电机与拖动课后习题答案（第二版）

第一章第一章 21.解：解：
N
N N N I U P h ´´´=-310
N N N N U kw P I h ´´=310)(A
9.9085.022*******=´´=
KW
I U P N
N
N
20109.9022010331=´´=´´=--
24．解：．解：
1）
W
I U P N N N 176********=´=´= 2）
W P P N N N 1496085.0176001=´=´=h 3）W P P p N N 26401=-=S 4）
W
R I p a
a
cu
6401.08022
=´==
5）
W R U R I p f f
f f cuf 05.5458.8822022
2
===
=
6) W P p N ad 6.14901.0=´=
7) W
p p p p p ad cu Fe m 1850=--S =+
25. 解：解：
1）
m N r n KW P m N T N N N ×=´
==×12.543000969550min)/()(9550)(
2）方法一）方法一
A R U I f f f 212.15
.181220
===
A I I I f N a 69.87212.19.88=-=-=
07
.03000114
.069.87220=´-=-=F N a a N N e n R I U C
m
N I C I C T a
N e
a
T
em ×=´´=F =F =62.5869.8707.055.955.9
方法二方法二
W P P p N N 2558170009.882201=-´=-=S =--S =++cuf cu ad Fe m p p P p p p f
f a a R I R I p 2
2--S
W 73.14145.181212.1114.069.8725582
2
=´-´-=
m
N n p p p n p m N T ad
Fe m ×=++==×5.455.955.9)(00
m N T
T T T T N
em
×=+=+=+=62.585.412.540
2
3）%92.86%1009.88220100017%10010)(3
=´´´=´´´=
N N N N I U kw P h
4）
min /1080
2036
.0220
0r C U n E e ==F =
28. 解：解：
1）
m N r n KW P m N T N N N ×=´==×6.1883500969550min)/()(9550)( 2）
A
I I I f
N
a
2505255=-=-=
841
.0500078
.0250440=´-=-=
F N a a N N e n R I U C
m N I C I C T a N e a T em ×=´´=F =F =88.2007250841.055.955.9
3）min
/2.523841
.04400r C U n N e N ==F = 4）min
/3.470250841.01.0078.02.523r I C R R C U n a N e ad a N e N =´+-=F +-F =
第二章第二章
15．
（1）W =÷÷øöççèæ´-´=÷÷øöççèæ-=571.01.201075.11.201102121232N N N N a I P I U R 068
.01450571
.01.20110N a N N N e =´-=-=n R I U C f 649.055.9N e N T ==f f C C min
/r 1618068.0110N e N 0===f C U n
m N 04.131.20649.0N N T emN ×=´==I C T f
固有特性两点坐标为：固有特性两点坐标为：
A 点（min /r 1618
,00em ==n T ） B 点（
min /r 1450,m N 04.13N em ==×=n n T ） （2）
min /r 15341.205.0068.0571.01618a
N e a 0=´´-=-
=I C R n n f （3）a
N e a
0I C R n n f -=
A 5.10571.0068.0)15301618()
(a N e 0a =-=-=R C n n I f 16．（1）W =÷÷øöççèæ´-´=÷÷øöççèæ-=315.07.5310107.53220212123
2N N
N N a I P I U R
0677.03000315
.07.53220N a N N N e =´-=-=n R I U C f 6465.055.9N e N T ==f f C C
min
/r 32500677.0220N e N 0===f C U n 197.76465.00677.0315.02N T e a =´==f b C C R
em em 0197.73250T T n n -=-=b
（2）9.526465.00677.02
315.02
N T e ad a '
=´+=+=f b C C R R
em em '09.523250T T n n -=-=b
（3） min /r 16250677.01102N e N
N e a '0====f f C U C U n
A
5.10571.0068
.0)1530
1618()
(a N
e 0a =-=-=R C n n I f
em 197.71625T n -=
（4）min /r 4063
8.00677.0220''e a 0=´==f C U n
25.118.06466.00677.0315
.0''2
2T e a =´´==
f b C C R
em 25.114063T n -=
17．
（1）N a N a
a
N
st 8.15A 6.3283
067.0220
I R U R E
U
I ====-=
（2）A 25.3115.2075.15.1N s t =´==I I
W
=-=-=64.0067.025
.311220a st N st R I U R 19．
V 8.20612.0110220a N N aN =´-=-=R I U E
A 2751105.25.2N max -=´-=-=I I
W
=---=--=632
.012.02758.206a max aN ad R I E R
20. 
