240_homework_1_soln

# 240_homework_1_soln - ECE 240 Electrical Engineering...

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ECE 240 – Electrical Engineering Fundamentals Solutions to Homework #1 1. i = dq dt = 2500 π cos(250 π t ) mA p ( t ) = 10000 π cos 2 (250 π t ) mW p (.001) = 10000 π cos 2 (.250 π ) mW = 15.71 W 2. a. v ( t ) = 4 6 e t 7 ( ) b. v = 7(10) = 70 c. v ( t ) = (5 + 12)sin4 t = 17sin4 t d. v ( t ) = (8)10cos3 t = 80cos3 t 3. a. i ( t ) = 6 e 2 t 3 = 2 e 2 t b. i ( t ) = 10 5cos t ( ) 7 c. i ( t ) = cos2 t 5 + cos2 t 2 = 7 10 cos2 t d. i ( t ) = sin t 10 4. For the resistors, each one has a voltage of cos 5t, thus: p to 6 Ω = cos 2 5 t 6 p to 4 Ω = cos 2 5 t 4 p to 3 Ω = cos 2 5 t 3 the current source also has a voltage of cos 5t (current into +), thus: p by 10 A = p to 10 A = 10cos5 t the voltage source has a current leaving the + terminal of: i = 10 + cos5 t 6 + cos5 t 4 + cos5 t 3 = 10 + 3 4 cos5 t p byVsource = cos5 t 10 + 3 4 cos5 t 5. The elements are all in series and so they all have a current of 6e 3t going clockwise. Thus: p to 5 Ω = 5 6 e 3 t ( ) 2 = 180 e 6 t p to 4 Ω = 4 6 e 3 t ( ) 2 = 144 e 6 t p by cos t = cos t 6 e 3 t ( ) = 6 e 3 t cos t p by sin t = sin t 6 e 3 t ( ) = 6 e 3 t sin

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