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1 PROBLEM SOLUTIONS: Chapter 1 Problem 1-1 Part (a): Rc = lc µAc = lc µrµ0Ac = 0 A/Wb Rg = g µ0Ac = 5.457× 106 A/Wb Part (b): Φ = NI Rc +Rg = 2.437 × 10−5 Wb Part (c): λ = NΦ = 2.315× 10−3 Wb Part (d): L = λ I = 1.654 mH Problem 1-2 Part (a): Rc = lc µAc = lc µrµ0Ac = 2.419× 105 A/Wb Rg = g µ0Ac = 5.457× 106 A/Wb c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 2 Part (b): Φ = NI Rc +Rg = 2.334 × 10 −5 Wb Part (c): λ = NΦ = 2.217× 10−3 Wb Part (d): L = λ I = 1.584 mH Problem 1-3 Part (a): N = √ Lg µ0Ac = 287 turns Part (b): I = Bcore µ0N/g = 7.68 A Problem 1-4 Part (a): N = √ L(g + lcµ0/µ) µ0Ac = √ L(g + lcµ0/(µrµ0)) µ0Ac = 129 turns Part (b): I = Bcore µ0N/(g + lcµ0/µ) = 20.78 A c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 3 Problem 1-5 Part (a): Part (b): Bg = Bm = 2.1 T For Bm = 2.1 T, µr = 37.88 and thus I = ( Bm µ0N )( g + lc µr ) = 158 A Part (c): Problem 1-6 Part (a): Bg = µ0NI 2g c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 4 Bc = Bg ( Ag Ac ) = ( µ0NI 2g )( 1− x X0 ) Part (b): Will assume lc is “large” and lp is relatively “small”. Thus, BgAg = BpAg = BcAc We can also write 2gHg +Hplp +Hclc = NI ; and Bg = µ0Hg; Bp = µHp Bc = µHc These equations can be combined to give Bg = µ0NI 2g + ( µ0 µ ) lp + ( µ0 µ ) ( Ag Ac ) lc = µ0NI 2g + ( µ0 µ ) lp + ( µ0 µ ) ( 1− x X0 ) lc and Bc = ( 1− x X0 ) Bg Problem 1-7 From Problem 1-6, the inductance can be found as L = NAcBc I = µ0N 2Ac 2g + µ0 µ (lp + (1− x/X0) lc) from which we can solve for µr c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 5 µr = µ µ0 = L ( lp + (1− x/X0) lc ) µ0N2Ac − 2gL = 88.5 Problem 1-8 Part (a): L = µ0(2N) 2Ac 2g and thus N = 0.5 √ 2gL Ac = 38.8 which rounds to N = 39 turns for which L = 12.33 mH. Part (b): g = 0.121 cm Part(c): Bc = Bg = 2µ0NI 2g and thus I = Bcg µ0N = 37.1 A Problem 1-9 Part (a): L = µ0N 2Ac 2g c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 6 and thus N = √ 2gL Ac = 77.6 which rounds to N = 78 turns for which L = 12.33 mH. Part (b): g = 0.121 cm Part(c): Bc = Bg = µ0(2N)(I/2) 2g and thus I = 2Bcg µ0N = 37.1 A Problem 1-10 Part (a): L = µ0(2N) 2Ac 2(g + (µ0 µ )lc) and thus N = 0.5 √√√√2(g + (µ0µ )lc)L Ac = 38.8 which rounds to N = 39 turns for which L = 12.33 mH. Part (b): g = 0.121 cm c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 7 Part(c): Bc = Bg = 2µ0NI 2(g + µ0 µ lc0) and thus I = Bc(g + µ0 µ lc) µ0N = 40.9 A Problem 1-11 Part (a): From the solution to Problem 1-6 with x = 0 I = Bg ( 2g + 2 ( µ0 µ ) (lp + lc) ) µ0N = 1.44 A Part (b): For Bm = 1.25 T, µr = 941 and thus I = 2.43 A Part (c): Problem 1-12 g = µ0N 2Ac L − ( µ0 µ ) lc = 7.8× 10−4 m c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8 Problem 1-13 Part (a): lc = 2pi ( Ri +Ro 2 ) − g = 22.8 cm Ac = h(Ro − Ri) = 1.62 cm2 Part (b): Rc = lc µAc = 0 Rg = g µ0Ac = 7.37× 106 H−1 Part (c): L = N2 Rc +Rg = 7.04× 10 −4 H Part (d): I = BgAc(Rc +Rg) N = 20.7 A Part (e): λ = LI = 1.46× 10−2 Wb c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 9 Problem 1-14 See solution to Problem 1-13 Part (a): lc = 22.8 cm Ac = 1.62 cm 2 Part (b): Rc = 1.37× 106 H−1 Rg = 7.37× 106 H−1 Part (c): L = 5.94× 10−4 H Part (d): I = 24.6 A Part (e): λ = 1.46× 10−2 Wb c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 10 Problem 1-15 µr must be greater than 2886. Problem 1-16 L = µ0N 2Ac g + lc/µr Problem 1-17 Part (a): L = µ0N 2Ac g + lc/µr = 36.6 mH Part (b): B = µ0N 2 g + lc/µr I = 0.77 T λ = LI = 4.40× 10−2 Wb Problem 1-18 Part (a): With ω = 120pi c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 11 Vrms = ωNAcBpeak√ 2 = 20.8 V Part (b): Using L from the solution to Problem 1-17 Ipeak = √ 2Vrms ωL = 1.66 A Wpeak = LI2peak 2 = 9.13× 10−2 J Problem 1-19 B = 0.81 T and λ = 46.5 mWb Problem 1-20 Part (a): R3 = √ (R21 +R 2 2) = 4.49 cm Part (b): For lc = 4l +R2 +R3 − 2h; and Ag = piR 2 1 L = µ0AgN 2 g + (µ0/µ)lc = 61.8 mH Part (c): For Bpeak = 0.6 T and ω = 2pi60 c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 12 λpeak = AgNBpeak Vrms = ωλpeak√ 2 = 23.2 V Irms = Vrms ωL = 0.99 A Wpeak = 1 2 LI2peak = 1 2 L( √ 2Irms) 2 = 61.0 mJ Part (d): For ω = 2pi50 Vrms = 19.3 V Irms = 0.99 A Wpeak = 61.0 mJ Problem 1-21 Part (a); c©2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 13 Part (b): Emax = 4fNAcBpeak = 118 V part (c): For µ = 1000µ0 Ipeak = lcBpeak