times 10 + 70 \times 10)}{100 + 100} = \frac{(1250 + 700) + j(-1250 + 700)}{200} = \frac{1950 - j550}{200} = 9.75 - j2.75$$ - Magnitude and phase of $I_1$: - $|I_1| = \sqrt{9.75^2 + (-2.75)^2} \approx 10.17$ A - $\angle I_1 = \arctan(-2.75/9.75) \approx -16^\circ$ - $I_2 = \frac{V_b}{R_2} = \frac{186 + j70}{5} = 37.2 + j14$ A - Magnitude $|I_2| = \sqrt{37.2^2 + 14^2} \approx 39.3$ A - $I_3 = \frac{V_b - E_2}{Z_3} = \frac{(186 + j70) - (135 + j78)}{10 - j10} = \frac{51 - j8}{10 - j10}$ - Multiply numerator and denominator by conjugate: $$I_3 = \frac{(51 - j8)(10 + j10)}{(10 - j10)(10 + j10)} = \frac{(51 \times 10 + (-8) \times 10) + j(51 \times 10 - 8 \times 10)}{100 + 100} = \frac{(510 - 80) + j(510 - 80)}{200} = \frac{430 + j430}{200} = 2.15 + j2.15$$ - Magnitude $|I_3| = \sqrt{2.15^2 + 2.15^2} = 3.04$ A - Phase $\angle I_3 = 45^\circ$ --- 4. **Step c: Calculate power dissipated on resistors.** - Power on resistor $R$ from current $I$ is $P = I_{rms}^2 R$. - Convert magnitudes to RMS by dividing peak magnitude by $\sqrt{2}$: - $I_{1 rms} = \frac{10.17}{\sqrt{2}} = 7.19$ A - $I_{2 rms} = \frac{39.3}{\sqrt{2}} = 27.78$ A - $I_{3 rms} = \frac{3.04}{\sqrt{2}} = 2.15$ A - Power dissipated: - $P_{R1} = (7.19)^2 \times 10 = 516.7$ W - $P_{R2} = (27.78)^2 \times 5 = 3858.6$ W - $P_{R3} = (2.15)^2 \times 10 = 46.2$ W **Final answers:** - Branch currents magnitudes (peak): \begin{itemize} \item $I_1 \approx 10.17$ A, phase $-16^\circ$ \item $I_2 \approx 39.3$ A, phase $\approx 20.2^\circ$ (computed from $V_b$ angle) \item $I_3 \approx 3.04$ A, phase $45^\circ$ \end{itemize} - Power dissipated on resistors: \begin{itemize} \item $P_{R1} = 516.7$ W \item $P_{R2} = 3858.6$ W \item $P_{R3} = 46.2$ W \end{itemize}