(1) 
a
N e a 0I C R n n f --= 172
.012008.206N aN N e =
==n E C f min
/r 12791723.0220N e N 0===f C U n
min
/r 135********.012.01279a N e a 0-=´--=--=I C R n n f (2) a
N
e ad
a 0I C R R n n f +--=
()()W
=--´+-=
--´+=
226.012.0110
1723.012771500a
N
N
e 0ad R I C n n R f
21． （1）
294
.0685296
.064220N a N N N e =´-=-=
n R I U C f
min /r 7671974864
3.029
4.0296
.0294.0220a N
e a
N e N a
N e a
0-=--=´´--=--=-
-=I C R C U I C R n n f f f ()()W =-´-´+´-=
--´+=837.0296.0643.0294.07486852.1a
N
N e 0R I C n n R ad f
(3) 可用串电阻方法可用串电阻方法
r/min
5.3425.0N =n
7.1793.0803.22935.0296.07485.342ad
´´´+-=R W =91.5ad R
22 
193
.010005
.054220N a N N N e =´-=-=n R I U C f
843.155.9N e N T =´=f f C C m N 522.99emN ×=T
(1) n 不能突变,
min /r 1000N ==n n A
R R n C U I ad a N N e N a 5.135.15.01000
193.0220=+´-=+-=f Nm
I C T a N T em
88.24==f
(2) L T 保持不变且L em T T =
稳定状态时，A 54N ==I I a
min /r 580522.99843.1193.02
90.1139emN 2
N T e ad a N e N =´´-=+-=
T C C R R C U n f f
（3）原%2.84542201000012N =´==
P P h
%
36.47542205.15410000%100212
12
=´´-=-=´=N ad a N I U R I P P P h
或：%
8.481000580
%2.84N N =´==n n h h
23．
（1）a e a
e N N I C R C U n
f f -=
A 2.1315.01000193.08.0220a N e N a
=´´-=-=R n C U I f m N 5.1932.131193.055.98.08.0a N T em ×=´´´==I C T f
（2）A
5.678.0548.08.0N N T N T N a =====I C T C T I f f
min /r 1206
5.678.0193.05.08.0193.0220a e a
e N =´´-´=
-=I C R C U n f
f
(3) 
12P P =
h kW
06.1210003
.120610N N 2=´==n n P P
kW 8514W 14850
567220a N 1
==´==I U P %2.81%10085.1406
.12%10012=´=
´=P P h
24. 
(1) 
a
N e a
N e N I C R C U
n f f -=
A 345.01000
193.0176a N N e a -=´-=-=R n C U I f m N 67.62)34(193.055.955.9a N e em ×-=-´´==I C T f
(2) 稳态时稳态时
A 54N a ==I I
min
/r 772193.0545.0176a
e a e N =´-=-=I C R C U n f
f
(3) W 9504
54176N 1=´==UI P W
7.771660772
14.325.9522=´´´
=W ×=T P
%2.8195047.7716%10012==´=
P P h
第三章第三章 4．
N 2N 2N 1N 1N 33750I U I U S === A
34.425
.10310750A
37.12353107503
N23
1N =´´==´´=I I
5. 
