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江苏2020年苏南五市单招二模卷答案(电子电工)

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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)

1 A

11 D

21 C 22 C 12 A 13 A 14 D 15 C 16 A 17 C 18 B 19 D 20 D 2 B 3 C 4 A 5 D 6 B 7 B 8 D 9 B 10 C 二、判断题(本大题共16小题,每小题2分,共32分)

23 A

33 A 34 A 35 B 36 A 37 B 38 A 24 B 25 B 26 B 27 B 28 A 29 A 30 B 31 B 32 A 三、填空题(本大题共18小题,30空,每空2分,共60分) 39.?17 40.9 41.16??127? 42.22 43.22 44.b 45.?3 46.24 47.2.64 48.+5 49.1

50.不小于2(或?2)

《电子电工专业综合理论》答案及评分参考 第1页(共6页)

4 72 512 30 5808 c

N沟道结型

202或(28.28)

地 非门

51.1.0?102 52.400 53.434

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)I1?2?0?A ................................................................................................... (1分) (2)U2?60?180?V............................................................................................ (1分)

(3)PL=20 W ...................................................................................................... (2分) 58.(4分) 答:(1)1 V ................................................................................................................. (2分) (2)1.63 V ............................................................................................................ (2分) 59.(4分) 答:(1)IO=54.5 mA .................................................................................................... (2分) (2)UO=15.9 V ..................................................................................................... (2分) 60.(9分)

答:(1)Y0?DA1A0 .................................................................................................. (1分)

Y1?DA1A0 ................................................................................................... (1分) Y2?DA1A0 .................................................................................................. (1分)

??Y0?DA1A0 .................................................................................................... (1分)

(2)真值表见答60表

答60表D00001111输入A100110011A0Y30010001000100011输出Y2Y10000000000011000Y000001000输入A1A000011011答60表输出Y3Y2Y100000D0D0D00Y0D000 或 ............. (3分)

《电子电工专业综合理论》答案及评分参考 第2页(共6页)

(3)4选1数据选择器

(或A1A0=00时,Y0=D;A1A0=01时,Y1=D;

A1A0=10时,Y2=D;A1A0=11时,Y3=D) ............................... (2分)

61.(8分) 答:(1)50 kΩ ............................................................................................................. (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 (?1??2 ) ................................................................................. (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:UOC? Ro? R?

《电子电工专业综合理论》答案及评分参考 第3页(共6页)

4V ............................................................................................................ (5分) 37? ................................................................................................................. (5分) 61................................................................................................................... (4分) ? 6

65.(15分)

解:US?10V .............................................................................................................. (6分) Ro?5k? .............................................................................................................. (6分)

a+?10 V5 kΩ b66.(15分)

.................................................................................................... (3分)

解:相量图如答66图所示:

U1I330°30°30°30°U3(U =U1+U2+U3)I1I2U2U2+U3答 66 图 .............................................. (3分)

XL=20 Ω .................................................................................................................. (3分) XC=20 Ω ................................................................................................................... (3分) Z1=j20 Ω .................................................................................................................. (3分) Z2?20??30?? .................................................................................................... (3分)

67.(14分)

解:(1)求Q点

Rb215VBQ??VCC??10?4.3V

Rb1?Rb220?15ICQ?IEQ?VBQ?UBEQRe?4.3?0.7?1.8mA ........................................................ (1分) 2IBQ?ICQ??1.8?18μA ........................................................................................ (1分) 100UCEQ?VCC?ICQ(Rc?Re)?10?1.8?(2?2)?2.8V ............................................ (1分)

《电子电工专业综合理论》答案及评分参考 第4页(共6页)

(2)

rbe?rbb??(1??)26mV26?200?(1?100)?1.66k? ...................................... (1分)

IEQ(mA)1.8Ri?Rb1//Rb2//[rbe?(1??)Re?20//15//[1.66?(1?100)?2]?8.2k? ................. (2分)

(3) Aus1?uo1uo1ui?RcRi100?28.2??????????0.79 ..... usuiusrbe?(1??)ReRi?Rs1.66?(1?100)?28.2?2 ................................................................................................................................. (2分) uu(1??)ReuRi(1?100)?28.2Aus2?o2?o2?i?????0.8 ..............

usuiusrbe?(1??)ReRi?Rs1.66?(1?100)?28.2?2 ................................................................................................................................. (2分) (4)

Ro1?Rc?2k? ...................................................................................................... (2分)

Ro2?Re//rbe?(Rb1//Rb2//Rs)1???31? ................................................................. (2分)

68.(9分)

RR解:uo1??(5ui?5uo) ......................................................................................... (3分)

R1R4uo??(R3R........................................................................................... (3分) ui?3uo1)

R2R6uo................................................................................................................. (3分) ??2

ui69.(9分)

解:(1)状态真值表如答69表所示:

答 69 表CP012Q300000Q2000Q100110Q00101034567890111100000110011000101010

100 ............................................................................. (2分)

《电子电工专业综合理论》答案及评分参考 第5页(共6页)

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