电工学概论答案
电工学概论习题答案_第二章

2-1. 列出以下节点的电流方程式,并求出x I 。
解:(a) 由KCL 得:1234()()()0x I I I I I +-+-+-+= ∴2341x I I I I I =++-;(b) 由KCL 得:39(5)()0x I ++-+-= ∴7x I A =(c) 由KCL 之推广得:69()0x I ++-= ∴15x I A =(d) 由KCL 得:1(2)(3)40x I +-++-+= ∴0x I =2-2. 列出以下回路的电压方程式,并标出回路循环方向,求出U 。
解:(a) 回路循环方向与I 方向相同,由KVL 得:U+(2S U -)+I R ⋅=0 ∴U = 2S U -I R ⋅(b) 回路循环方向与3I 、4I 方向相同,由KVL 得:44I R ⋅+33I R ⋅+(U -)=0 ∴U = 44I R ⋅+33I R ⋅(c) 回路循环方向为逆时针方向,由KVL 得:U+(9-)+7=0 ∴U = 2V(d) 设回路电流大小为I ,方向为逆时针方向,由KVL 得:2S U +I ⋅4+I ⋅6+(1S U -)+I ⋅3+I ⋅7=0 ∴I = 0.3A因此,U=2S U +I ⋅4 =2-3. 求出下列电路中的电流I 。
解:(a) 由KCL ,流过2Ω电阻的电流大小为(9+I),对2Ω电阻与10V 电压源构成的回路,由KVL ,得:(9)I +⋅2+(10-) =0, 因此,I =-4A ;(b) (2)I -⋅3+6=0; ∴I = 4A;(c) 流过3Ω电阻的电流 1I =93÷=3A流过6Ω电阻的电流 2I =(6+9) ÷6=2.5 A(d) 由KCL ,I + 3-3=0 ∴I =0;2-4. 求出下列电路中的电压U 。
解:(a) (U/10 + U/10) ⨯5 + U = (4-U/10-U/10) ⨯10 ∴U=10V ;(b) 10⨯2 + 10 + U =0 ∴U=-30V(c) 1I =10⨯(64)(46)(64)++++=5A; 2I =10⨯(46)(46)(64)++++=5A; 4⨯1I + U + 6⨯(-2I )=0; ∴U = 10V(d) U + 18⨯393+-=0; ∴U = 02-5. 有50个彩色白炽灯接在24V 的交流电源上,每个白炽灯为60W ,求每个白炽灯的电流及总电流,另外消耗的总功率为多少 答:每个白炽灯上消耗的功率为2U P UI R==,而U =24V ,P =60W , 因此每个白炽灯的电流为P I U ==2.5A ,总电流为I 总=I ⨯ 50=125A 消耗的总功率为P 总=P ⨯ 50=3000W2-6. 在220V 单相交流电源上,接有两台电阻热水器,一台为,一台为3kW ,分别求这两台电热器的电阻。
电气工程学概论第二章答案

第2章 电路问题习题参考解答2.6 某电源具有线性的外特性,如题2.6所示。
当输出电流为0A 和400mA 时,其端电压分别为1.48V 和1.39V 。
请给出该电源的电压源模型和电流源模型的参数。
解:线性外特性为: S S U U R I =-⎪⎩⎪⎨⎧-==SS 4.048.139.148.1R U 解得:⎪⎩⎪⎨⎧=-==Ω225.04.039.148.148.1SS R V U ⎪⎩⎪⎨⎧===A 58.6Ω225.0SS SS R U I R2.8 考察题2.8图所示几个电路的特点,用最简单的方法计算电路中各个支路的电流、各个元件上的电压和功率。
[提示1 别忘设定电压电流的参考方向];[提示2 充分利用电路中理想电压源和理想电流源的特性和连接特点]; [提示3 注意判断各个元件是消耗还是产生电功率]。
解: (a)A I 2 510 2 = =;1S 24A I I I =+=;510V U Ω=; 4248V U Ω=⨯= 2A 1082V U =-= 244216W P Ω=⨯=; 255220WP Ω=⨯=(消耗功率); 2A 224W P =⨯=(吸收功率); 10V41040W P =⨯=(供出功率);(d ) 416 1.5A 4I I Ω===;6V 22.5A I I ==;电流源模型+_U电压源模型+_5Ω(a)2(d)1.481.392uRCX LX题 2.16 图3339V U Ω=⨯=;3A 693V U =-=-;2A 6212W P =⨯=(供出功率); 3A 339W P =-⨯=-(供出功率); 6V 6 2.515W P =⨯=(供出功率);244 1.59W P Ω=⨯=(消耗功率); 233327W P Ω=⨯=(消耗功率);2.13 已知交流电压V )511000sin(10 +=t u ,交流电流2sin(1000)A 6i t π=+。
试问它们的最大值、有效值、频率、周期、初相位以及两者之间的相位差各是多少。
电工学概论答案

电工学概论答案【篇一:电工学概论习题答案_第二章(下)】?)与u2?173sin(314t?120?)之和u?u1?u2,并写出它们的相量表示式。
?解:u1??30??2?12i)??u2??120???(?12?2)??u?u1?u2??90?因此,u?u1?u2?200sin(314t?90?)2-32. 已知某负载的电流的有效值及初相为2a、45?,电压的有效值及初相为100v、?45?,频率为50hz,写出它们的向量表达式,并判断该负载是什么元件,元件参数为多少?解:i?2?45????45??2??50iu?100??z?因此,该负载是电容,元件阻抗大小为50。
??220?900v i??10?900a,(b) 2-33. 在50hz的单相交流电路中,若(a) u??220?900v i??10?450a,(c) u??220?900v i??10?1200a求三种情况下电u路中的r及x,并写出电路阻抗z的复数式。
解:(a) z?220i10i?22因此,电路中的r?22?,x?0(b) z??因此,电路中的r?,x?(c) z??11i因此,电路中的r?,x?11?2-34. 一个电感线圈接到20v的直流电源时,通过电流为0.5a。
接到50hz、100v的交流电源时通过电流为1.25a,求线圈的r和l。
解:r?200.5?40??1.25?xl?42 又?xl?2?f?l ?l?0.2hz?50?88.6i设u?220?0?,则??i?220?0?50?88.6i?2.2(12?2)?2.2??602-36. 把上述的rl串联电路与一个电容器并联,若并联后的电路总电流为原来rl串联电路的电流,求电容电流及电容量,作出相量图。
解:如图所示?ic?2.2?90?90??xc?1?而xc?2?f?c,故c?12?f?xc?55.2uf?1?c?12?f?c??50?设u?220?0,则?ir?220?050220?0?50i???225225???0??ic????225?90?i?ir?ic=225i45?2-38. 把上述的r与c并联电路与一电感元件串联,接在u?220v,50hz的交流电源上,测得rc并联电路的端电压正好也是220v,求电感元件两端的电压及电感量,作出向量图。
