hwk 2 solution - ECE472 HOMEWORK#2 SOLUTIONS FALL 2013...

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ECE472 HOMEWORK #2 – SOLUTIONS FALL 2013 Problem 2.1 + - i(t) v(t) LOAD Given, v ( t ) = 120 + 40 2 cos(100 t + 30 o ) + 30 2 sin(300 t + 16 o ) + 10 2 cos(500 t ) V = 120 + 40 2 sin(100 t + 120 o ) + 30 2 sin(300 t + 16 o ) + 10 2 sin(500 t + 90 o ) V i ( t ) = 10 + 4 2 cos(100 t - 30 o ) + 3 2 cos(300 t - 74 o ) + 2 2 sin(500 t ) A = 10 + 4 2 sin(100 t + 60 o ) + 3 2 sin(300 t + 16 o ) + 2 2 sin(500 t ) A find the average power P (W) dissipated in the load. P = V o I o + V 1 I 1 cos( θ v 1 - θ i 1 ) + V 3 I 3 cos( θ v 3 - θ i 3 ) + V 5 I 5 cos( θ v 5 - θ i 5 ) = (120)(10) + (40)(4) cos(120 o - 60 o ) + (30)(3) cos(16 o - 16 o ) + (10)(2) cos(90 o - 0 o ) = 1200 + 80 + 90 + 0 = 1370 W Problem 2.2 R = 10 + - π v(t)=100 2|sin 120 t| V i(t) (a) Find V rms , I rms , and the (exact) average power dissipated in the 10-ohm resistor. (b) Using the first four non-zero terms of the Fourier series expansion of v ( t ), obtain an approxi- mation of the average power dissipated in the 10-ohm resistor. 1
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(a) V rms = V m 2 = 100 2 2 = 100 V I rms = V rms R = V m / 2 R = 100 2 / 2 10 = 10 A P = RI 2 rms = V 2 rms R = V rms I rms = 100 × 10 = 1000 W (b) V rms = ( 2 V m π ) 2 + ( 4 V m 3 π 2 ) 2 + ( 4 V m 15 π 2 ) 2 + ( 4 V m 35 π 2 ) 2 = v u u t ( 200 2 π ) 2 + ( 400 3 π ) 2 + ( 400 15 π ) 2 + ( 400 35 π ) 2 = 99 . 96 V P = V 2 rms R = 99 . 96 2 10 = 999 . 2 W Problem 2.3 1 1 V + - i(t) Given that i ( t ) = 24 + 7 2 sin 120 πt : (a) Find the average power dissipated in the 1-Ω resistor.
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