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Solutions Manual Electric Machines

D P Kothari I J Nagrath

Tata McGraw Hill Education Private Limited NEW DELHI McGraw-Hill Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

Electric Machines

1

CHAPTER 2: MAGNETIC CIRCUITS AND INDUCTION Note Unless otherwise specified, leakage and fringing are neglected. 2.1 A square loop of side 2 d is placed with two of its sides parallel to an infinitely long conductor carrying current I. The centre line of the square is at distance b from the conductor. Determine the expression for the total flux passing through the loop. What would be the loop flux if the loop is placed such that the conductor is normal to the plane of the loop? Does the loop flux in this case depend upon the relative location of the loop with respect to the conductor? Solution 2d At distance r from conductor H=

I A/m 2p r

I

b –d

df = B dA =

m 0I ¥ 2d dr 2p r

f=

=

m 0 Id p

Ú

b+ d b-d

2d

b b+ d

Fig. P2.1

Ê m Id ˆ dr =Á 0 ˜ Ë p ¯ r

Hence

dr

r

µ 0I T B = m0 H = 2π r Flux passing through elemental strip

dr r

m 0 Id Ê b + dˆ ln Á Wb Ë b - d ˜¯ p

If conductor is normal to the plane of the loop, flux through loop is zero, independent of its relative location. 2.2 For the magnetic circuit of Fig. P2.2, find the flux density and flux in each of the outer limbs and the central limbs. Assume mr for iron of the core to be (a) • (b) 4500. Solution Mean flux path

A

0.5 A

1000 turns

1 mm 5 cm

30 cm

f1

10 cm

B

Fig. P2.2

30 cm

Core thickness = 5 cm

40 cm

f2

2 mm 5 cm

2 Solutions Manual (a) mr = • . The corresponding electrical analog of magnetic circuit is shown in Fig. P2.2(a). Ni = 1000 ¥ 0.5 = 500 AT

R g1

f2 A f

2 ¥ 10 -3 = 0.6366 ¥ 106 AT/Wb = 4p ¥ 10 -7 ¥ 25 ¥ 10 -4

R g2 =

f1

R g2

Ni

1 ¥ 10 -3 = 0.3183 ¥ 106 AT/Wb 4p ¥ 10 -7 ¥ 25 ¥ 10 -4

R g1

B

500 = 0.785 mWb f1 = 0. 6366 ¥ 10 6

Fig. P2.2(a)

B1 =

0 . 785 ¥ 10 -3 = 0.314 T 25 ¥ 10 -4

f2 =

500 = 1571 mWb 0. 3183 ¥ 10 6

B2 =

157 ¥10 -3 = 0.628 T 25 ¥ 10 -4

f = f1 + f2 = 2.356 mWb B=

2. 356 ¥ 10 -3 50 ¥ 10 -4

= 0.471 T

(b) mr = 4,500. The corresponding analogous electrical circuit is given in Fig. P2.2(b). Effect of air-gaps on iron path length is negligible. A

f2

f1 f

R c2

Ni

R c1

R g2

R c3

R g1

B

Fig. P2.2(b)

l c1 = lc2 = (40 + 5) + 2 ¥ (30 + 5 + 2.5) = 120 cm

R c1 = R c2 =

120 ¥ 10 -2 = 0.085 ¥ 106 AT/Wb 4p ¥ 10 -7 ¥ 4, 500 ¥ 25 ¥ 10 -4

l c3 = 40 + 5 = 45 cm

Electric Machines

R c3 =

45 ¥ 10 -2 4p ¥ 10 -7 ¥ 4, 500 ¥ 50 ¥ 10

-4

3

= 0.016 ¥ 106 AT/Wb

R eq = [(R c1 + R g1 ) || (R c2 + R g2)] + R c3 R c1 + R g1 = 0.6366 + 0.084 = 0.7206 ¥ 106 R c2 + R g2 = 0.3183 + 0.085 = 0.4033 ¥ 106 È 0.7206 ¥ 0.4033 ˘ + 0.016 ˙ ¥ 106 = 0.2746 ¥ 106 AT/Wb R eq = Í Î (0.7206 + 0.4033) ˚ f=