电气工程学概论答案(第十九章)

第十五章习题 参考解答Ⅰ思考题1.有a 、b 、c 三个几何尺寸相同的环形磁路,均绕有N 匝线圈,线圈通入相同电流。
磁路a 由铸铁材料构成,磁路b 由铸钢材料构成,磁路c 由铸铁材料构成但留有一小段气隙,试问:(1) 磁路a 与磁路b 中的磁感应强度是否相等?磁场强度是否相等?(2) 磁路c 铁心中的磁场强度与气隙中的磁场强度是否相等?磁感应强度是否相等? (3) 沿三个磁路的中心线闭合环路对磁场强度的积分值各为何值?它们是否相等?答:(1)磁感应强度b a B B <,磁场强度b a H H =, (2)磁路C 中,磁感应强度SB CC φ=,各处相等;铁心磁场强度CCC B H μ=,气隙磁场强度00μCB H =;不相等。
(3)沿三个磁路对磁场强度的积分值均为:NI l d H =⎰,它们完全相等。
2.如图15.1所示磁路,保持励磁电流的频率和电压不变,试问: (1)励磁匝数不变,磁路的截面增大一倍,励磁电流有何变化? (2)励磁线圈匝数增加一倍,铁心中的磁通有何变化? (3)匝数不变,衔铁拉开一个空气隙δ,励磁电流如何变化?答:(1)交流电磁铁S fNB fN U m m 44.444.4==φ,当f 、U 、N 一定时,Φm 一定; 当S 增大一倍时,磁阻SlR m μ=减少一倍;因为磁动势m m m R NI F φ==,所以励磁电流i 减小一倍心铁1.15图UU U(2)当f 、U 一定时,N 增加一倍,Φm 减小一倍;(3)当f 、U 、N 一定时,Φm 一定;衔铁拉开一个空气隙δ,磁阻R m 迅速增大,磁动势m m m R NI F φ==亦增 大,所以,励磁电流i 迅速增大。
3.上述磁路如用直流电励磁,保持励磁电压和线圈匝数不变,当磁路的截面增大一倍,试问:(1) 铁心的磁通有何变化?(2)若磁路的铁心与衔铁之间原有较大的空气隙,磁通的变化与问(1)有何异同?答:(1)直流电磁铁m m m R NI F φ==,当U 、N 一定时,励磁电流I 、磁动势F m 一定; 当S 增大一倍时,磁阻SlR m μ=减少一倍;磁通Φm 增大一倍;(2)当f 、U 一定时,N 增加一倍,Φm 减小一倍;(3)磁路磁阻等于铁心磁阻1m R 和空气气隙磁阻0m R 之和,即10m m m R R R +=,空气气隙气磁阻0m R >>1m R ,有气隙时,S 增大,导致m R 变化很小,所以磁通Φm 微量增大,几乎不变。
电工学概论习题答案第七章——电工习题及答案资料文档

7-1 填空题(1) 在多级放大电路中,常见的耦合方式有(阻容)耦合、(直接)耦合和(变压器)耦合三种方式。
(2) 阻容耦合放大电路,只能放大(交流)信号,不能放大(直流)信号;直接耦合放大电路,既能放大(直流)信号,也能放大(交流)信号。
(3) 直接耦合放大电路存在的主要问题是(各级静态工作点相互影响和零点漂移严重),解决的有效措施是采用(差分)放大电路。
(4) 共集电极放大电路的主要特点是电压放大倍数(小于1但接近于1)、输入电阻(高)、输出电阻(低)。
(5) 共集电极放大电路,因为输入电阻高,所以常用在多级放大器的(第一)级,以减小信号在(内阻)上的损失,因为输出电阻低,所以常用于多级放大器的(最后)级,以提高(带负载)的能力。
(6) 差模信号是指两输入端的信号(大小相等、极性相反)的信号,共模信号是指两输入端的信号(大小相等、极性相同)的信号。
(7) 差模信号按其输入和输出方式可分为(双端输入双端输出)、(单端输入双端输出)、(双端输入单端输出)和(单端输入单端输出)四种类型。
(8) 交流负反馈有(电压串联负反馈)、(电流串联负反馈)、(电压并联负反馈)和(电流并联负反馈)四种类型。
(9) 对于一个放大器,欲稳定输出电压应采用(电压)负反馈、欲稳定输出电流应采用(电流)负反馈、欲提高输入电阻应采用(串联)负反馈、欲降低输出电阻应采用(电压)负反馈、欲降低输入电阻而提高输出电阻应采用(电流并联)负反馈。
(10) 正弦波振荡器振幅平衡条件是(||1AF =)、相位平衡条件是(2A F n ϕϕπ+= 0,1,2,n =±±);正弦波振荡器四个基本组成部分是(基本放大器A)、(正反馈网络F)、(选频网络)和(稳幅环节)。
(11) 逻辑变量只有(逻辑1)和(逻辑0)两种对立的状态。
(12) 三种基本的逻辑运算是指(与)、(或)和(非)。
(13) 1+1=1是(或)运算,1+1=10是(二进制加法)。
电工学概论习题答案_第三章

3-1. 何谓电力系统?采用电力系统传输和分配电能比由发电厂直接向用户供电由什么优点?答:电力系统是由发电厂、变电站、配电所直到各个用户等环节所组成的电能生产消费系统。
实践证明独立运行的发电厂通过电力网联接成电力系统后,将在技术经济上具有以下优点:(1)减少系统中的总装机容量(2)合理使用动力资源,充分发挥水力发电厂的作用(3)提高供电的可靠性(4)提高运行的经济性3-2. 为什么要采用高压传输电能?我国目前远距离输电所采用的最高电压等级是多少?在城市内所采用高压配电的电压等级是多少?答:采用高压传输电能能够明显减少传输过程中的电能损耗。
我国远距离输电所采用的最高电压等级是500kV。
在城市内所采用的高压配电的电压等级是35 kV 或10 kV。
3-3. 输电线路在什么情况下采用架空线?在什么情况下采用电力电缆?答:交流输电线可分为架空线路与电缆线路两大类,前者应用于地区间的输电,一般电压较高、距离较长。
后者应用于城市内的输电,电缆线路一般埋设在地下,线路电压为35kV或10kV,也有380V/220V的低压,在大型工厂企业内部也采用电缆输电。
与架空线路相比,电缆线路的铺设成本要高许多,所以在电力系统中只有在一些不适于架空线路的地方如过江、跨海或严重污染区才考虑使用电缆输电。
3-4. 架空输电线为什么一般采用钢芯铝绞线?答:架空输电线用以传输电流、必须具有足够的截面以保持合理的电流密度及比较小的电能损耗,同时又必须有足够的机械强度和抗大气化学腐蚀能力。
一般采用钢芯铝绞线它能兼顾机械强度、导电能力、散热面积等要求。
3-5. 何谓高压走廊?在高压走廊的范围内应注意什么安全问题?答:高压架空输电线路所通过的路径必须占用一定的土地面积和空间区域,称为线路走廊或高压走廊,在该走廊内除杆塔基础占用一定土地外,其余土地可用于耕作和绿化,但不能用于建设居住用房,人应避免长期在强电磁场的环境下生活,因为强电磁场会引起人生理上发生一些不良反应,这些反应可能会引起心情上的变化甚至会引起某些慢性不可预知的疾患。
电气工程学概论 第三章部分习题解答

2 按照以下步骤求解图示电路中5Ω电阻中的电流i 。
(1)将右侧支路等效成一个电阻之路后;用网孔电流法求解3Ω电阻中的电流; (2)用分流公式求解电流i 。
