东南大学信息学院10-11-2电路基础期末考试试卷
东南大学电子学院《电路基础》期末考试样卷及答案
Solve the following problems. (100 points)1、( 6 points) Find U ab in the circuit in Figure 1.Figure 1a b2、( 8 points) Find u o and i o in the circuit in Figure 2.u o Figure 2-3、( 6 points) In the circuit of Figure 3, readings of voltmeter ○V1，○V2 and ○V3 are 10V , 18V and 6V , respectively. Please determinate the reading of the voltmeter ○V. aFigure 34、( 8 points) The resonant or tuner circuit of a radio is portrayed in Figure 4, where u s1 represents a broadcast signal, given that R =10Ω，L =200μH , U s1rms =1.5mV ，f 1=1008kHz. If the circuit is resonant with signal u s1, please determinate: (1) the value of C ; (2) the quality factor Q of the circuit; (3) the current I rms ; (4) the voltage U c rms .Figure 4u5、( 8 points) A balanced three-phase circuit is shown in Figure 5. Calculate the phase currents and voltages in the delta-connected load Z, if V U A ︒∠=•0220, Z =(24+j15)Ω，Z L =(1+j1)Ω.Figure 5•6、( 8 points) Find the yparameters for the circuit in Figure 6.••Figure 67、( 15 points) For the circuit in figure 7, find the rms value of the current i and the average power absorbed by the circuit, given that Ω10=L ω,Ω51=R ,Ω202=R ,Ω20/1=C ω,V )302cos(50cos 10080)(︒+++=t t t u s ωω.Figure 72u s8、( 12 points) In the circuit of Figure 8, determinate the value of Z that will absorb the maximum power and calculate the value of the maximum power.Figure 8V ︒∠050ω=103•rms9、( 14 points) If the switch in Figure 9 has been closed for a long time and is opened at t = 0, find u c , i L , u k for t ≥ 0.20Figure 9u c +-10、( 15 points) The switch in Figure 10 has been in position a for a long time. At t = 0, it moves to position b. Find i 2(t) for t ≥ 0 and sketch it. (Attention: Please markthe key data in your graph.)Figure 1010ΩSolve the following problems. (100 points)1、( 6 points) Find U ab in the circuit in Figure 1.Figure 1a bSolution:U ab = 9V2、( 8 points) Find u o and i o in the circuit in Figure 2.u o Figure 2-Solution:V 4.2=o u , 0.256mA -=o i3、( 6 points) In the circuit of Figure 3, readings of voltmeter ○V1，○V2 and ○V3 are 10V , 18V and 6V , respectively. Please determinate the reading of the voltmeter ○V. aFigure 3Solution:The reading of the voltmeter ○V is 10V .4、( 8 points) The resonant or tuner circuit of a radio is portrayed in Figure 4, where u s1 represents a broadcast signal, given that R =10Ω，L =200μH , U s1rms =1.5mV ，f 1=1008kHz. If the circuit is resonant with signal u s1, please determinate: (1) the value of C ; (2) the quality factor Q of the circuit; (3) the current I rms ; (4) the voltage U c rms .Figure 4uSolution:pF 5.12)1(≈C 6.126)2(≈Q μA 150)3(=rms I mV 190)4(=Crms U5、( 8 points) A balanced three-phase circuit is shown in Figure 5. Calculate the phase currents and voltages in the delta-connected load Z, if V U A ︒∠=•0220, Z =(24+j15)Ω，Z L =(1+j1)Ω.Figure 5•Solution:A 3.69-11.74''︒∠=•B A I , V 31.28332''︒∠=•B A U6、( 8 points) Find the y parameters for the circuit in Figure 6.