500 = 1823 mWb 0. 2742 ¥ 10 6

B=

1.823 ¥ 10 -3 = 0.365 T 50 ¥ 10 - 4

f1 = 1.823 ¥ B1 =

0 .4033 = 0.654 mWb 1. 1239

0. 653 ¥ 10 -3 = 0.261 T 25 ¥ 10 -4

. f2 = 1.823 ¥ 0 7206 = 1.17 mWb 1. 1239

B2 =

1. 17 ¥ 10 -3 = 0.468 T 25¥ 10 -4

2.3 For the magnetic circuit shown in Fig. P2.3, calculate the exciting current required to establish a flux of 2 mWb in the air-gap. Take fringing into account empirically. Use the B–H curve of Fig. 2.15. 20 cm

200 turns

15 cm

Ac = 5 cm ¥ 4 cm

0.1 cm f = 2 mWb

20 cm

Fig. P2.3

4 Solutions Manual 2.0

Crgos*

1.8

B(T)

1.6

1.4

1.2 1.0 0.05

0.1

0.2 0.3 0.5 H(kAT/m) *Cold rolled grain oriented steel

1

2

3

5

Fig. P2.3(a)

Solution

Taking fringing into account empirically Ag = (5 + 0.1) (4 + 0.1) = 20.91 ¥ 10 –4 m2 Bg =

2 ¥ 10 -3 = 0.957 T 20 . 91 ¥ 10 -4

Hg =

0. 957 = 7.616 ¥ 105 AT 4p ¥ 10 - 7

ATg = 7.616 ¥ 105 ¥ 0.1 ¥ 10–2 = 761.6 Bc =

2 ¥ 10 - 3 =1T 20 ¥ 10 -4

Corresponding Hc is obtained from B–H curve of Fig. 2.15 Hc = 0.06 kAT/m = 60 AT/m ATc = 60 ¥ 55 ¥ 10–2 = 33 1 (33 + 761.6) = 3.973 A 200 2.4 A steel ring has a mean diameter of 20 cm, a cross-section of 25 cm2 and a radial air-gap of 0.8 mm cut across it. When excited by a current of 1A through a coil of 1000 turns wound on the ring core, it produces an air-gap flux of 1 mWb. Neglecting leakage and fringing, calculate (a) relative permeability of steel, and (b) total reluctance of the magnetic circuit. Solution lc = p ¥ 20 – 0.08 = 62.75 cm; l g = 0.08 cm; Ag = 25 cm2

i=

Rg =

0 . 8 ¥ 10 -3 = 0.255 ¥ 106 4p ¥ 10 -7 ¥ 25 ¥ 10 -4

F = 1000 ¥ 1 = 1000 AT f = 1 mWb

Electric Machines

(a)

R (core) = (1 – 0.255) ¥ 106 = 0.745 ¥ 106

(b)

R (total) =

1000 ¥ 10 3 = 1 ¥ 106 1

5 (i) (ii)

From (i) 0.745 ¥ 106 =

62 . 75 ¥ 10 - 2 4p ¥ 10 -7 ¥ m rc ¥ 25 ¥ 10 -4

\ mrc = 268 2.5 The core made of cold-rolled silicon steel (B–H curve of Fig. 2.15) is shown in Fig. P2.5. It has a uniform cross-section (not iron) of 5.9 cm2 and a mean length of 30 cm. Coils A, B and C carry 0.4, 0.8 and 1 A respectively in the directions shown. Coils A and B have 250 and 500 turns respectively. How many turns must coil C have to establish a flux of 1 mWb in the core?

A

C B

Fig. P2.5

Solution NAiA = 250 ¥ 0.4 = 100 AT NBiB = 500 ¥ 0.8 = 400 AT B=

1 ¥ 10 -3 = 1.695 T 5. 9 ¥ 10 -4

Corresponding H from B–H curve of Fig. 2.15 is H = 0.5 kAT/m = 500 AT/m (AT)net = 500 ¥ 30 ¥ 10–2 = 150 = (AT)A + (AT)B – (AT)C or (AT)C = 100 + 400 – 150 = 350 NCiC = 350 \ NC = 350 2.6 In the magnetic circuit shown in Fig. P2.6, the coil F1 is supplying 4000 AT in the direction indicated. Find the AT of coil F2 and current direction to produce air-gap flux of 4 mWb from top to bottom. The relative permeability of iron may be taken as 2500.