解:(1)网孔电流法:⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡--480136612II I i i ; ⎩⎨⎧=+-=-4813602II I II I i i i i A 4.2I =i A 8.4II =i (2)分流公式:A 4.221II ==i i3用网孔电流法求解图示电路中的电压U o 。
解:网孔电流()()⎩⎨⎧-=⨯-++-=⨯--++5010340104013683404082II I II I i i i i⎩⎨⎧-=+-=-2050401604050II I II I i i i i A 8I =i ,A 6II =i , ()V 8040II I o =⨯-=i i U7用节点电压法求解图示电路中各支路电流。
解:节点电压法⎪⎪⎩⎪⎪⎨⎧=-⎪⎭⎫ ⎝⎛+=-⎪⎭⎫ ⎝⎛++221211011021215111221U U U U⎪⎩⎪⎨⎧=-=-221106102110171221U U U U V 111001=U V 111202=U A 9.01=i A 9.02-=i A 82.1513==U i A 9.101024==Ui题3-2图Ω 4题3.3图50V题 3.7 图2 A11 求图示电路在ab 端口的戴维南等效电路。
解:方法1:()V515100cbdc ad abK =⨯-++=++=U U U UΩ=+=105201.R方法2:利用电源等效替换求戴维南等效电路原电路 ⇒⇒⇒13 在图示电路中,试问:(1)电阻R 为多大时,它吸收的功率最大?试求出最大功率。
(2)在a 、b间并联一理想电流源,以使R 中的电流为0,试求该理想电流源的方向和参数。
解:(1) 原电路的戴维南等效电路如图(a )所示: 可通过网孔电流法和节点电压法求取U abK ,其中:节点电压法:2020//205020502020//201201201c ++=⎪⎭⎫⎝⎛+++U V 4125c =U ,V 5.37V 275abK a ===U U网孔电流法()⎩⎨⎧-=+++-=-502020//202020502040II III I i i i i A 85I -=i ,A 1615II =i ,V 5.375020II abK =+=i U等效电阻: R ab = [20//20+20//20] // 20 = 10Ω 当R =R ab =10Ω时,它的吸收功率最大,W 16.3510 205.372max=⨯⎪⎭⎫ ⎝⎛=P3-11图b ada babababb 题3-13图(a)3.15 应用叠加原理求解电路中的电压u 。
电工学概论习题答案_第二章(上)

2-1. 列出以下节点的电流方程式,并求出x I 。
解:(a) 由KCL 得:1234()()()0x I I I I I +-+-+-+= ∴2341x I I I I I =++-; (b) 由KCL 得: 39(5)()0x I ++-+-=∴7x I A = (c) 由KCL 之推广得:69()0x I ++-= ∴15x I A = (d) 由KCL 得: 1(2)(3)4xI +-++-+=∴0x I =2-2. 列出以下回路的电压方程式,并标出回路循环方向,求出U 。
解:(a) 回路循环方向与I 方向相同,由KVL 得: U+(2S U -)+I R ⋅=0 ∴U = 2S U -I R ⋅(b) 回路循环方向与3I 、4I 方向相同,由KVL 得: 44I R ⋅+33I R ⋅+(U -)=0 ∴U = 44I R ⋅+33I R ⋅ (c) 回路循环方向为逆时针方向,由KVL 得: U+(9-)+7=0 ∴U = 2V(d) 设回路电流大小为I ,方向为逆时针方向,由KVL 得: 2S U +I ⋅4+I ⋅6+(1S U -)+I ⋅3+I ⋅7=0 ∴I = 0.3A 因此,U=2S U +I ⋅4 = 13.2V2-3. 求出下列电路中的电流I 。
解:(a) 由KCL ,流过2Ω电阻的电流大小为(9+I),对2Ω电阻与10V 电压源构成的回路,由KVL ,得:(9)I +⋅2+(10-) =0, 因此,I =-4A ; (b) (2)I -⋅3+6=0; ∴I = 4A; (c) 流过3Ω电阻的电流 1I =93÷=3A 流过6Ω电阻的电流 2I =(6+9) ÷6=2.5 A (d) 由KCL ,I + 3-3=0 ∴I =0;2-4. 求出下列电路中的电压U 。
解:(a) (U/10 + U/10) ⨯5 + U = (4-U/10-U/10) ⨯10 ∴U=10V ;(b) 10⨯2 + 10 + U =0 ∴U=-30V(c) 1I =10⨯(64)(46)(64)++++=5A; 2I =10⨯(46)(46)(64)++++=5A;4⨯1I + U + 6⨯(-2I )=0; ∴U = 10V (d) U + 18⨯393+-4.5=0; ∴U = 02-5. 有50个彩色白炽灯接在24V 的交流电源上,每个白炽灯为60W ,求每个白炽灯的电流及总电流,另外消耗的总功率为多少?答:每个白炽灯上消耗的功率为2U P UI R==,而U =24V ,P =60W ,因此每个白炽灯的电流为PI U==2.5A ,总电流为I 总=I ⨯ 50=125A消耗的总功率为P 总=P ⨯ 50=3000W2-6. 在220V 单相交流电源上,接有两台电阻热水器,一台为1.5kW ,一台为3kW ,分别求这两台电热器的电阻。
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电工学概论答案【篇一:电工学概论习题答案_第二章(下)】?)与u2?173sin(314t?120?)之和u?u1?u2,并写出它们的相量表示式。
?解:u1??30??2?12i)??u2??120???(?12?2)??u?u1?u2??90?因此,u?u1?u2?200sin(314t?90?)2-32. 已知某负载的电流的有效值及初相为2a、45?,电压的有效值及初相为100v、?45?,频率为50hz,写出它们的向量表达式,并判断该负载是什么元件,元件参数为多少?解:i?2?45????45??2??50iu?100??z?因此,该负载是电容,元件阻抗大小为50。
??220?900v i??10?900a,(b) 2-33. 在50hz的单相交流电路中,若(a) u??220?900v i??10?450a,(c) u??220?900v i??10?1200a求三种情况下电u路中的r及x,并写出电路阻抗z的复数式。
解:(a) z?220i10i?22因此,电路中的r?22?,x?0(b) z??因此,电路中的r?,x?(c) z??11i因此,电路中的r?,x?11?2-34. 一个电感线圈接到20v的直流电源时,通过电流为0.5a。
