Let's solve **Problem 6** step by step: --- ## **Problem Statement** Find \( I_R \) in the circuit, given \( V_s = 25\,V \). **Circuit Elements:** - Series branch 1: \( 13\,\Omega \) and \( -j5\,\Omega \) (capacitor) - Series branch 2: \( j12\,\Omega \) (inductor) and \( 10\,\Omega \) (resistor) - Both branches are in parallel across \( V_s \). --- ## **Step 1: Find Equivalent Impedance of Each Branch** ### **Left Branch (Z1):** \[ Z_1 = 13 - j5 \quad \Omega \] ### **Right Branch (Z2):** \[ Z_2 = (j12 + 10) \quad \Omega = 10 + j12 \quad \Omega \] --- ## **Step 2: Find Total Impedance \( Z_{total} \) (Parallel Combination)** \[ \frac{1}{Z_{total}} = \frac{1}{Z_1} + \frac{1}{Z_2} \] \[ Z_{total} = \frac{Z_1 Z_2}{Z_1 + Z_2} \] First, calculate \( Z_1 + Z_2 \): \[ Z_1 + Z_2 = (13 - j5) + (10 + j12) = (23 + j7)\,\Omega \] Now, calculate \( Z_1 Z_2 \): \[ Z_1 Z_2 = (13 - j5)(10 + j12) \] \[ = 13 \times 10 + 13 \times j12 - j5 \times 10 - j5 \times j12 \] \[ = 130 + j156 - j50 + 60 \] \[ = (130 + 60) + (j156 - j50) \] \[ = 190 + j106 \] So, \[ Z_{total} = \frac{190 + j106}{23 + j7} \] --- ### **Divide Complex Numbers** Multiply numerator and denominator by the conjugate of denominator: Conjugate of \( 23 + j7 \) is \( 23 - j7 \): \[ Z_{total} = \frac{(190 + j106)(23 - j7)}{(23 + j7)(23 - j7)} \] \[ = \frac{190 \times 23 + 190 \times (-j7) + j106 \times 23 + j106 \times (-j7)}{23^2 - (j7)^2} \] \[ = \frac{437 - j133 + j2438 - j^2 742}{529 - (-49)} \] Recall: \( j^2 = -1 \), so \( -j^2 742 = +742 \): \[ = \frac{437 + 742 + (-j133 + j2438)}{578} \] \[ = \frac{5112 + j1108}{578} \] \[ Z_{total} \approx 8.845 + j1.917\ \Omega \] --- ## **Step 3: Find Current through \( 10\,\Omega \) Resistor (\( I_R \))** The current \( I_R \) is the current through the right branch (\( Z_2 \)): \[ I_R = \frac{V_s}{Z_2} \] \[ Z_2 = 10 + j12 \] \[ |Z_2| = \sqrt{10^2 + 12^2} = \sqrt{100 + 144} = \sqrt{244} \approx 15.62\,\Omega \] \[ \theta = \arctan\left(\frac{12}{10}\right) \approx 50.19^\circ \] \[ I_R = \frac{25}{15.62} \angle -50.19^\circ = 1.601 \angle -50.19^\circ \text{ A} \] --- ### **Express in Rectangular Form** \[ I_R = 1.601 \times (\cos(-50.19^\circ) + j \sin(-50.19^\circ)) \] \[ \cos(-50.19^\circ) \approx .6402, \quad \sin(-50.19^\circ) \approx -.7682 \] \[ I_R \approx 1.601 \times .6402 + j (1.601 \times -.7682) \] \[ I_R \approx 1.025 - j1.229\ \text{A} \] --- ## **Final Answers** \[ \boxed{ I_R = 1.60 \angle -50.2^\circ\ \text{A} \quad \text{or} \quad I_R = 1.03 - j1.23\ \text{A} } \] --- **Summary of Steps:** 1. Identify and label all impedances. 2. Combine impedances for parallel circuit. 3. Compute total impedance. 4. Calculate branch current \( I_R \) using Ohm's Law. 5. Give final answer in both polar and rectangular form. If you need the answers to the other problems, let me!