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Unformatted text preview: Ch. 24 HW Solutions 7.) The average power lost in a resistor is given by P = I 2 R where I is the rms value of current, given by I = i o / √ 2 , and R is the resistance of the resistor. The current in the problem has amplitude i o = 5 . A and the resistor has R = 15 Ω . So P = I 2 R = ( i o / √ 2) 2 R = i 2 o R/ 2 = 187 . 5 W . 9.) (a.) A V = 80 V AC power source and a resistor in series yield a current I = V/R = 80 / 30 = 2 . 67 A. (b.) An 80 V power source has an rms voltage of 80 V, though the amplitude of the voltage oscillations will be greater by a factor of √ 2 . (c.) v o = √ 2 V = 113 . 14 V (d.) With an rms current of I = 2 . 67 A, the peak current is i o = √ 2 I = 3 . 77 A. 13.) For discharging a capacitor, q ( t ) = q o e ( t/τ C ) . (a.) After four time constants, t = 4 τ C so q (4 τ ) = q o e ( 4 τ C /τ C ) = q o e 4 = q o (0 . 37) 4 = (0 . 0187) q o . The fraction is 0.0187. (b.) Initially, q o = CV o . Finally, q f = CV f . Dividing the latter by the former,....
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 Summer '08
 Lee
 Applied Physics, Current, Inductance, Resistance, Power, Alternating Current, Inductor, Electrical impedance

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