接到50hz、100v的交流电源时通过电流为1.25a,求线圈的r和l。
解:r?200.5?40??1.25?xl?42 又?xl?2?f?l ?l?0.2hz?50?88.6i设u?220?0?,则??i?220?0?50?88.6i?2.2(12?2)?2.2??602-36. 把上述的rl串联电路与一个电容器并联,若并联后的电路总电流为原来rl串联电路的电流,求电容电流及电容量,作出相量图。
解:如图所示?ic?2.2?90?90??xc?1?而xc?2?f?c,故c?12?f?xc?55.2uf?1?c?12?f?c??50?设u?220?0,则?ir?220?050220?0?50i???225225???0??ic????225?90?i?ir?ic=225i45?2-38. 把上述的r与c并联电路与一电感元件串联,接在u?220v,50hz的交流电源上,测得rc并联电路的端电压正好也是220v,求电感元件两端的电压及电感量,作出向量图。
解:z?220225?225i?25?25i由向量图可知,xl?50?,又xl??l?2?f?l,故l?2-39. 有两台单相交流电动机(电感性负载)并联在u?220v的电源上所消耗的功率及功率因数为p1?1kw,cos?1?0.8,p2?0.75kw,cos?2?0.6,求每台电动机的电流,无功功率、视在功率,以及总有功功率,总无功功率,总视在功率,总电流、总功率因数。
?2解:由于p1?ui1cos?1p2?ui2cos8 因此电流 i1?5.68a i2?5.6axl2?f?0.159h?2?1000 var无功功率 q1?ui1sin?1?750varq2?ui2sin视在功率 s1?ui1?1250v?as2?ui2?1250v?a 总有功功率p?p1?p2?1750w 总无功功率 q?q1?q2?1750var 总视在功率s??2474.87v?aa总电流i?s/u?11.25总功率因素 cos??0.7072-40. 在40w日光灯电路中已知电源为220v、50hz灯管端电压为100v,功率消耗为40w,镇流器功率消耗为8w,灯管电流为0.4a,求(1)电路的总有功功率,视在功率和无功功率,功率因数。
(2)镇流器的无功功率,视在功率及端电压。
(3)作出相量图。
解:l电路如上图所示,其中l、r为镇流器电阻电感,r1为日光灯电阻(1) 电路的总有功功率为 p = 40 + 8 = 48w总视在功率为s?ui?220?0.4?88v?a总无功功率为q??73.76var功率因素为cos??p/s?0.54 (2) 镇流器的无功功率qz?q?73.76var视在功率为sz???76.86v?a端电压uz?sz/i?192.15v(3) 向量图uli2-41. 上述电路中欲使功率因数提高到0.9,应并联多大的电容?解:并联电容后无功变化,有功不变变化后的无功功率为q?48?0.9?23.25var 变化量为q变?q?q?50.51var xc?u/q变?958.2c?12?f?xc?3.32uf2解:该负载需要的无功功率及视在功率分别为:p?uicos??90kwq?uisin??90000?0.6?0.8?120kw s??150kv?a若要求该电网的功率因数提高到0.9,则需要并联电容,其大小为c?p?u2(tan?1?tan?2)?900002??50?2202(1.333?0.484)?5.03mf2-43. 解释rlc串联电路中的串联谐振现象,串联谐振有哪几个特征?通过哪些参数调节可以出现串联谐振?解:在rlc串联电路中,当xl?xc时,电路两端的电压与电流同相,即为串联谐振串联谐振有以下特征: (1)电路的阻抗|z|?|z|为最小值。
(2) 由于电源电压与电路中的电流同相,因此电路对电源呈电阻性。
(3) 由于xl?xc,有ul?uc。
(4) 串联谐振电路对频率有选择性,可在无线电选择接收信号频率时加以利用。
由谐振条件xl?xc有2?fl?12?fc,因此通过调节l、c、f可出现串联谐振。
2-44. 解释电感线圈与电容并联电路中出现的并联谐振现象?此种谐振现象的最主要特征是什么?其谐振频率与哪些参数有关?解:在电感线圈与电容并联电路中,当电路端电压与电流相位相同,则电路出现并联谐振现象。
并联谐振的主要特征有:(1) 并联谐振时,电路的阻抗z?流i接近最小。
(2) 并联谐振时,由于电源电压与电流相同时(??0),因此,电路对电源呈电阻性,即|z|相当于一个电阻。
(3) 在谐振时电感线圈和电容器的电流近似相等,而比总电流大许多倍。
(4) 如并联谐振电流改由恒流源供电,当电源为某一频率时电路发生谐振,电路的阻抗为最大,这样起到了选频的作用。
lrc接近最大,在电源电压u一定时,电路中的总电【篇二:英语学习电工学概论习题答案_第四章】cage form and the winding pattern from the structural characteristics of the three-phase asynchronous motor?answer: rotor winding induced electromotive force is generated, currents and electromagnetic torque, the structure type with cage and wound rotor two, squirrel cage rotor ofeach rotor slot insert a piece of copper conducting bar, in an extended at both ends of the core at the notch, with two short circuit copper rings respectively, the all guide bar ends are welded together. if the core is removed, the overall shape ofthe winding is like a cage, so it is called a cage rotor. the rotor winding and the stator winding is similar to that of theinsulated conductor embedded in the rotor slot and connected into a star three-phase symmetric winding, winding threeoutlet terminals are respectively connected to the rotor shaft three slip ring (ring and ring, the ring and the shaft areinsulated from each other), the carbon