江苏2020年苏南五市单招二模卷答案(电子电工)
2020年苏南五市职业学校对口单招第二次调研性统测电子电工专业综合理论试卷答案及评分参考
一、单项选择题(本大题共22小题,每小题4分,共88分)
二、判断题(本大题共16小题,每小题2分,共32分)
三、填空题(本大题共18小题,30空,每空2分,共60分)
39.?17 4
40.9 72
41.?
16512
-
∠127
42.2
230
43.22 5808
44.b c
45.?3 N沟道结型
46.24 2
20或(28.28)
47.2.64
48.+5 地
49.1 非门
50.不小于2(或2
≥)
51.2100.1? 52.400 53.4
34
0.01mV
54.阴极射线示波管 55.时间
56.48.63
103.34
四、简答题(本大题共7小题,共45分) 57.(4分)
答:(1)A 021?∠=?
I ................................................................................................... (1分) (2)V 180602?∠=?U ............................................................................................ (1分)
(3)P L =20 W ...................................................................................................... (2分) 58.(4分) 答:(1)1 V ................................................................................................................. (2分) (2)1.63 V ............................................................................................................ (2分) 59.(4分) 答:(1)I O =54.5 mA .................................................................................................... (2分) (2)U O =15.9 V ..................................................................................................... (2分) 60.(9分)
答:(1)010A A D Y = .................................................................................................. (1分)
011A A D Y = ................................................................................................... (1分) 012A DA Y = .................................................................................................. (1分)
010A DA Y = .................................................................................................... (1分)
(2)真值表见答60表
A 1A 0输入输出Y 3Y 2Y 1Y 000000D 0100D 0100D 0011
D 00
答60表
或 A 1A 0输入输出Y 3Y 2Y 1Y 0000000010000100000110000答60表
D
00000000010100101001001
11
00
1111 ............. (3分)
(3)4选1数据选择器
(或A 1A 0=00时,Y 0=D ;A 1A 0=01时,Y 1=D ;
A 1A 0=10时,Y 2=D ;A 1A 0=11时,Y 3=D ) ............................... (2分)
61.(8分) 答:(1)50 k Ω ............................................................................................................. (2分)
250 V ........................................................................................................... (2分) (2)4000 V ........................................................................................................... (2分) (3)1800 V ........................................................................................................... (2分) 62.(8分) (1)分频或倍频器 ..................................................................................................... (1分)
控制电路(逻辑控制电路) ............................................................................. (1分) A 通道 ................................................................................................................. (1分) B 通道.................................................................................................................. (1分)
u 1 超前u 2 (21??> ) ................................................................................. (1分) (2)100 ....................................................................................................................... (1分) (3) + ....................................................................................................................... (1分)
50% .................................................................................................................... (1分) 63.(9分)
答:(1)能耗制动 ....................................................................................................... (2分) (2)热继电器 ....................................................................................................... (1分)
过载保护作用 ............................................................................................... (1分) (3)电气互锁环节,防止交流电源短路事故 .................................................... (2分) (4)电动机起动后,KV 的常开触头闭合,为能耗制动停车作好准备;
当电动机转速下降到速度继电器的释放值时,KV 触点释放,切断KM2线圈,电动机能耗制动结束 ................................................................................. (2分)
五、计算题(本大题共6小题,共75分) 64.(14分) 解:V 3
4
OC =U ............................................................................................................ (5分) Ω=6
7
o R ................................................................................................................. (5分) Ω=6
1
R ................................................................................................................... (4分)
65.(15分)
解:V 10S =U .............................................................................................................. (6分) Ω=k 5o R .............................................................................................................. (6分)
Ω
.................................................................................................... (3分)
66.(15分)
解:相量图如答66图所示:
3=3U 2+1+( )
答 66 图
.............................................. (3分)
X L =20 Ω .................................................................................................................. (3分) X C =20 Ω ................................................................................................................... (3分) Z 1=j20 Ω .................................................................................................................. (3分) Ω?-∠=30202Z .................................................................................................... (3分)
67.(14分)
解:(1)求Q 点
V 3.41015
2015
CC b2b1b2BQ ≈?+=?+=V R R R V
mA 8.12
7
.03.4e
BEQ
BQ EQ CQ =-=
-=
≈R U V I I ........................................................ (1分) μA 18100
8
.1CQ
BQ ==
=
β
I I ........................................................................................ (1分) V 8.2)22(8.110)(e c CQ CC CEQ =+?-=+-=R R I V U ............................................ (1分)
(2)
Ω≈++=++='k 66.18
.126
)1001(200(mA)mV 26)
1(EQ b b be I r r β ...................................... (1分)
Ω≈?++=++=k 2.8]2)1001(66.1//[15//20)1(//[//e be b2b1i R r R R R β ................. (2分)
(3) 79.02
2.82.82)1001(66.12100)1(s i i e be c s i i o1s o11s -≈+??++?-=+?++-=?==
R R R R r R u u u u u u A u ββ ..... ................................................................................................................................. (2分) 8.022.82.82)1001(66.12)1001()1()1(s i i e be e s i i o2s o2s2≈+??++?+=+?+++=?==R R R R r R u u u u u u A u ββ ..............
................................................................................................................................. (2分) (4)
Ω==k 2c o1R R ...................................................................................................... (2分)
Ω≈++=311////(//
s)
b2b1be e o2β
R R R r R R ................................................................. (2分)
68.(9分)
解:)(o 4
5i 15o1u R R
u R R u +-= ......................................................................................... (3分)
)(
o16
3i 23o u R R
u R R u +-=
........................................................................................... (3分) 2i
o
-=u u
................................................................................................................. (3分) 69.(9分)
解:(1)状态真值表如答69表所示:
CP Q 3Q 2Q 1Q 0012
3456789
10答 69 表
00000
000110
000
0111100
0010
011000
01010
101010
1 ............................................................................. (2分)
(2)Q0、Q1、Q2、Q3的波形如答69图所示:
CP
Q0
Q1
Q2
Q3
答 69 图
.... (4分)(3)C的作用:进位信号.......................................................................................... (1分)(4)电路的逻辑功能:8421编码的异步十进制加法计数器................................. (2分)