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please solve this problem step by step6.5 Find it) in the crcut in Exgure P55 V(t) = 12 cos (377t + 40°) & 10 mH FIGURE P55 6 Deion 4) he nebcck Eu P50 it V(t) = 12 cos (3771 - 60°) V 500 uF

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please solve this problem step by stepUploaded Image6.5 Find it) in the crcut in Exgure P55 V(t) = 12 cos (377t + 40°) & 10 mH FIGURE P55 6 Deion 4) he nebcck Eu P50 it V(t) = 12 cos (3771 - 60°) V 500 uF

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Asked by: jack
Created at: 2025-11-20 22:31:36

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Let's solve both problems step by step: --- ## **Problem 1: Inductor Circuit** ### **Given:** - Source: \( v_s(t) = 12 \cos(377t + 40^\circ) \) V - Inductor: \( L = 10\,\text{mH} = .01\,\text{H} \) ### **Step 1: Phasor Representation** - Convert \( v_s(t) \) to phasor: \[ \vec{V}_s = 12\angle40^\circ \text{ V} \] ### **Step 2: Inductive Reactance** - Angular frequency: \( \omega = 377\,\text{rad/s} \) - Inductive reactance: \[ X_L = \omega L = 377 \times .01 = 3.77\,\Omega \] - Impedance of inductor: \[ Z_L = jX_L = j3.77\,\Omega \] ### **Step 3: Apply Ohm's Law (Phasor Domain)** \[ \vec{I} = \frac{\vec{V}_s}{Z_L} = \frac{12\angle40^\circ}{3.77\angle90^\circ} \] \[ = \frac{12}{3.77}\angle(40^\circ - 90^\circ) \] \[ = 3.18\angle(-50^\circ) \text{ A} \] ### **Step 4: Convert Back to Time Domain** \[ i(t) = 3.18\cos(377t - 50^\circ) \ \text{A} \] --- ## **Problem 2: Capacitor Circuit** ### **Given:** - Source: \( v_s(t) = 12\cos(377t - 60^\circ) \) V - Capacitor: \( C = 500\,\mu\text{F} = 500 \times 10^{-6}\,\text{F} \) ### **Step 1: Phasor Representation** - Convert \( v_s(t) \) to phasor: \[ \vec{V}_s = 12\angle{-60^\circ} \text{ V} \] ### **Step 2: Capacitive Reactance** - Angular frequency: \( \omega = 377\,\text{rad/s} \) - Capacitive reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{377 \times 500 \times 10^{-6}} \] \[ = \frac{1}{.1885} = 5.31\,\Omega \] - Impedance of capacitor: \[ Z_C = -jX_C = -j5.31\,\Omega \] (or \( 5.31\angle{-90^\circ} \,\Omega \)) ### **Step 3: Apply Ohm's Law (Phasor Domain)** \[ \vec{I} = \frac{\vec{V}_s}{Z_C} = \frac{12\angle{-60^\circ}}{5.31\angle{-90^\circ}} \] \[ = \frac{12}{5.31}\angle(-60^\circ - (-90^\circ)) \] \[ = 2.26\angle(30^\circ)\ \text{A} \] ### **Step 4: Convert Back to Time Domain** \[ i(t) = 2.26\cos(377t + 30^\circ) \ \text{A} \] --- ## **Final Answers** ### **1. Inductor Circuit** \[ \boxed{i(t) = 3.18\cos(377t - 50^\circ)\ \text{A}} \] ### **2. Capacitor Circuit** \[ \boxed{i(t) = 2.26\cos(377t + 30^\circ)\ \text{A}} \] Let me know if you'd like a more detailed explanation for any step!

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