6 Solutions Manual F ¢2

50 cm

50 cm

Ac = 40 cm2 20 cm 2 mm

F1

Solution 2.6(a).

The equivalent electric circuit is shown in Fig. l c1 = l c3 = 0.5 m

0. 5 4 p ¥10 -7 ¥ 2 , 500 ¥ 40 ¥10 = 3.98 ¥ 104 AT/Wb

R c1 = R c3 =

-4

4000 AT = (Ni )1

Fig. P2.6 f1

A

f2

f

(Ni )2

R c2 R c3

R c1 Rg

-

R c2 =

-4

B

= 1.59 ¥ 10 AT/Wb 4

Rg =

Fig. P2.6(a)

2 ¥ 10 -3 = 39.79 ¥ 104 AT/Wb 4 p ¥ 10 -7 ¥ 40 ¥ 10 -4

f = 4 mWb (AT)AB = f ( R c2 + R g) = 4 ¥ 10 –3 (159 + 39.79) ¥ 10 4 = 1,655 f1 =

4, 000 - 1, 655 3. 98 ¥ 10 4

= 58.9 mWb

f2 = f1 – f = 58.9 – 4 = 54.9 mWb 1, 655 + (Ni ) 2 = 54.9 ¥ 10–3 3 .98 ¥ 10 4

1,655 + (Ni)2 = 2,185 or (Ni)2 = 530 AT 2.7 For the magnetic circuit shown in Fig. P2.7, the air-gap flux is 0.24 mWb and the number of turns of the coil wound on the central limb is 1000. Calculate (a) the flux in the central limb and (b) the current required. The magnetization curve of the core is as follows: H (AT/m) 200 400 500 600 800 1060 1400 B (T) 0.4 0.8 1.0 1.1 1.2 1.3 1.4

Electric Machines 10 cm

10 cm

2 cm Air gap, 1 mm

2

1 15 cm

2 cm

2 cm 4 cm 2 cm

2 cm

B

Core thickness = 3 cm

1.4 1.2

1.0

B(T)

0.8

0.6

0.4

0.2

0

200

400

600

800

1000

1200

1400

H(AT/m) (a)

Fig. P2.7

Solution Bg =

0. 24 ¥ 10 -3 = 0.4 T ( 2 ¥ 3) ¥ 10 - 4

Hg =

0.4 = 31.83 ¥ 104 AT/m 4p ¥ 10 - 7

Bc = Bg = 0.4 T (no fringing) From the B–H data given, H1 = 200 AT/m (AT)AB = H1l1 + Hgl g = 200 ¥ (2 ¥ 10 + 15) ¥ 10–2 + 31.83 ¥ 104 ¥ 1 ¥ 10–3 = 388.3

7

8 Solutions Manual For the left limbs H2 =

(AT) AB 388. 3 = = 1,109 AT/m l2 35 ¥ 10 -2

From the B–H curve given B2 = 1.31 T f2 = 1.31 ¥ (2 ¥ 3 ¥ 10 –4) = 0.786 mWb fc (central limb) = 0.24 + 0.786 = 1.026 mWb Bc (central limb) =

1. 026 ¥ 10-3 4 ¥ 3 ¥ 10 - 4

= 0.855 T

Hc (central limb) = 430 AT/m ATC (central limb) = 430 ¥ 15 ¥ 10–2 = 64.5 (AT)total = (AT)AB + 64.5 = 388.3 + 64.5 = 452.8 i=

452. 8 = 0.453 A 1, 000

2.8 The magnetic circuit shown in Fig. P2.8 has a coil of 500 turns wound on the central limb which has an air-gap of 1 mm. The magnetic path from A to B via each outer limb is 100 cm and via the central limb 25 cm (air-gap length excluded). The cross-sectional area of the central limb is 5 cm ¥ 3 cm and each outer limb is 2.5 cm ¥ 3 cm. A current of 0.5 A in the coil produces an air-gap flux of 0.35 mWb. Find the relative permeability of the medium. 1 mm, f = 0.35 mWb

A 0.5A

A c = 15 cm2 500 Turns

100 cm

Ac = 7.5 cm2

25 cm

B

Fig. P2.8

Solution

Figure P2.8(a) gives the electric equivalent of the magnetic circuit of Fig. P2.8. f1

A

f2

f Rg R c1

R c3 R c2

B

Fig. P2.8(a)

