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An AC sine wave has an instantaneous value of current of 11.917A at 37°. What is the RMS value of current? A. 14A В. 13А С. 12A D. 11A

An AC sine wave has an instantaneous value of current of 11.917A at 37°. What is the RMS value of current? A. 14A В. 13А С. 12A D. 11A

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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### AC Sine Wave RMS Value Calculation **Question:** An AC sine wave has an instantaneous value of current of 11.917A at 37°. What is the RMS value of current? **Options:** A. 14A B. 13A C. 12A D. 11A --- **Explanation:** To solve this problem, we need to understand the relationship between the instantaneous value, peak value, and RMS (Root Mean Square) value of a sine wave. For a pure sine wave, the relationship between the instantaneous value (I), peak value (I_peak), and RMS value (I_RMS) is given by: \[ I = I_{\text{peak}} \cdot \sin(\theta) \] where: - \( I \) is the instantaneous current, - \( I_{\text{peak}} \) is the peak current, - \( \theta \) is the phase angle. In this case: - \( I = 11.917 \text{A} \), - \( \theta = 37^\circ \). So, \[ 11.917 = I_{\text{peak}} \cdot \sin(37^\circ) \] The sine of 37 degrees (sin 37°) is approximately 0.6018. \[ 11.917 = I_{\text{peak}} \cdot 0.6018 \] Solving for \( I_{\text{peak}} \), \[ I_{\text{peak}} = \frac{11.917}{0.6018} \] \[ I_{\text{peak}} \approx 19.80 \text{A} \] The RMS value of the current for a sine wave is: \[ I_{\text{RMS}} = \frac{I_{\text{peak}}}{\sqrt{2}} \] Substituting \( I_{\text{peak}} \approx 19.80 \text{A} \), \[ I_{\text{RMS}} = \frac{19.80}{\sqrt{2}} \] \[ I_{\text{RMS}} \approx 14 \text{A} \] Therefore, the correct answer is: #### **A. 14A**
Transcribed Image Text:### AC Sine Wave RMS Value Calculation **Question:** An AC sine wave has an instantaneous value of current of 11.917A at 37°. What is the RMS value of current? **Options:** A. 14A B. 13A C. 12A D. 11A --- **Explanation:** To solve this problem, we need to understand the relationship between the instantaneous value, peak value, and RMS (Root Mean Square) value of a sine wave. For a pure sine wave, the relationship between the instantaneous value (I), peak value (I_peak), and RMS value (I_RMS) is given by: \[ I = I_{\text{peak}} \cdot \sin(\theta) \] where: - \( I \) is the instantaneous current, - \( I_{\text{peak}} \) is the peak current, - \( \theta \) is the phase angle. In this case: - \( I = 11.917 \text{A} \), - \( \theta = 37^\circ \). So, \[ 11.917 = I_{\text{peak}} \cdot \sin(37^\circ) \] The sine of 37 degrees (sin 37°) is approximately 0.6018. \[ 11.917 = I_{\text{peak}} \cdot 0.6018 \] Solving for \( I_{\text{peak}} \), \[ I_{\text{peak}} = \frac{11.917}{0.6018} \] \[ I_{\text{peak}} \approx 19.80 \text{A} \] The RMS value of the current for a sine wave is: \[ I_{\text{RMS}} = \frac{I_{\text{peak}}}{\sqrt{2}} \] Substituting \( I_{\text{peak}} \approx 19.80 \text{A} \), \[ I_{\text{RMS}} = \frac{19.80}{\sqrt{2}} \] \[ I_{\text{RMS}} \approx 14 \text{A} \] Therefore, the correct answer is: #### **A. 14A**
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