Figure B16
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B27 For the circuit of Figure B16, where i = 2A, = 1 10 3 Wb, cross-sectional
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F = R Also, the magnetomotive force is related to the current i by F = N i = 100 2 = 200 A t which means that the reluctance is R= 200 A t in2 F = = 200,000 3 1 10 Wb
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Therefore, the answer is b
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The nonelectrical engineer will be frequently exposed to electrical machinery This subject is covered in s 17 and 18 The most popular electric machines in engineering applications are the DC motor and the AC induction motor The
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Appendix B
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Fundamentals of Engineering (FE) Examination
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former is discussed in the rst half of 17, while the latter is discussed in 17 (three-phase motors) and in 18 (single-phase motors)
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Check Your Understanding
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B28 A four-pole synchronous motor operating from a 60-Hz supply will have a synchronous speed, in rev/min, of a 900 b 1,800 c 1,200 d 3,600 e 4,800
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Solution: Frequency = (no of poles/2) rev/min/60) Therefore, 60 = (4/2) (rev/min/60) or rev/min = 1,800 Thus, b is the answer
B29 The armature resistance of a 55-hp, 525-V, DC shunt wound motor is 04 The full-load armature current of this motor is 80 A If the initial starting current is 175 percent of the full-load value, the resistance of the starting coil in should be nearest to a 66125 b 375 c 335 d 415 e 275
Solution: Istart (starting current) = 175 80 = 140 A Therefore, the total resistance is R= 525 V = 375 = IS 140 The correct answer is c
Thus, the resistance of the starter is 375 04 = 335
B30 The speed of an AC electric motor
a b c d e Is independent of the frequency Is directly proportional to the square of the frequency Varies directly with the number of poles Varies inversely with the number of poles None of the above
Solution: The speed varies inversely with the number of poles and directly with the frequency Thus, d is the correct answer
A P P E N D I X
Answers to Selected Problems
2
24 212 a 360,000 C b 2247 1022 Element A: 300 W (dissipating) Element B: 375 W (dissipating) Element C: 675 W (supplying) 288 , 192 ; 1152 R = 18 k , v = 16 V, v1 = 2 V, I = 05 mA b 05 A c 3 W d ex b x = 908 cm a vout (x) = 2203 a rB = 0061 b rB = 841 a i 1 mA
With meter in circuit a b c d 861 mA 396 mA 619 mA 656 mA
215 219 228 249 254 256 260
2 V
b ra = 928
Without meter in circuit 892 mA 472 mA 826 mA 893 mA
Appendix C
Answers to Selected Problems
3
33 313 317 320 332 336 339 346 358 i1 = 0143A v = 066V i = 829A v2 = 004 AV = v1 RTH = 4 vTH = 214 V IN = iSC = 125 A IN = 305 A io = 2i = 0 RN = a I = 522 mA; V = 457 V b Rinc = 438 c I = 73 mA; V = 540 V; Rinc = 37 RN = RTH = 057 i2 = 0856A
4
41 47 418 420 423 438 443 448 450 454 457 vL (t) = 377 sin 377t w1F = 72 J w1H xrms = 287 V vrms = 640 V irms = 11547 A i(t) = 212 cos t vout (t) = 90 cos( t) V a 8 A , VT = 10 0 V
5 V 6 = 8 J w2H = 4 J
w2F = 36 J
ZT = 500 + j 1001
ZT = 500 j 1001
, VT = 10 0 V
i(t) = 02357 cos(2t + 45 ) A ZT = 2 VT = 1414 45 V
5
521 523 526 528 539 542 545 548 a vc (0 ) = vc (0+ ) = 0 V b = 48 s c vc (t) = 8e 1/48t + 8 V t > 0 d vc (0) = 0 V; vc ( ) = 506 V; vc (2 ) = 69 V; vc (5 ) = 795 V; vc (10 ) = 80 V a vc (0 ) = 1167 V b Vc (t) = 1109e 3/70t + 1167 V t > 0 = 0923 ms; = 1333 ms a = 5005 ms L = 16 mH, R = 560 v(t) = 12e 2t 12te 2t + 12 V for t > 0 v(t) = 18e t 3e 6t V t >0 b = 5 s i(t) = 2 + e 0041t { 3641 cos[0220(t 5)] + 177 sin[0220(t 5)]} A For t > 5s, the inductor current is:
