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PROBLEM SOLUTIONS: Chapter 1
Problem 1. Part (a): Rc =
lc μAc
lc μrμ 0 Ac
= 0 A/Wb
Rg =
g μ 0 Ac = 1. 017 × 106 A/Wb
part (b):
N I
Rc + Rg
= 1. 224 × 10 −^4 Wb
part (c):
λ = N Φ = 1. 016 × 10 −^2 Wb
part (d):
L =
λ I
= 6. 775 mH
Problem 1. part (a): Rc =
lc μAc
lc μrμ 0 Ac
= 1. 591 × 105 A/Wb
Rg =
g μ 0 Ac
= 1. 017 × 106 A/Wb
part (b):
N I
Rc + Rg
= 1. 059 × 10 −^4 Wb
part (c):
λ = N Φ = 8. 787 × 10 −^3 Wb
part (d):
L =
λ I = 5. 858 mH
part (c):
Problem 1. part (a):
Hg =
N I
2 g ; Bc =
Ag Ac
Bg = Bg
x X 0
part (b): Equations
2 gHg + Hclc = N I; Bg Ag = BcAc
and
Bg = μ 0 Hg; Bc = μHc
can be combined to give
Bg =
N I
2 g +
μ 0 μ
Ag Ac
(lc + lp)
N I
2 g +
μ 0 μ
1 − (^) Xx 0
(lc + lp)
Problem 1. part (a):
I = B
g +
μ 0 μ
(lc + lp) μ 0 N
= 2. 15 A
part (b):
μ = μ 0
1 + 0. 05 B^8
= 1012 μ 0
I = B
g +
μ 0 μ
(lc + lp) μ 0 N
= 3. 02 A
part (c):
Problem 1.
g =
μ 0 N 2 Ac L
μ 0 μ
lc = 0. 353 mm
Problem 1. part (a):
lc = 2π(Ro − Ri) − g = 3. 57 cm; Ac = (Ro − Ri)h = 1. 2 cm^2
part (b):
Rg = g μ 0 Ac
= 1. 33 × 107 A/Wb; Rc = 0 A/Wb;
part (c):
L =
N 2
Rg + Rg
= 0. 319 mH
part (d):
I =
Bg(Rc + Rg )Ac N
= 3 3. 1 A
part (e):
λ = N BgAc = 10. 5 mWb
Problem 1. part (a): Same as Problem 1. part (b):
Rg =
g μ 0 Ac
= 1. 33 × 107 A/Wb; Rc =
lc μAc
= 3. 16 × 105 A/Wb
Problem 1. part (a):
R 3 =
R^21 + R^22 = 4. 27 cm
part (b):
L =
μ 0 AgN 2 g +
μ 0 μ
lc
= 251 mH
part (c): For ω = 2π60 rad/sec and λpeak = N Ag Bpeak = 0.452 Wb:
(i) Vrms = ωλpeak = 171 V rms
(ii) Irms =
Vrms ωL
= 1. 81 A rms
(iii) Wpeak = 0. 5 L(
2 Irms)^2 = 0. 817 J
part (d): For ω = 2π50 rad/sec and λpeak = N AgBpeak = 0.452 Wb:
(i) Vrms = ωλpeak = 142 V rms
(ii) Irms = Vrms ωL
= 1. 81 A rms
(iii) Wpeak = 0. 5 L(
2 Irms)^2 = 0. 817 J
Problem 1. part (a):
part (b):
Emax = 4f N AcBpeak = 3 45 V
Problem 1. part (a):
N =
LI
AcBsat
= 99 turns; g =
μ 0 N I Bsat
μ 0 lc μ
= 0. 3 6 mm
part (b): From Eq.3.
Wgap =
AcgB^2 sat 2 μ 0
= 0. 207 J; Wcore =
AclcB^2 sat 2 μ
= 0. 045 J
Thus Wtot = Wgap + Wcore = 0.252 J. From Eq. 1.47, (1/2)LI^2 = 0.252 J. Q.E.D.
Problem 1. part (a): Minimum inductance = 4 mH, for which g = 0.0627 mm, N = 20 turns and Vrms = 6.78 V part (b): Maximum inductance = 144 mH, for which g = 4.99 mm, N = 1078 turns and Vrms = 224 V
Problem 1. part (a):
L =
μ 0 πa^2 N 2 2 πr
= 56. 0 mH
part (b): Core volume Vcore ≈ (2πr)πa^2 = 40.0 m^3. Thus
W = Vcore
B^2
2 μ 0
= 4. 87 J
part (c): For T = 30 sec,
di dt
(2πrB)/(μ 0 N ) T
= 2. 92 × 103 A/sec
v = L di dt
= 163V
Problem 1. part (a):
Acu = fwab; Volcu = 2ab(w + h + 2a)
part (b):
part (b):
(i) B 1 = 0; B 2 =
μ 0 N 2 I 2 g 2
(ii) λ 1 = N 1 A 2 B 2 = μ 0 N 1 N 2
A 2
g 2
I 2
(iii) λ 2 = N 2 A 2 B 2 = μ 0 N 22
A 2
g 2
I 2
part (c):
(i) B 1 =
μ 0 N 1 I 1 g 1
; B 2 =
μ 0 N 1 I 1 g 2
μ 0 N 2 I 2 g 2
(ii) λ 1 = N 1 (A 1 B 1 + A 2 B 2 ) = μ 0 N 12
A 1
g 1
A 2
g 2
I 1 + μ 0 N 1 N 2
A 2
g 2
I 2
(iii) λ 2 = N 2 A 2 B 2 = μ 0 N 1 N 2
A 2
g 2
I 1 + μ 0 N 22
A 2
g 2
I 2
part (d):
L 11 = N 12
A 1
g 1
A 2
g 2
; L 22 = μ 0 N 22
A 2
g 2
; L 12 = μ 0 N 1 N 2
A 2
g 2
Problem 1.
RA =
lA μAc
; R 1 =
l 1 μAc
; R 2 =
l 2 μAc
; Rg =
g μ 0 Ac
part (a):
L 11 =
N 12
R 1 + R 2 + Rg + RA/ 2
N 12 μAc l 1 + l 2 + lA/2 + g (μ/μ 0 )
LAA = LBB =
N 2
RA + RA||(R 1 + R 2 + Rg )
N 2 μAc lA
[
lA + l 1 + l 2 + g (μ/μ 0 ) lA + 2(l 1 + l 2 + g (μ/μ 0 ))
]
part (b):
LAB = LBA =
N 2 (R 1 + R 2 + Rg ) RA(RA + 2(R 1 + R 2 + Rg ))
N 2 μAc lA
[
l 1 + l 2 + g (μ/μ 0 ) lA + 2(l 1 + l 2 + g (μ/μ 0 ))
]
LA1 = L1A = −LB1 = −L1B =
−N N 1
RA + 2(R 1 + R 2 + Rg )
−N N 1 μAc lA + 2(l 1 + l 2 + g (μ/μ 0 ))
part (c):
v 1 = d dt
[LA1iA + LB1iB] = LA d dt
[iA − iB]
Q.E.D.
Problem 1. part (a):
L 12 =
μ 0 N 1 N 2 2 g
[D(w − x)]
part (b):
v 2 = dλ 2 dt
= I 0
dL 12 dt
N 1 N 2 μ 0 D 2 g
dx dt
= −
N 1 N 2 μ 0 D 2 g
% ωw 2
cos ωt
Problem 1. part (a):
H =
N 1 i 1 2 π(Ro + Ri)/ 2
N 1 i 1 π(Ro + Ri)
part (b):
v 2 =
d dt [N 2 (tn∆)B] = N 2 tn∆
dB dt
part (c):
vo = G
v 2 dt = GN 2 tn∆B
part (b): Area = 191 Joules part (c): Core loss = 1.50 W/kg.
Problem 1. Brms = 1.1 T and f = 60 Hz,
Vrms = ωN AcBrms = 46. 7 V
Core volume = Aclc = 1. 05 × 10 −^3 m^3. Mass density = 7. 65 × 103 kg/m^3. Thus, the core mass = (1. 05 × 10 −^3 )(7. 65 × 103 ) = 8.03kg. At B = 1.1 T rms = 1.56 T peak, core loss density = 1.3W/kg and rms VA density is 2.0 VA/kg. Thus, the core loss = 1. 3 × 8 .03= 10 .4 W. The total exciting VA for the core is 2. 0 × 8 .03= 16 .0 VA. Thus, its reactive component is given by
16. 02 − 10. 42 = 12.2 VAR.
The rms energy storage in the air gap is
Wgap =
gAcB^2 rms μ 0 = 3. 61 Joules
corresponding to an rms reactive power of
VARgap = ωWgap = 1361 Joules
Thus, the total rms exciting VA for the magnetic circuit is
VArms = sqrt 10. 42 + (1361 + 12.2)^2 = 1373 VA
and the rms current is Irms = VArms/Vrms = 29.4 A.
Problem 1. part(a): Area increases by a factor of 4. Thus the voltage increases by a factor of 4 to e = 1096cos 377 t. part (b): lc doubles therefore so does the current. Thus I = 0.26 A. part (c): Volume increases by a factor of 8 and voltage increases by a factor of 4. There Iφ,rms doubles to 0.20 A. part (d): Volume increases by a factor of 8 as does the core loss. Thus Pc = 128 W.