（1）此时,
V 110,
V 220ax A X =
=
U U 由于加于绕组两端的电压不变，所以m m F =F ¢不变不变
磁势不变磁势不变
设原边匝数为N 1 
0'
0'
1
'
0101322·
····
=+
=I I I N I N I N （2）,
V 110,V 220ax A X ==U U m m F =F ¢不变不变
磁势不变磁势不变
0'
0'
01'
010122
·
·
···=-=I I I N I N I N
8．
空载变比43.144.03
10
==
k
W
===42.22193
65400
320202m k I U k Z
W =÷øöçèæ=÷øöçè
æ=35.182********
33
2
2
200
2
m k I P k
R
W =-=91.22112
m 2m m R Z X
W ==
=42.7353
450sh
sh sh I U Z
W
===W ===02.12104.23537500
3sh '
212
2sh sh
sh R R R I P R W =-===57.321212sh
2sh sh '2σ1σR Z X X X
9．（1）
W =W =´+=+=W =´÷ø
öçèæ
+=+=+=
1.604.60137.096.287
2.000194.023********.0sh 2221sh 2
221sh 2sh
2sh
sh Z k X k X X R k R R R
X
Z
(3) 
()()()%
97.1%10034506.004.68.0872.0188.23%),(8.0cos %
9.2%1003450
6.004.68.0872.0188.23%),(8.0cos %
586.0%1003450
872
.0188.23%,1cos A
188.233450
1080%100sin cos %2223
N11
N 2sh 2sh N1-=´´-´´=D ==´´+´´=D ==´´=D ==´=´+=D u u u I U X R I u 超前滞后时j j j j j b
10.(1) 
()W Ð=+=+=W
+===´´=´==== 58.28735.315.636A
47.1151000310200100035
.23
40031000
'L sh L 2'L 3
N
1N 2
N 1N j Z Z Z j Z k Z S I U U k W -Ð=ÐÐ==··
58.2848.8258.2870310001N 1Z U I V
3.38307.12.20633A
2.206A
48.82L 22L 121=´´=====Z I U kI I I (2) 
()
kW
5.12558.28cos 48.8210003cos 31111=-´´´==
j I U P
878.0)58.28cos(cos
1=-= j
(3) 
()
()
%
100sin cos %100sin cos %N 1
2sh 2sh 1N 1
221´+=
´+=
D U X R I U X R I u K K N j j j j b 894.096.048.0cos cos cos 12=÷øöçè
æ
==-tg L j j ()%
2.4%1003
1000
449
.035.0894.015.048.82%=´´+´´=
D u
7143
.047.11548
.82==
b %5.97%100105.125894.02.2063.3833%100cos 33122212=´´´´´=´==P I U P P j h
11.(1) 
66
.8400
3
6000==
k
96.30746.85.45.7L 2
'L Ð=+==j z k Z
W =´´==064.09
.5383105632
32sh
shN sh I p R
A
9.5386000310
560033
N 1N N 1sh =´´=
=U S I I
W ==
=
3.09
.5383280
3sh
sh
sh I U Z
W
=-=293.02sh 2sh sh R z X
W Ð=W +=+= 36.32955.8793.4564.7'
L s h j Z Z Z
36.328.38636.32955.80
3
6000
3
11-Ð=ÐÐ=
=
·
·
z
U I N
A 8.3861=I A 5802312==kI I V
6.3901166.03
25
.58023L 2
2L =´=
×=Z I U
96.301166.006.01.0L Ð=+=j Z
(2) 
shN 202N 0
shN 2cos 1p p S p p b j b b h +++-=
718.09
.5388.386N 1
1=
=
=I I b
858
.096.30cos cos
2=
=
j
%
66.9810
181056718.0858.010*******.010*********.013
3
2
3
3
32=´+´´+´´´´+´´-
=h
567.010561018,3
3
shN 0m max =´´==p p b h 时
%
7.9810181056567.0858.010*******.010*********.0133233
32max
=´+´´+´´´´+´´-=h
或
%
7.982cos 210
2N 0
max
=+-
=p S p j b h
第四章第四章
24. 解：
05.01000950
10001N 1N =-=-=
n n n s
em N cu2P s p = ∴5%的电磁功率消耗在转子电阻上。

的电磁功率消耗在转子电阻上。

em N m )1(P s P -=
∴95%的电磁功率转化为机械功率。

的电磁功率转化为机械功率。

25. 解：2min r 1500min r 14501N =\=\=p n n
033.015001450
15001N 1N =-=-=
\n n n s
3min r 1000min r 9601=\=\=p n n e
04
.0100096010001N 1N =-=-=\n n n s 26. 解：A
97.9985.0894.038031050cos 33
N N N N N =´´´´==j h U P I
27. 解：033.015001450
15001N 1N =-=-=
n n n s
（1）
V 63.3110033.02N 2s =´==E s E （2）
A
8.35)
5.0033.0(1.063.3)
(2
2
2
σ222
2S 2=´+=
+=
sX R E I S
（3）Hz 65.150033.01N N 2=´==f s f
29. 解：解：
（1）
3
min
r 1000min r 9621N =\=
\=p n n
（2）038.01000962
10001N 1N =-=-=\n n n s Hz 9.150038.01N N 2=´==f s f （3）W 7625
80457500ad m 2m =++=++=p p P P W 2.7926
1N
m
em =-=
s P P
W 2.301em N cu2==P s p
（4）
%9.862344702.79267500%100Fe cu1em N 12
N
=++=++=´=p p P P P P h （5）A 86.15869.0827.038037500
cos 3N N N N N 1=´´´==j h U P I
30. 解：（1）
05.01000950
10001N 1N =-=-=
n n n s
（2）kW 53.1W 58.1531em N cu2===P s p
kW 63.30W 58.30631
05.0111002800011N m N N m
em ==-+=-+=-=
s p P s P P
（3）%3.85220058.3063128000
%100Fe cu1em N 12N =+=++=´=p p P P P P h
（4）A 67.56853
.088.0380328000
cos 3N N N N 1N =´´´==j h U P I （5） Hz 5.25005.01N N 2=´==f s f
31. 解：（1）0133.015001480
15001N 1N =-=-=
n n n s
（2）
96
.8353.142675.20380m 1m -Ð=+Ð==j Z U I 073
.00133.00133
.0177.0073.0404.0088.00380122112´-++++Ð=¢-+¢+=¢-j j R s s Z Z U I
89.1168.66-Ð= 88.2249.72m 21-Ð=+¢-=I I I
1
I 1
U m
I m
R
1Z 2Z ¢
2
I ¢ 2
1R s s
¢- m X
(3）W 2.732120133.0073.068.66332N 22
2em =´´=¢¢=s R I P m N 1.46655.91em em ×==n P T
或W 71100
110070000ad m 2m =+=++=p p P P m N 8.45855.9m em ×==n P T （4）921.0cos cos 1N ==j j
%94.91921.0)49.723(380370000cos 3%100N N N N 12N =´´´´==´=
j h I U P P P
第五章第五章 19 解：解：
04.075072075011=-=-=
n n n s e e m N n P T e e e .4.132672010*100*55.955.93==== 217.0)18.28.2(04.0)1(2
22=-+=-+=T
T e m K K s s
m st st m m st s s s s T T +=2=m N .35.15391217.0217.014.1326*8.2*2=+ 20 解：（1）04.075072075011=-=-=n n n s e e
183.0)14.24.2(04.0)1(2
22=-+=-+=T T e m K K s s m N n P T e e e .79.99472010*750*55.955.93==== m N T K T e T m .5.2387==
min 75.612)1(1r n s n m m =-=
m st st m m
st s s s s T T +=
2=m N .51.8451183.0183.015.2387*2=+
（2） W ==0224.03222e e e I E s R 221212)(s s X X R R s m ¢++¢= 221212)(s s X X R R R s m ¢++¢
+¢=¢ 2222R R R R R R s s m m +=¢¢+¢=¢\ =+=¢\22R R R s s m m 549.00224.00448.00224.0*183.0=+ min 25.338750*)549.01()1(1
r n s n m m =-=-= （3）
4.1750)300(75011=--=-=
n n n s
T 55)12-T K s s T m *0746.0==m 5r 100050010001=-=
-=n s s s T m *6.779*22¢¢(s m -¢-0746.05.2(1000)500(10001=--=-=n s s s T m *6.779*22¢¢(s m -¢-0746.05.7(
25 解：033.015001450150011=-=-=
n n n s e e
=-+=)1(22T T e m K K s s 0.1244 
m N n P T e e e .4.490.14510*5.7*55.955.93
==== m N T K T e T m .8.98== ===033.0*7.0e e L s T T s 0.231 带负载时的转速降为：==D 1sn n 0.0231*1500=34.65r/min 
(1) 65.93465.349001
=+=D +=¢n n n r/min Hz
n p f 15.31601
1=¢=¢
(2) 
V f f U U e e 9.23611=¢=。