EE110 HW3 Sol

# EE110 HW3 Sol - EE110 Homework 3 Solution Problem 1(a 1 2 1...

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EE110 Homework 3 Solution Problem 1. (a) (b) Problem 2. , 2 4, 10 2 10, 5(10 5) 4 4 2.5 5 6.5 5 10 100 12.194 37.57 A 6.5 5 1 P 100 12.194cos37.57 483.3W 2 1 P (12.194) 4 297.4W, 2 P 0 100 5 6.097 52.43 so 6.5 5 10 1 P (6.097) 10 185.87 2 in s s abs abs cabs abs j j j j j j j = + = + + = + Ω = = ∠ − ° + = − × × ° = − = = = = = ° + = × = Z I I W P 0 ( 0) L = Σ = 2 2 4 2 4 2 4 2 1 1 4H : 2 1 4(4 ) 16 , 4(4 4 1) 2 2 8 8 2 (3) (1) 8 3 8 3 2 8 1 8 1 2 576 J L L L L i t v Li t t w Li t t w t t w w = − ∴ = = = = = × + = + ∴ = × − × + − × + × − = 2 3 3 1 1 1 2 2 2 0.2F: (2 1) 2 5 2 5 5 1 2 0.2 3 3 3 10 10 61 61 (2) 8 10 5 2 V P (2) 7 142.33W 3 3 3 3 t t c c c v t dt t t t t v = + = + = + = × − + + = = × =

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Problem 3. 20 2 [1] 2 3 0 [2] 2 3 which simplify to 5 14 = 60 [1] and 2 + (3 2) = 0 [2] Solving, 9.233 83.88 V and 5.122 140.2 V 1 P 9. 2 x x c c c c x x c x c x c gen and j j j + = = + = ∠ − ° = ∠− ° = × V V V V V V V V V V V V V ( ) 233 2 5.122 cos( 83.88 140.2 ) 26.22 W × × °+ ° = Problem 4. 2 2 ,max 2 10(20) 10 100 20 40, 4 8 20 10 20 10 R R 8.944 1 20 40 P 8.944 38.63W 2 (4 8.944) 64 th th L th L L j j j j j j = = + = = + Ω + + = = Ω + = × = + + V Z Z Problem 5. (a) (b) 2 2 2 1 F (10 20 10 ) 150 12.247 4 eff = + + = = (c) F avg = 10 4 40 4 ) 1 )( 10 ( ) 1 )( 20 ( ) 1 )( 10 ( = = + + 10 9cos100 6sin100 1 1 V 100 81 36 158.5 12.590V 2 2 eff v t t = + + = + × + × = =
Problem 6. (a)

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EE110 HW3 Sol - EE110 Homework 3 Solution Problem 1(a 1 2 1...

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