Equivalent Electric Circuit of Fig. P2.18

Electric Machines

R c1 = R c3 = R c2 = Rg =

4p ¥

9

100 ¥ 10 -2 1061 ¥ 10 9 = AT/Wb 4 mr ¥ m r ¥ 7. 5 ¥ 10

10 -7

25 ¥10 -2 0 .133 = ¥ 109 AT/Wb mr 4p ¥ 10 ¥ m r ¥ 15 ¥ 10 -4 -7

1 ¥ 10 - 3 = 0.5305 ¥ 106 AT/Wb 4p ¥ 10 -7 ¥ 15 ¥ 10 -4

Ni = f(R c1 || R c3 + R g + R

c2)

Ê 0.531 ¥ 109 0.133 ¥ 109 ˆ 500 ¥ 0.5 = 0.35 ¥ 10–3 Á + + 0.5305 ¥ 10 6 ˜ mr mr Ë ¯ mr = 3,612 2.9 A cast steel ring has an external diameter of 32 cm and a square cross-section of 4 cm side. Inside and across the ring, a cast steel bar 24 ¥ 4 ¥ 2 cm is fitted, the butt-joints being equivalent to a total air-gap of 1 mm. Calculate the ampere-turns required on half of the ring to produce a flux density of 1 T in the other half. Given: H (AT/m) 0 200 400 600 800 1000 1200 1400 1600 B (T) 0 0.11 0.32 0.6 0.8 1.0 1.18 1.27 1.32 Solution Figure P2.9(a) shows the sketch of the cast steel ring. B1 = 1T; H1 = 1,000 AT/m (From graph of Fig. P2.9(b)) 32 cm dia. Cross-section = 4 cm ¥ 4 cm

A f1

B1 = 1T

fr fc r

AT 14 cm 24 cm ¥ 4 cm ¥ 2 cm Air gap equivalent of 1 mm

B

(a) Fig. P2.9(a)

(AT)AB = AT1 = 1,000 ¥ p ¥ 14 ¥ 10–2 = 439.82 f1 = 1 ¥ 16 ¥ 10 –4 = 1.6 mWb ATC = (AT)AB = 439.82 439.82 =

Bc ¥ 1 ¥ 10 –3 + Hc ¥ 0.28 4p ¥ 10 - 7

439.82 = 795.77 Bc + 2.28 Hc

10 Solutions Manual 1.4 1.2

B(T)

1.0 0.8 0.6 0.4 0.2 0

200

400

600

800 1000 1200 1400 1600 1800 H(AT/m) (b)

Fig. P2.9(b)

Intersection of the above straight line with B–H curve gives Bc = 0.39 T; fc = 0.39 ¥ 8 ¥ 10 –4 = 0.312 mWb fr = f1 + fc = (1.6 + 0.312) mWb = 1.912 mWb Br =

1. 912 ¥ 10 -3 = 1195 T 16 ¥ 10 -4

From graph, Hr = 1230 AT/m \ ATr = 1230 ¥ p ¥ 14 ¥ 10–2 = 540.98 Total AT = 439.82 + 540.98 = 980.8 2.10 In Prob. 2.2 the B-H curve of the core material is characterized by the data given below. Find the flux and flux densities in the three limbs of the core. H (AT/m) 50 100 150 200 250 300 350 B (T) 0.14 0.36 0.66 1.00 1.22 1.32 1.39 Hint This problem can be solved by the graphical-cum-iterative technique. Solution The B–H curve as per the data is drawn in Fig. P2.10. Using the solution of 2.1(a) as a starting point: B1 = 0.34 Æ H1 = 90; H1l1 = 90 ¥ 1.2 = 108 B2 = 0.628 Æ H2 = 145; H2l2 = 145 ¥ 1.2 = 174 B = 0.471 Æ H = 120; Hl = 120 ¥ 0.45 = 54 ATg1 = 500 – 108 – 54 = 338

B1 (new) =

440 ¥ 4p ¥ 10 -7 = 0.212 2 ¥ 10 -3

ATg2 = 500 – 174 – 54 = 272

B2 (new) =

272 ¥ 4p ¥ 10 -7 = 0.342 1¥ 10 -3

B(new) =

0 . 212 ¥ 0 .342 = 0.277 2

Electric Machines

11

1.4 1.2

B(T)

