Solutions to Homework Problems in Electrical and Computer Engineering Course - Prof. Said , Assignments of Electrical and Electronics Engineering

Download Solutions to Homework Problems in Electrical and Computer Engineering Course - Prof. Said and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! ECE473 HOMEWORK #2 – SOLUTIONS FALL 2008 Problem 2.1 o 208 150 o 8.64Ω j11.52Ω 8.64Ω j11.52Ω 8.64Ω j11.52Ω j7Ω I ~ A I ~ 208 30 Ω −j54Ω −j54Ω a b 208 −90 Ωj7 j7Ω j7Ω + − n N 8.64Ω Ωj11.52 o A2 AN Per−Phase Circuit an E ~ jX C + − + − AB V ~ V ~ I ~ C a A CB ~ I ~ A1 I ~ AB I A ~ I ~ A2 I A1 ~+ − V AN ~ + −V −j54 + − + − + − c N A B C (a) Ẽan = ṼAB√ 3 6 30o = 208 6 30o√ 3 6 30o ∼= 120 6 0o V jXC = jXC,Y = jXC,∆ 3 = −j54 3 = − j18 Ω (b) Z̄tot = j7 + (8.64 + j11.52)(−j18) 8.64 + j11.52− j18 = 24 + j7 = 25 6 16.26o Ω ĨA = Ẽan Z̄tot = 120 6 0o 25 6 16.26o = 4.8 6 − 16.26o A ṼAN = (8.64 + j11.52)(−j18) 8.64 + j11.52− j18 × ĨA = (24 + j0)(4.8 6 − 16.26o) = 115.2 6 − 16.26o V ĨA1 = ṼAN −j18 = 115.2 6 − 16.26o 18 6 − 90o = 6.4 6 73.74o A ĨA2 = ṼAN 8.64 + j11.52 = 115.2 6 − 16.26o 14.4 6 53.13o = 8 6 − 69.39o A 1 (c) ṼAB = ( √ 3 6 30o)ṼAN = ( √ 3 6 30o)(115.2 6 − 16.26o) ∼= 200 6 13.74o V ṼBC = (1 6 − 120o)ṼAB = 200 6 13.74o − 120o = 200 6 − 106.26o V ṼCB = −ṼBC = − 200 6 − 106.26o = 200 6 − 106.26o + 180o = 200 6 73.74o V ĨC = (1 6 120o)ĨA = (1 6 120o)(4.8 6 − 16.26o) = 4.8 6 103.74o A (d) WA = |ṼAB||ĨA| cos(6 ṼAB − 6 ĨA) = (200)(4.8) cos(13.74o − (−16.26o)) ∼= 831 W WC = |ṼCB||ĨC | cos(6 ṼCB − 6 ĨC) = (200)(4.8) cos(73.74o − 103.74o) ∼= 831 W (e) P3ph = WA + WC = 831 + 831 ∼= 1660 W Q3ph = √ 3(WC −WA) = √ 3(0− 0) = 0 VAr (f) P3ph = 3× 8.64× |ĨA2|2 = 3× 8.64× 82 ∼= 1660 W Q3ph = 3× (−18)× |ĨA1|2 + 3× 11.52× |ĨA2|2 = 3× (−18)× 6.42 + 3× 11.52× 82 = 0 VAr Problem 2.2 (Problem 2.40 p. 78) V ~ BC V CA ~ AB V ~ ~ I ~ AB I A 30 −60 o o CN V AN ~ V BN ~ V ~ (a) ṼAN = 120 6 0o V ṼBN = 120 6 − 120o V ṼCN = 120 6 120o V ṼAB = (√ 3 6 30o ) ṼAN = 208 6 30o V ṼBC = 208 6 − 90o V =⇒ ĨA = 10 6 − 90o A ṼCA = 208 6 150o V Z̄Y = ṼAN ĨA = 120 6 0o 10 6 − 90o = 12 6 90o = j12 Ω (b) ĨAB = ( 1√ 3 6 30o ) ĨA = 10√ 3 6 (−90o + 30o) = 10√ 3 6 − 60o A Z̄∆ = ṼAB ĨAB = 208 6 30o 10/ √ 3 6 − 60o = 36 6 90o = j36 Ω or Z̄∆ = 3Z̄Y = 36 6 90o = j36 Ω 2