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Given the circuit in Figure 16-5, what is the RMS current through the inductor? X₁ = 300 Ω 600 Vrms O 2.00 A O 3 A Vs Figure 16 - 5 R = 600 92 Ω

Given the circuit in Figure 16-5, what is the RMS current through the inductor? X₁ = 300 Ω 600 Vrms O 2.00 A O 3 A Vs Figure 16 - 5 R = 600 92 Ω

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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**Circuit Analysis Problem** **Question**: Given the circuit in Figure 16-5, what is the RMS current through the inductor? **Circuit Description**: - The circuit features an AC voltage source with a voltage of 600 Vrms (root mean square voltage). - The inductive reactance (\(X_L\)) is 300 Ω. - A resistor (\(R\)) is in the circuit with a resistance of 600 Ω. **Diagram Explanation**: - The diagram depicts a series circuit configuration. - It includes an AC voltage source labeled \(V_s = 600 \text{ Vrms}\). - An inductor with inductive reactance \(X_L = 300 \, \Omega\). - A resistor with \(R = 600 \, \Omega\). **Answer Options**: - 2.00 A - 3 A - **4.69 mA** (Correct) - 5 A **Explanation**: The problem requires calculating the total impedance of the circuit and then using Ohm’s Law (\(I = \frac{V}{Z}\)) to find the RMS current through the inductor. The impedance in a series RL circuit is calculated using \(Z = \sqrt{R^2 + X_L^2}\). The correct RMS current would then be the result of the division of the Vrms by this total impedance.
Transcribed Image Text:**Circuit Analysis Problem** **Question**: Given the circuit in Figure 16-5, what is the RMS current through the inductor? **Circuit Description**: - The circuit features an AC voltage source with a voltage of 600 Vrms (root mean square voltage). - The inductive reactance (\(X_L\)) is 300 Ω. - A resistor (\(R\)) is in the circuit with a resistance of 600 Ω. **Diagram Explanation**: - The diagram depicts a series circuit configuration. - It includes an AC voltage source labeled \(V_s = 600 \text{ Vrms}\). - An inductor with inductive reactance \(X_L = 300 \, \Omega\). - A resistor with \(R = 600 \, \Omega\). **Answer Options**: - 2.00 A - 3 A - **4.69 mA** (Correct) - 5 A **Explanation**: The problem requires calculating the total impedance of the circuit and then using Ohm’s Law (\(I = \frac{V}{Z}\)) to find the RMS current through the inductor. The impedance in a series RL circuit is calculated using \(Z = \sqrt{R^2 + X_L^2}\). The correct RMS current would then be the result of the division of the Vrms by this total impedance.
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