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opitts2k
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#10
Feb5-10, 09:00 PM
P: 1
Schaums - Theory and Problems of Electric Circuits (4nd ed) 1.3- Solution [tex]\copyright[/tex]

Sin squared = 1 - cosine 2x

1.3 A certain circuit element has a current i = 2.5 sin [tex]\omega[/tex]t (mA), where [tex]\omega[/tex] is the angular frequency in rad/s, and a voltage difference v = 45 sin [tex]\omega[/tex]t (V) between terminals. Find the average power [tex] P_{avg}[/tex] and the energy [tex]W_{T}[/tex] transferred in one period of the sine function.


RECALL THAT : VOLTAGE(v) X CURRENT(i) = WORK([tex]W_{T}[/tex])

2.5 X 45 = 112.5 so....

[tex]W_{T} = \int_{0}^{\frac{2\pi}{\omega}} vi dt = W_{T} = 112.5 \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt [/tex]


RECALL TRIGONMETRY IDENTITY [tex] sin^2x = 1/2(1- cos2x) [/tex]

[tex]W_{T} = \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt = W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt [/tex]

WE NOW HAVE ....

[tex] W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt [/tex]

CALCULUS IDENTITY FOR [tex]\int\cos(x) dx = \sin x + C[/tex]
PROOF with U Substitution : Let u = 2x THEN
[tex]\frac{du}{dx} = 2 [/tex] AND dx = [tex]\frac{1}{2}du[/tex]
[tex]\int\cos\2x [/tex] = [tex]\int\cos u\frac{1}{2}du[/tex]
[tex]\frac{1}{2}\int\cos u du[/tex]
[tex]\frac{1}{2}\sin u + C[/tex]
[tex]\frac{1}{2}\sin2x + C[/tex]


SO NOW ....

[tex] \left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{2\pi}{\omega}} [/tex]


NOW UPPER AND LOWER INTEGRATION GIVES

[tex] \left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{}} [/tex]


SUBSTITUTION HERE .... AFTER SUBSTITUTION RECALL TRIGONOMETRY SINE [tex]\frac{\pi}{\theta} [/tex] = 0

[tex] \left[ \frac{1}{2} ( ( \frac{2\pi}{\omega} ) - \frac{1}{2} \sin 2 ( \frac{2\pi}{\omega} ) ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( ( 0 ) - \frac{1}{2} \sin 2 ( 0 ) ) \right]_{0}^{\frac{}} [/tex]

OK ... THIS IS WHAT'S LEFT

[tex] W_{T} = 112.5 \left[ ( \frac{1}{2} ) ( \frac{2\pi}{\omega} ) - 0 \right] [/tex]

THEREFORE:

[tex] W_{T} = 112.5 \left[ ( \frac{\pi}{\omega} ) \right] = ( \frac{ 112.5\pi}{\omega} ) \right] [/tex]

FINALLY:


[tex] P_{avg} = ( \frac{ W_{T} }{ \frac { 2\pi}{\omega}} ) = 56.25mW [/tex]