CH01 - Problem 1.1 Part. [a]: . IO 2.: , _, RC — “A:...

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Unformatted text preview: Problem 1.1 Part. [a]: . IO 2.: , _, RC — “A: — [Lil-#0516 — 0 Afl’R-b . 9 - -y - 6 " ." P = — 21.01: < 10 Aflfl-b Kg .UIJAC E part. [1)]: NI q» = ,—, = 1.224 < 10“1 um RC + 12% part. {c}: A = M) = 1.016 < 10—2 Wb part. [(1 ): A L = T = 6.775 mH Problem 1.6 part. [a]: NI A- :E =—: BC: —% B =3 1—— g 29' (Ac) g g( X0) part. [1)]: Equations H QQHg + HCEC = NI; Bgr’lg = BCAC and Bg = runI-Ig; BC 2 gtHc can be combined to give NI NI Bg = — = 29 + (%) (if) we + 1p] 29+ (%) (1— 1%) “a Hp} Problem 1.7 part. [a]: g + {it + (1ij “UN = 2.1;. A part. [1)]: 1+ —1199 1012 1: f. —.,_ = i g m V1 +0.1.r530 ‘0 g + {at +3131: _ = 3.02 A £5096 Problem 11 17 part. [a]: . N . . \F = .qf-Bisat = turns; 9 2 “firm; — ruff: = nun part. [11]: From Eq.3.21 A 9.32 A I. B? I r.— IéH : C hat, 2 I —r : {rem-e : E' c hat. 2 I ,. 3- 1% 1_ —2#0 0 201 J_ 1 —2# 0 0—1:: J Tth 1-1-10: = H":ng + I'll-"001.8 = 0.25? J. From Eq. 1.41", [1}:"2JI4-I2 = 11.252 .1. QED. Problem 122 part {a}: ,. N I N I m 51:91.1 1 1; Ba =91] 1 1 91 92 .. _ . 1 _1 A A [11] A1 = 3"1I‘14131 + 14232} =1101N12 (—1 + I1 91 92 T I. I. .43 [111] A3 = 1392.4.ng = #1099199; — I1 92 part. [b]: N- \ [1] Bl = 0- 32 = 90 2172 92 _ _ _ .42 (11] A] = A’lr’lng = fitgfl’lfl’g — I2 9‘2 . .— .T-2 142 [111] A; = 39.4ng = 1103-2 — I2 93 part. {C}: N I N I N» I - 'm 3129011; 32:90 11+,U0 22 91 92 92 - , _. A A _ _ A- 1111 A] = 311.4151 + 1423-2] = 1me (—1 + i) 11 + #11313? (—2) I2 91 92 92 _ _ _ A _. .4- {llii} A2 = A"2-4an = 9031952 Jr1 + 91133 I2 92 92 part. [(1]: -1 A A; _. A- _ _ .4.» L11 = (—1 + J) 1. L22 = 91.19%2 ; L12 = #03’13’2 91 92 92 92 Problem 1.23 fax {.1 5-2 9 P = P = : P = P = \ 31 [HAG ‘1 “‘4: - ‘2 [ii-4c \g .U'Cl—qc part. [3.]: L _ _ N'fgtr’lc “ _ R1 +132 + R9 + RAIEQ _ 31 + [3 + $9152 + g (“Ix-1m} N“? ngélc 3,; +51 + 33 —l— g whim} LAA = L131} = = RA + RAIHR1+ R2 + Ry] 3A {A + 2(51+12 +9 [Minn] part {b}: NQE'Rq + '13; + R9} N2ur’1c 3.1 + 32 + g [pfpthfi LAB = LEA = = R.I51[RI51+2|IIR1+ RQ+R9IIIJ 5A EI51+2|II31 +Eg+guiggtgjj — NNl — NAN-71 Iur’lc L =L =—L =—L =—=— A1 1A B] “3 ’12,», + mm + 112 + 729} 5A + 2m + £2 + g {pg-“pajn part {c}: (I I I d I I 1-51 = a [LAN-A + 15131113] — LAI E [?-A — 113] QED. Problem 1.24 part {a}: N A 1:12 = rug 2 [D['LL — Bil] part (b): (3A2 13 3'71 3'72 IUQD (3;? U2 2 _=f0 =_ — _ (it (It 29 (It N1}ngth € w‘w — — ( ) cos wt 29 2 Problem 1.35 From Fig. 1.19, the maximum energy product for neodymium—iron—boron oc— curs at (approximately) Bm = 0.63 T and H m 2 —47’0 kAfm. The magnetization curve for neodymium-iron-boron can be represented as Bm = JAR-Him + Br where Br 2 1.26 T and “R = 1.06Tuo. The magnetic circuit must. satisfy Hmd+ H59 = Ni; BmAm = 33.4g part [a]: For 1' = U and Bg = 0.5 T, the minimum magnet volume will occur when the magnet is operating at. the maximum energy point. Am = (Ba ) Ag = 4.?6 cm2 H. (I: — g 21.69 mm part (b): lull-411: “U N wee—e] 5,: For Bg = 0.?5, i = 1?.9 A. For Bg = 0.25, i = 6.0 A. Because the neodymium—iron—boron magnet. is essentially linear over the op— erating range of this problem. the system is linear and hence a sinusoidal flux variation will correspond to a sinusoidal current. variation. ...
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CH01 - Problem 1.1 Part. [a]: . IO 2.: , _, RC — “A:...

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