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Derive this relationship.

Derive this relationship.

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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"Problem 1.

In your textbook, Equation 11-5 defines Vrms = 0.707 VPEAK . Derive this relationship."

See image of Equation 11-5. I don't remember how to do this. Please help! Please make legible, typed in equation form would be great. Thanks!

 

Equation 11-5 RMS (Effective) Value of a Sine Wave The abbreviation "rms" stands for the root mean square process by which this value is derived. In the process, we first square the equation of a sinusoidal voltage wave. v² = V₂2² sin²0 Next, we obtain the mean or average value of ² by dividing the area under a half-cycle of the curve by (see Figure B-10). The area is found by integration and trigonometric identities. Figure B-1 V2 avg = area π V² Vp = 7T V2 - (1 - 006 20) 40 - 1 0 - 2 (-00620) de 201 = cos do ²5 -cos 2п Jo 2π 2T 1/V² sin²0 de V2 1 V2 22 (0-20) = (1-0)-2 sin = 2T 2π 0 Area under the half-cycle squared of a sinusoidal voltage wave. Finally, the square root of Vis Vrms. Vrms= /v² Vavg= Vp V2/2 = = 0.707V₂ √2
Transcribed Image Text:Equation 11-5 RMS (Effective) Value of a Sine Wave The abbreviation "rms" stands for the root mean square process by which this value is derived. In the process, we first square the equation of a sinusoidal voltage wave. v² = V₂2² sin²0 Next, we obtain the mean or average value of ² by dividing the area under a half-cycle of the curve by (see Figure B-10). The area is found by integration and trigonometric identities. Figure B-1 V2 avg = area π V² Vp = 7T V2 - (1 - 006 20) 40 - 1 0 - 2 (-00620) de 201 = cos do ²5 -cos 2п Jo 2π 2T 1/V² sin²0 de V2 1 V2 22 (0-20) = (1-0)-2 sin = 2T 2π 0 Area under the half-cycle squared of a sinusoidal voltage wave. Finally, the square root of Vis Vrms. Vrms= /v² Vavg= Vp V2/2 = = 0.707V₂ √2
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