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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

The circuit shown has a source voltage of \(440 \mathrm{V}\) ms, resistance \(R=250 \Omega\), inductance \(L=0.800 \mathrm{H},\) and capacitance $C=2.22 \mu \mathrm{F} .\( (a) Find the angular frequency \)\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: \(V_{a b}, V_{b c}, V_{c d}, V_{b d},\) and \(V_{c d},\) (d) The resistor is replaced with one of \(R=125 \Omega\). Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?
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 variable inductor with negligible resistance is connected to an ac voltage source. How does the current in the inductor change if the inductance is increased by a factor of 3.0 and the driving frequency is increased by a factor of \(2.0 ?\)
An x-ray machine uses \(240 \mathrm{kV}\) rms at \(60.0 \mathrm{mA}\) ms when it is operating. If the power source is a \(420-\mathrm{V}\) rms line, (a) what must be the turns ratio of the transformer? (b) What is the rms current in the primary? (c) What is the average power used by the \(x\) -ray tube?
A computer draws an rms current of \(2.80 \mathrm{A}\) at an \(\mathrm{rms}\) voltage of \(120 \mathrm{V}\). The average power consumption is 240 W. (a) What is the power factor? (b) What is the phase difference between the voltage and current?
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