TEST BANK FOR Fitzgerald Kingsley Electric Machinery 7th Edition By Stephen D. Umans

TEST BANK FOR Fitzgerald Kingsley Electric Machinery 7th Edition By Stephen D. Umans 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):  = NI Rc + Rg = 2.437 × 10−5 Wb Part (c):  = N = 2.315 × 10−3 Wb Part (d): L =  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):  = NI Rc + Rg = 2.334 × 10−5 Wb Part (c):  = N = 2.217 × 10−3 Wb Part (d): L =  I = 1.584 mH Problem 1-3 Part (a): N = s Lg μ0Ac = 287 turns Part (b): I = Bcore μ0N/g = 7.68 A Problem 1-4 Part (a): N = s L(g + lcμ0/μ) μ0Ac = s 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 = 0 @ μ0NI 2g +  μ0 μ  lp +  μ0 μ   Ag Ac  lc 1 A = 0 @ μ0NI 2g +  μ0 μ  lp +  μ0 μ   1 − x X0  lc 1 A and Bc =  1 − x X0  Bg Problem 1-7 From Problem 1-6, the inductance can be found as L = NAcBc I = μ0N2Ac 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 s 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 = μ0N2Ac 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 = s 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 vuut 2(g + (μ0 μ )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 = μ0N2Ac 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 = 2  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):  = 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 cm2 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):  = 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 = μ0N2Ac g + lc/μr Problem 1-17 Part (a): L = μ0N2Ac g + lc/μr = 36.6 mH Part (b): B = μ0N2 g + lc/μr I = 0.77 T  = LI = 4.40 × 10−2 Wb Problem 1-18 Part (a): With ! = 120 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 = !NAcBpeak p2 = 20.8 V Part (b): Using L from the solution to Problem 1-17 Ipeak = p2Vrms !L = 1.66 A Wpeak = LI2 peak 2 = 9.13 × 10−2 J Problem 1-19 B = 0.81 T and  = 46.5 mWb Problem 1-20 Part (a): R3 = q (R21 + R22 ) = 4.49 cm Part (b): For lc = 4l + R2 + R3 − 2h; and Ag = R2 1 L = μ0AgN2 g + (μ0/μ)lc = 61.8 mH Part (c): For Bpeak = 0.6 T and ! = 260 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 peak = AgNBpeak Vrms = !peak p2 = 23.2 V Irms = Vrms !L = 0.99 A Wpeak = 1 2 LI2 peak = 1 2 L(p2Irms)2 = 61.0 mJ Part (d): For ! = 250 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 μN = 0.46 A Problem 1-22 Part (a); Part (b): Ipeak = 0.6 A Part (c): Ipeak = 4.0 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. 14 Problem 1-23 For part (b), Ipeak = 11.9 A. For part (c), Ipeak = 27.2 A. Problem 1-24 L =