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 Schaums - Theory and Problems of Electric Circuits (4nd ed) 1.3- Solution $$\copyright$$ Sin squared = 1 - cosine 2x 1.3 A certain circuit element has a current i = 2.5 sin $$\omega$$t (mA), where $$\omega$$ is the angular frequency in rad/s, and a voltage difference v = 45 sin $$\omega$$t (V) between terminals. Find the average power $$P_{avg}$$ and the energy $$W_{T}$$ transferred in one period of the sine function. RECALL THAT : VOLTAGE(v) X CURRENT(i) = WORK($$W_{T}$$) 2.5 X 45 = 112.5 so.... $$W_{T} = \int_{0}^{\frac{2\pi}{\omega}} vi dt = W_{T} = 112.5 \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt$$ RECALL TRIGONMETRY IDENTITY $$sin^2x = 1/2(1- cos2x)$$ $$W_{T} = \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt = W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt$$ WE NOW HAVE .... $$W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt$$ CALCULUS IDENTITY FOR $$\int\cos(x) dx = \sin x + C$$ PROOF with U Substitution : Let u = 2x THEN $$\frac{du}{dx} = 2$$ AND dx = $$\frac{1}{2}du$$ $$\int\cos\2x$$ = $$\int\cos u\frac{1}{2}du$$ $$\frac{1}{2}\int\cos u du$$ $$\frac{1}{2}\sin u + C$$ $$\frac{1}{2}\sin2x + C$$ SO NOW .... $$\left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{2\pi}{\omega}}$$ NOW UPPER AND LOWER INTEGRATION GIVES $$\left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{}}$$ SUBSTITUTION HERE .... AFTER SUBSTITUTION RECALL TRIGONOMETRY SINE $$\frac{\pi}{\theta}$$ = 0 $$\left[ \frac{1}{2} ( ( \frac{2\pi}{\omega} ) - \frac{1}{2} \sin 2 ( \frac{2\pi}{\omega} ) ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( ( 0 ) - \frac{1}{2} \sin 2 ( 0 ) ) \right]_{0}^{\frac{}}$$ OK ... THIS IS WHAT'S LEFT $$W_{T} = 112.5 \left[ ( \frac{1}{2} ) ( \frac{2\pi}{\omega} ) - 0 \right]$$ THEREFORE: $$W_{T} = 112.5 \left[ ( \frac{\pi}{\omega} ) \right] = ( \frac{ 112.5\pi}{\omega} ) \right]$$ FINALLY: $$P_{avg} = ( \frac{ W_{T} }{ \frac { 2\pi}{\omega}} ) = 56.25mW$$