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Chapter 21: Problem 11

Show that over one complete cycle, the average value of a sine function squared is \(\frac{1}{2} .\) [Hint: Use the following trigonometric identities: \(\sin ^{2} a+\cos ^{2} a=1 ; \cos 2 a=\) \(\left.\cos ^{2} a-\sin ^{2} a .\right]\)

Short Answer

Expert verified
Answer: The average value of a sine function squared over one complete cycle is \(\frac{1}{2}\).

Step by step solution

01

Express sine squared using given trigonometric identities

Using the given identities, we can express sine squared in a more suitable form for integration. We will use the identity \(\cos 2a = \cos^2a - \sin^2a\). To express \(\sin^2a\), we rearrange the equation: \(\sin^2a = \cos^2a - \cos2a\) Now, we have \(\sin^2a\) in terms of cosines which makes integration easier.
02

Integrate sine squared over one complete cycle

To find the average value over one complete cycle, we need to integrate \(\sin^2a\) over the interval \([0, 2\pi]\) (since one cycle is \(2\pi\) long). We integrate the expression we found in step 1: \(\int_0^{2\pi} (\cos^2a - \cos2a) \; da\) Since \(\cos^2a\) and \(\cos2a\) are individual functions, we can separate the integrals: \(\int_0^{2\pi} \cos^2a \; da - \int_0^{2\pi} \cos2a \; da\) Now, we can integrate each part: \(\left[\frac{a}{2} + \frac{\sin 2a}{4}\right]_0^{2\pi} - \left[\frac{\sin 2a}{2}\right]_0^{2\pi}\)
03

Evaluate the integral

Now, we need to evaluate the values of our integrals at our integration limits. Plugging in \(2\pi\) and \(0\): \(\left(\frac{2\pi}{2} + \frac{\sin 4\pi}{4}\right) - \left(\frac{0}{2} + \frac{\sin 0}{4}\right) - \left(\frac{\sin 4\pi}{2}\right) + \left(\frac{\sin 0}{2}\right)\) Simplifying and calculating the values: \(\left(\pi + 0\right) - 0 - 0 + 0\) \(\pi\)
04

Calculate the average value

To find the average value of sine squared over one complete cycle, we need to divide the integral value we found in step 3 by the length of that cycle (\(2\pi\)): Average value = \(\frac{\pi}{2\pi}\) Average value = \(\boxed{\frac{1}{2}}\) Now, we have shown that the average value of a sine function squared over one complete cycle is \(\frac{1}{2}\).

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Most popular questions from this chapter

A series combination of a resistor and a capacitor are connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) ac source. If the capacitance is \(0.80 \mu \mathrm{F}\) and the rms current in the circuit is $28.4 \mathrm{mA},$ what is the resistance?
A capacitor (capacitance \(=C\) ) is connected to an ac power supply with peak voltage \(V\) and angular frequency \(\alpha\) (a) During a quarter cycle when the capacitor goes from being uncharged to fully charged, what is the average current (in terms of \(C . V\), and \(\omega\) )? [Hint: \(\left.i_{\mathrm{av}}=\Delta Q / \Delta t\right]\) (b) What is the rms current? (c) Explain why the average and rms currents are not the same.
A \(0.250-\mu \mathrm{F}\) capacitor is connected to a \(220-\mathrm{V}\) rms ac source at \(50.0 \mathrm{Hz}\) (a) Find the reactance of the capacitor. (b) What is the rms current through the capacitor?
A \(1500-\) W electric hair dryer is designed to work in the United States, where the ac voltage is \(120 \mathrm{V}\) rms. What power is dissipated in the hair dryer when it is plugged into a \(240-\mathrm{V}\) rms socket in Europe? What may happen to the hair dryer in this case?
A 6.20 -m \(H\) inductor is one of the elements in a simple RLC series circuit. When this circuit is connected to a \(1.60-\mathrm{kHz}\) sinusoidal source with an \(\mathrm{rms}\) voltage of \(960.0 \mathrm{V},\) an rms current of $2.50 \mathrm{A}\( lags behind the voltage by \)52.0^{\circ} .$ (a) What is the impedance of this circuit? (b) What is the resistance of this circuit? (c) What is the average power dissipated in this circuit?
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