1.0 0.8 0.6 0.4 0.2 0

50

100

150

200 250 AT/m

300

350

400

Fig. P2.10

B1 = 0.212

H1 = 70

H1l1 = 70 ¥ 1.2 = 84

B2 = 0.342 B1 = 0.277

H2 = 95 H = 83

H2l2 = 95 ¥ 1.2 = 114 Hl = 83 ¥ 0.45 = 38

ATg1 = 500 – 84 – 38 = 378

B1 =

ATg2 = 500 – 114 – 38 = 348

B2 =

B1 = 0.237 B2 = 0.437 B4 = 0.337

H1 = 70 H2 = 120 H = 95

378 ¥ 4p ¥10 -7 = 0.237 2 ¥ 10 -3 348 ¥ 4p ¥ 10

-7

= 0.437

1¥ 10 -3

B = (0.237 + 0.437)/2 = 0.337 T H1l1 = 70 ¥ 1.2 = 84 H2l2 = 120 ¥ 1.2 = 144 Hl = 95 ¥ 0.43 = 40.85

373 ¥ 4 p ¥ 10 2 ¥ 10 -3

-7

= 0.234

ATg1 = 500 – 84 – 43 = 373

B1 =

ATg2 = 500 – 144 – 43 = 313

B2 = 313 ¥ 4p ¥ 10–4 = 0.4

0 . 634 = 0.317 T 2 (Almost converged) 2.11 A ring of magnetic material has a rectangular cross-section. The inner diameter of the ring is 20 cm and the outer diameter is 25 cm, its thickness being 2 cm. An air-gap of 1 mm length is cut across the ring. The ring is wound with 500 turns and when carrying a current of 3 A produces a flux density of 1.2 T in the air-gap. Find (a) magnetic field intensity in the magnetic material and in the air-gap (b) relative permeability of the magnetic material, and (c) total reluctance of the magnetic circuit and component values.

B=

12 Solutions Manual Solution tic ring.

Figure P2.11 gives the sketch of the magne-

3A

Ni = 500 ¥ 3 = 1,500 AT Bc = Bg = 1.2 T (no fringing) (a)

Hg =

1. 2 = 9.549 ¥ 105 AT/m 4p ¥ 10 -7

500 ¥ 3 = 9.549 ¥ 10 ¥ 1 ¥ 10 ¥ 11.25 ¥ 10–2 \ Hc = 771.16 AT/m 5

–3

Hc =

Bc m0 mr

\

mr =

1.2 = 1,238.3 4p ¥ 10 -7 ¥ 771.16

(c)

R total = R

2.12 For (a) (b) (c) Solution (a)

g

1 mm

11.25 cm 12.5 cm 2.5 cm

Fig. P2.11

+ Rc

1 ¥ 10 -3 2p ¥ 11. 25 ¥ 10- 2 + 4 p ¥ 10 -7 ¥ 2 ¥ 2. 5 ¥ 10 -4 4p ¥ 10 -7 ¥ 1, 238. 3 ¥ 2 ¥ 2. 5 ¥ 10 -4

= 1.592 ¥ 106 + 0.909 ¥ 10 6 = 2.5 ¥ 106 AT/Wb the magnetic ring of Prob. 2.11, the exciting current is again 3 A. Find the following: Inductance of the coil energy stored in the magnetic material and in the air-gap rms emf induced in the coil when it carries alternating current of 3 sin 314 t. Refer to Fig. P2.11. f = BA = 1.2 ¥ 2.5 ¥ 2 ¥ 10 –4 = 0.6 mWb l = Nf = 500 ¥ 0.6 ¥ 10–3 = 0.3 WbT L=

l 0 .3 = = 0.1H 3 i Bc

(b)

10 cm

+ Hc ¥ 2p

(b)

