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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. \[ v^2 = V_p^2 \sin^2 \theta \] Next, we obtain the mean or average value of \( v^2 \) by dividing the area under a half-cycle of the curve by \( \pi \) (see Figure B–1). The area is found by integration and trigonometric identities. \[ V^2_{\text{avg}} = \frac{\text{area}}{\pi} = \frac{1}{\pi} \int_0^{\pi} V_p^2 \sin^2 \theta \, d\theta \] \[ = \frac{V_p^2}{2\pi} \int_0^{\pi} (1 - \cos 2\theta) \, d\theta = \frac{V_p^2}{2\pi} \int_0^{\pi} 1 \, d\theta - \frac{V_p^2}{2\pi} \int_0^{\pi} (-\cos 2\theta) \, d\theta \] \[ = \frac{V_p^2}{2\pi} \left( \theta - \frac{1}{2} \sin 2\theta \right) \bigg|_0^{\pi} = \frac{V_p^2}{2\pi} \left( \pi - 0 \right) = \frac{V_p^2}{2} \] **Figure B–1** - The figure shows the area under the half-cycle squared of a sinusoidal voltage wave. It is depicted as a shaded region under the curve from 0 to \(\pi\) on the \( \theta \)-axis, with a peak at \( V_p^2 \). Finally, the square root of \( V^2_{\text{avg}} \) is \( V_{\text{rms}} \). \[ V_{\text{rms}} = \sqrt{V^2_{\text{avg}}} = \sqrt{\frac{V_p^2}{2}} = \frac{V_p}{
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. \[ v^2 = V_p^2 \sin^2 \theta \] Next, we obtain the mean or average value of \( v^2 \) by dividing the area under a half-cycle of the curve by \( \pi \) (see Figure B–1). The area is found by integration and trigonometric identities. \[ V^2_{\text{avg}} = \frac{\text{area}}{\pi} = \frac{1}{\pi} \int_0^{\pi} V_p^2 \sin^2 \theta \, d\theta \] \[ = \frac{V_p^2}{2\pi} \int_0^{\pi} (1 - \cos 2\theta) \, d\theta = \frac{V_p^2}{2\pi} \int_0^{\pi} 1 \, d\theta - \frac{V_p^2}{2\pi} \int_0^{\pi} (-\cos 2\theta) \, d\theta \] \[ = \frac{V_p^2}{2\pi} \left( \theta - \frac{1}{2} \sin 2\theta \right) \bigg|_0^{\pi} = \frac{V_p^2}{2\pi} \left( \pi - 0 \right) = \frac{V_p^2}{2} \] **Figure B–1** - The figure shows the area under the half-cycle squared of a sinusoidal voltage wave. It is depicted as a shaded region under the curve from 0 to \(\pi\) on the \( \theta \)-axis, with a peak at \( V_p^2 \). Finally, the square root of \( V^2_{\text{avg}} \) is \( V_{\text{rms}} \). \[ V_{\text{rms}} = \sqrt{V^2_{\text{avg}}} = \sqrt{\frac{V_p^2}{2}} = \frac{V_p}{
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