brush to draw out the current.4-2. how to change the three-phase induction motor to change?answer: any of the two will be connected with three-phase power supply connected three wire switch, three-phase asynchronous motor direction change.4-3. known a three-phase cage induction motor rated power = 3kw, rated speed = 2880r/min. find the number of pole pairs (1);(2) when the rated slip ratio; (3) rated torque.solution: (1) synchronous speed, so the logarithm of the motor pole p is 1;(2)(3) =9.95technical data of 4-4. known y112m-4 type asynchronous motor = 4kw, delta connection method, rated voltage = 380v, = 1440r / min, rated current = 8.8 a, power factor = 0.82,efficiency = 84.5%. find the number of pole pairs (1); (2) input power during normal operation; (3) when the rated slip ratio; (4) rated torque.solution: (1) synchronous speed, so the logarithm of the motor pole p is 2;(2)(3)(4) =26.54-5. known y132m-4 type asynchronous motor rated power7.5kw, rated current = 15.4a, rated speed = 1440r / min, rated voltage = 380v, rated power factor of 0.85 and rated efficiency = 0.87, starting torque and rated torque = 2.2, the starting current and rated current = 7.0, the maximum torque / rated torque = 2.2. tries (1) rated power input; (2) rated torque; (3) rated slip; (4) thestarting current; (5) starting torque; (6) the maximum torque.solution: (1) rated input power(2) =49.74(3)(4) =7=107.8a(5) =2.2=109.43(6) =2.2=109.434-6. in the three-phase asynchronous motor start instant (s = 1), why the rotor current is large, and the power factor of the rotor circuit is small?a: the electric motor is connected to the power supply, the induction electromotive force and the induced current of the rotor circuit are the largest, which is called the starting current or the blocking current. in general, the starting current of the small and medium sized three phase induction motor is 5 to 7 times of the rated current. although the motor starting current is very large, but the power factor of the rotor is very low, in fact, the starting torque of the motor is not large.4-7. if there is a large number of asynchronous motors in the three-phase power grid at the same time, will be the impact of the grid voltage? why?a large number of asynchronous motors start at the same time, the power grid will have a larger voltage drop. because asynchronous motor starting current is 5 ~ 7 times the rated current, if in the three-phase power grid has a large number of asynchronous motor start at the same time, will produce a large starting current, large starting current will in power line produced larger voltage landing, transformer power supplywith other work load.4-8. three-phase induction motor torque characteristics of the shape of the curve by what factors?answer: electromagnetic torque can be expressed aswhen the s is very small, can be neglected, t is proportional to s; when the s is closer to 1, can be omitted, t is inversely proportional