=

Thickness = 2 cm

N = 500

Wfc = Aclc

Ú

Hc dBc =

0

Ac l c m0 m r

Bc

ÚB

0

c

ÊA l ˆ dB c = 1 Á c c ˜ B 2c 2 Ë m 0 mr ¯

mr (as determined in Prob. 2.9) = 1,238.3 Wfc =

1 5 ¥ 10 -4 ¥ 22 .5 ¥ p ¥ 10 -2 ¥ ¥ (1.2)2 2 4 p ¥ 10 -7 ¥ 1, 238. 3

= 0.1635 J Bg = Bc = 1.2 T (no fringing) Wfg =

1 5 ¥ 10 -4 ¥ 10 -3 (1.2)2 ¥ 2 4 p ¥ 10 - 7

= 0.2865 J

Electric Machines

13

dl d = 0.3 sin 314 t d t dt = 94.2 cos 314 t V 2.13 Assume that the core of the magnetic circuit of Fig. P2.3 has mr = 2500. (a) Calculate the energy stored in the core and in the air-gap for an excitation current of 5 A. What will be these values if mr = •? (b) What will be the excitation current to produce a sinusoidally varying flux of 0.5 sin 314 t mWb in the air-gap? (c) Calculate the inductance of the coil. What will be the inductance if mr = •? Solution (c)

e=

(a)

Rc =

70 ¥ 10 -2 7 p 4 ¥ 10 ¥ 2, 500 ¥ 20 ¥ 10

Rg =

0 .1 ¥ 10 -2 = 0.381 ¥ 106 AT/Wb 4p ¥ 10 -7 ¥ 20. 91 ¥ 10 -4

-4

= 0.111 ¥ 106 AT/Wb

R total = 0.492 ¥ 106 AT/Wb 200 ¥ 5 = f R total or

Let

f=

200 ¥ 5 0. 492 ¥ 10 6

= 2.033 mWb

Wf = (1/2) f2 R Wf (core) = (1/2) ¥ (2.033 ¥ 10–3)2 ¥ 0.111 ¥ 10 6 = 0.23 J Wf (air-gap) = (1/2) ¥ (2.033 ¥ 10–3)2 ¥ 0.381 ¥ 10 6 = 0.787 J mr = • f iR c = 0 R total = R g = 0.381 ¥ 106 AT/Wb f=

200 ¥ 5 = 2.625 mWb 0. 381 ¥ 10 6

Wf (air-gap) = (1/2) ¥ (2.625 ¥ 10–3) ¥ 0.38 ¥ 106 = 1.312 J Wf (core) = 0 (as R c = 0) (b)

N R

= 0.5 ¥ 10 –3 sin 314 t

total

Ê 0.492 ¥ 106 ¥ 0.5 ¥ 10 -3 ˆ i= Á ˜¯ sin 314 t 200 Ë = 1.23 sin 314 t A (c) If

\

L = N2 P =

(200 ) 2 = 0.0813 H 0. 492 ¥ 10 6

mr = • R total = R g = 0.381 ¥ 106 AT/Wb L=

( 200) 2 = 0.105 H 0. 381 ¥ 10 6

14 Solutions Manual 2.14 The magnetic circuit of Fig. P2.14 has a magnetic core of relative permeability 1600 and is wound with a coil of 1500 turns excited with sinusoidal ac voltage, as shown. Calculate the maximum flux density of the core and the peak value of the exciting current. What is the peak value of the energy stored in the magnetic system and what percentage of it resides in the air-gap? f

i + 200 V

E

0.15 mm

f = 50 Hz

Cross-sectional area = 5 cm2

20 cm

Fig. P2.14

Solution or

E ª V = 200 = 4.44 ¥ 50 ¥ 1500 ¥ fmax 200 fmax = = 0.6 mWb 4. 44 ¥ 50 ¥ 1500 Bmax =

6 0 .6 = = 1.2 T 1000 ¥ 5 ¥10 -4 5

Rc =

20 ¥ 10 - 2 4p ¥ 10 -7 ¥ 1600 ¥ 5 ¥ 10

Rg =

0 .15 ¥ 10 -3 = 0.239 ¥ 106 4p ¥ 10 -7 ¥ 5 ¥ 10 -4

-4

= 0.2 ¥ 106

R (total) = 0.439 ¥ 106 Fmax = imax ¥ 1500 = 0.439 ¥ 10 6 ¥ 0.6 ¥ 10 –3 imax = 0.176 A 1 1 2 (0.6 ¥ 10 –3)2 ¥ 0.439 ¥ 106 R (total) = W(peak) = ÊÁ ˆ˜ fmax Ë2 ¯ 2

= 0.079 J

0. 239 ¥ 100 = 54.4 0. 439 2.15 The material of the core of Fig. P2.15 wound with two coils as shown, is sheet steel (B-H curve of Fig. 2.15. Coi...