Problem 1. From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately) B = 0.47 T and H = -360 kA/m. Thus the maximum energy product is 1. 69 × 105 J/m^3. Thus,
Am =
2 cm^2 = 3.40 cm^2
and
lm = − 0 .2 cm
μ 0 (− 3. 60 × 105 )
= 0.3 5 cm
Thus the volume is 3. 40 × 0 .3 5 = 1.20 cm^3 , which is a reduction by a factor of 5.09/1.21 = 4.9.
Problem 1. From Fig. 1.19, the maximum energy product for neodymium-iron-boron occurs at (approximately) B = 0.63 T and H = -470 kA/m. Thus the maximum energy product is 2. 90 × 105 J/m^3. Thus,
Am =
2 cm^2 = 2.54 cm^2
and
lm =^ −^0 .2 cm
μ 0 (− 4. 70 × 105 )
= 0.27 cm
Thus the volume is 2. 54 × 0 .25 = 0.688 cm^3 , which is a reduction by a factor of 5.09/0.688 = 7.4.
Problem 1. From Fig. 1.19, the maximum energy product for samarium-cobalt occurs at (approximately) B = 0.47 T and H = -360 kA/m. Thus the maximum energy product is 1. 69 × 105 J/m^3. Thus, we want Bg = 1.2 T, Bm = 0.47 T and Hm = −360 kA/m.
hm = −g
Hg Hm
= −g
Bg μ 0 Hm
= 2. 65 mm
Am = Ag
Bg Bm
= 2πRh
Bg Bm
= 26. 0 cm^2
Rm =
Am π
= 2. 87 cm
Problem 1. From Fig. 1.19, the maximum energy product for neodymium-iron-boron oc- curs at (approximately) Bm = 0.63 T and Hm = -470 kA/m. The magnetization curve for neodymium-iron-boron can be represented as
Bm = μRHm + Br
where Br = 1.26 T and μR = 1. 067 μ 0. The magnetic circuit must satisfy
PROBLEM SOLUTIONS: Chapter 2
Problem 2. At 60 Hz, ω = 120π.
primary: (Vrms)max = N 1 ωAc(Brms)max = 2755 V, rms
secondary: (Vrms)max = N 2 ωAc(Brms)max = 172 V, rms
At 50 Hz, ω = 100π. Primary voltage is 2295 V, rms and secondary voltage is 143 V, rms.
Problem 2.
N =
2 Vrms ωAcBpeak
= 167 turns
Problem 2.
N =
= 3 turns
Problem 2. Resistance seen at primary is R 1 = (N 1 /N 2 )^2 R 2 = 6.25Ω. Thus
I 1 =
V 1
R 1
= 1. 6 A
and
V 2 =
N 2
N 1
V 1 = 40 V
Problem 2. The maximum power will be supplied to the load resistor when its im- pedance, as reflected to the primary of the idealtransformer, equals that of the source (2 kΩ). Thus the transformer turns ratio N to give maximum power must be
N =
Rs Rload
Under these conditions, the source voltage will see a total resistance of Rtot = 4 kΩ and the current will thus equal I = Vs/Rtot = 2 mA. Thus, the power delivered to the load will equal
Pload = I^2 (N 2 Rload) = 8 mW
Here is the desired MATLAB plot:
Problem 2. The maximum power will be supplied to the load resistor when its im- pedance, as reflected to the primary of the idealtransformer, equals that of the source (2 kΩ). Thus the transformer turns ratio N to give maximum power must be
N =
Rs Rload
Under these conditions, the source voltage will see a total impedance of Ztot = 2 + j2 kΩ whose magnitude is 2
2 kΩ. The current will thus equal I = Vs/|Ztot| = 2
2 mA. Thus, the power delivered to the load will equal
Pload = I^2 (N 2 Rload) = 16 mW
Here is the desired MATLAB plot:
Problem 2. part (a):
part (b):
Iˆload = 30 kW 230 V
ejφ^ = 93. 8 ejφ^ A
where φ is the power-factor angle. Referred to the high voltage side, IˆH =
- 38 ejφA.
VˆH = ZH IˆH
Thus, (i) for a power factor of 0.85 lagging, VH = 2413 V and (ii) for a power factor of 0.85 leading, VH = 2199 V. part (c):
Problem 2. part (a):
part (b): Following methodology of Problem 2.11, (i) for a power factor of 0.85 lagging, VH = 4956 V and (ii) for a power factor of 0.85 leading, VH = 4000 V. part (c):
Problem 2. part (a): Iload = 160 kW/2340 V = 68.4 A at = cos−^1 (0.89) = 27. 1 ◦
V^ ˆt,H = N ( VˆL + ZtIL)
which gives VH = 33.7 kV. part (b):
Vˆsend = N ( VˆL + (Zt + Zf )IL)