to s. so on the torque characteristic curve (t ~ s) as shown in figure.t ~ s curveat that time, said the motor at the rated operating state, at this time the rotor speed for the rated speed, torque for the rated torque, the output of the mechanical power for the rated power. at that time, the motor is in critical condition. at the same time, the electromagnetic torque is the largest value can by calculating t ~ s curve extremum method, to obtain the critical slip. visible and directly proportional to, but has nothing to do with; regardless of is proportional to. only in this way can by changing (wound rotor circuit external rheostat) and reducedto change t ~ s curve shape, as shown in 4.2.3 and 4.2.4.figure t increases the 4.2.3 ~ s curves of 4.2.4 to reduce the t ~ s curve4-9. why should adopt the measure of reducing blood pressure? what is the effect of reducing the starting torque of the motor? under what circumstances can the buck start?answer: reduced voltage starting of the purpose is to reduce the motor starting when starting the effects of electric current on the grid, the method is in starting lower voltage in the stator winding of the motor, to be motor speed is close to the stable, then the voltage returns to normal value. due to the motor torque and voltage is proportional to the square, so reduced voltage starting torque will also decrease accordingly, current step-down start the main method for star - delta (y - delta) change to connect to start and its applicable condition is normal operation of the stator winding is delta connected squirrel cage asynchronous motor.what are some of the main methods of 4-10. reduction?answer: the start method (1) direct starting, (2) y - delta change is connected with a step-down starting, (3) soft starting method. (4) rotor series resistance starting. starting mainly y - delta switching starting and soft start method.why does the series resistance in the rotor circuit of 4-11. wound rotor type induction motor improve the starting performance and speed performance of the induction motor?answer: winding type motor can be used in the rotor circuit in series resistance starting method. this can not only limit the starting current, but also increase the starting torque. so we need to start production of mechanical torque is large, such as cranes, forging machines and other commonly used linear motor drive around. after the motor is started, the starting resistance is removed by the step with the increase of the rotating speed.4-12. three-phase induction motor in the normal operation, if the rotor suddenly stuck, what will be the consequences of the motor?answer: three-phase asynchronous motor in normal operation, if the rotor suddenly stuck (stall), motor current increased immediately for several times of the rated current, if no protective measures promptly cut off the power supply, the motor will be burnt.4-13. three-phase asynchronous motor in normal operation, if suddenly there is a phase power, the motor can run? if the motor is under heavy load, what is the change of the motor current?answer: three-phase asynchronous motor in normal operation, if suddenly there is a phase power, the motor can run. if the motor is overloaded in case of slip increases, the induced electromotive force increases, the motor current increases. long run may burn the motor.4-14. three phase induction motor which are several speed control method? which method is the best performance of the speed control?answer: (1) the number of magnetic poles of the speed change, (2) change of the rotating slip speed, (3) variable frequency speed regulation are the main speed control method of three-phase asynchronous motor. in many speed regulation methods of ac induction motor, the performance of variable frequency speed regulation is the best, which is characterized by high speed range, good stability and high efficiency.4-15. constant supply voltage in the case, if the rating for the error connected y type connection delta connection, what are the consequences? and as the amount of time for a false delta connection type y connection, and what are the consequences? answer: y type equivalent connecting resistance of the same size three resistor delta after connection for direct y connection type size 1/3. therefore, at a supply voltage of constant, if rated delta connected mistakenly connected type y connection, actual larger resistance, smaller current in the circuit, resulting in lack of power. if the amount?a delta connection error word type fast connection, the actual resistance is small, the current in the circuit becomes larger, it is possible to burn out resistance.4-16. thermal relay in the three-phase asynchronous motor control circuit to play what role? can it play the role of short circuit protection?answer: thermal relay is used for ac asynchronous motor overload protection, it uses the thermal effect of current and action. now the production of heat relay has three groups of heating elements and has a phase failure protection function, namely in the current serious imbalance or even disconnectionof a phase, the mechanical structure can also make a trip ofthe gusset plate will move off contact disconnect and protects the motor because of the phase interruption caused by phase current overload and off the ring.4-17. test to be able to directly control the control circuit of a three-phase cage asynchronous motor at two locations.answer:small capacity cage type motor direct starting control circuit, which used the air switch q, accontactor km, button sb, thermal relay fr and fuse fu and other control electrical appliances. work first air q switch closed,into the power, when the press the start button (dynamic close contact points), the coil of the ac contactor km energized, moving core is attracted, the three main contacts (contact closed. the speed of the motor through electric starting. when the release button contact restored to original broken position, but due to and parallel ac contactor km of auxiliary contacts (contact) and main contact is closed at the same time, so contactor coil circuit is switched on, and make contacts of the contactor is maintained in a closed position. the auxiliary contact is called the self locking contact, which has the function of self locking for the motor to run for a long time. if the stop button is stopped, the circuit of the contact coil is cut off, the moving iron core and the contact point are restored to the open position, and the motor is stopped.the control circuit has the functions of short circuit protection, overload protection, pressure loss and under voltage protection, besides the starting and stopping control functions of the motor.4-18. test to be able to make a point of control, but also can be used as a direct starting control (two switch conversion) control circuit.answer:4-19. test of the positive inversion control circuit in the contact of the contacts of the contactor what role?answer: if the two ac contactor at the same time, their main contacts are closed, will cause short circuit. in order to ensure the two contacts by not working at the same time, in thecontrol circuit, connected in series with the other contactor move off of auxiliary contact points and namely positiverotation contactor of a dynamic fault of auxiliary contact points on connected in reverse rotation contactor coil circuit, and the overturning contactor of a dynamic breaking auxiliary point connected in positive rotation contactor coil circuit. these two are called interlocking contacts. in this way, when you press a positive starting button, positive rotation contactor coils, and the main contact, motor forward. at the same time, the interlocking contacts are disconnected from the coil circuit of the reverse contactor. therefore, even if the reverse starting button is mistaken, the contactor can not be moved. at thesame time the two contactor only allows one to work in a controlled role called interlocking or interlocking.4-20. programmable controller has what function? what is the advantage of the traditional relay contactor control method when it is used in the motor control?answer: the relay - contactor control is the traditionalindustrial control mode, it relay contacts and coils accordingto the control logic relation of certain connected control circuit, control contactor on-off, followed by the motor or solenoid device drive mechanism motion. however, because the relay itself accounts for a certain size and power consumption, often a failure, coupled with the fixed wiring, so that change control logic is difficult. so the application of complex control system reliability and flexibility are relatively poor.【篇三:电工学概论习题答案第四章】txt>4-1. 怎样从三相异步电动机的结构特征来区别笼型和绕线型?答:转子绕组的作用是产生感应电动势、流过电流和产生电磁转矩,其结构型式有笼型和绕线型两种,笼型转子的每个转子槽中插入一根铜导条,在伸出铁心两端的槽口处,用两个短路铜环分别把所有导条的两端都焊接起来。