••Figure 6Solution:[]S y ⎥⎦⎤⎢⎣⎡=0.5 40.02 2.07、( 15 points) For the circuit in figure 7, find the rms value of the current i and theaverage power absorbed by the circuit, given that Ω10=L ω,Ω51=R ,Ω202=R ,Ω20/1=C ω,V)302cos(50cos 10080)(︒+++=t t t u s ωω.Figure 72u sSolution:A 17.6357.2714.42.3222222120=++=++=I I I I rmsP =P 0+P 1+P 2=256+333.28+50=639.28W8、( 12 points) In the circuit of Figure 8, determinate the value of Z that will absorbthe maximum power and calculate the value of the maximum power.Figure 8V ︒∠050ω=103•rmsSolution:If Ω135jZ Z eq+==*, Z will absorb the maximum average power. The maximum average power is, W 7.2491356.1941412max=⋅=⋅=*eq OC R U P9、( 14 points) If the switch in Figure 9 has been closed for a long time and isopened at t = 0, find u c , i L , u k for t ≥ 0.20Figure 9u c +-Solution:0 A,5.15.1)5.10(5.14040>-=-+=--t e e i t t L ， 0 V,60120)12060(12020/20/>-=-+=--t e e u t t C 0 ,V 6030902020/40>+--=-=--t e e u i u t t C L K10、( 15 points) The switch in Figure 10 has been in position a for a long time. At t = 0, it moves to position b. Find i 2(t) for t ≥ 0 and sketch it. (Attention: Please mark the key data in your graph.)Figure 1010ΩSolution: A 25.125.132t t e e i ---= 打发第三方.。
电路期末考试题及答案
电路期末考试题及答案一、选择题（每题2分，共20分）1. 在串联电路中，各电阻上的电压关系是：A. 电压相等B. 电压之和等于总电压C. 电压与电阻成正比D. 电压与电阻成反比答案：C2. 欧姆定律表达式为：A. V = IRB. V = R/IC. I = V/RD. I = R * V答案：A3. 电容单位是：A. 法拉（F）B. 欧姆（Ω）C. 亨利（H）D. 安培（A）答案：A4. 电流的国际单位是：A. 伏特（V）B. 安培（A）C. 欧姆（Ω）D. 法拉（F）答案：B5. 一个电路的总电阻为10Ω，当电路中增加一个5Ω的电阻后，总电阻变为：A. 5ΩB. 15ΩC. 20ΩD. 无法确定答案：B6. 电感器的单位是：A. 法拉（F）B. 欧姆（Ω）C. 亨利（H）D. 安培（A）答案：C7. 基尔霍夫电流定律表明：A. 电路中任意节点的电流之和为零B. 电路中任意回路的电压之和为零C. 电路中任意节点的电压之和为零D. 电路中任意回路的电流之和为零答案：A8. 电阻串联时，总电阻等于：A. 各电阻的倒数之和B. 各电阻的差值C. 各电阻之和D. 各电阻的乘积答案：C9. 电路中，功率（P）与电压（V）和电流（I）的关系是：A. P = V * IB. P = V / IC. P = I / VD. P = V + I答案：A10. 交流电的频率是指：A. 每秒电流方向改变的次数B. 电流的峰值C. 电流的均方根值D. 电流的周期答案：A二、填空题（每空2分，共20分）11. 电路中，当电阻为定值时，电流与电压成________关系。
答案：正比12. 串联电路中，总电阻等于各个电阻的________。
答案：和13. 并联电路中，总电阻的倒数等于各个电阻倒数的________。
答案：和14. 电容器在直流电路中表现为________。
答案：开路15. 电感器在直流电路中表现为________。
电路期末考试题及答案
电路期末考试题及答案1. 选择题1) 一个电路中有一个电源，一个电阻和一个电容器，如果电源的电压稳定为3V，电阻的阻值为2Ω，电容器的电容为6μF，则此电路的时间常数（τ）为多少？a. 2μsb. 3μsc. 6μsd. 12μs答案：c. 6μs2) 在一个RLC谐振电路中，若电感L为0.3H，电容C为2μF，则当谐振频率为10kHz时，电路的阻抗Z为多少？a. 5Ωb. 10Ωc. 15Ωd. 20Ω答案：d. 20Ω3) 在一个串联电路中，有一个电源、一个电阻R1和一个电感L，当开关S刚刚闭合瞬间，电路中的电流为零。
此时以下哪个说法是正确的？a. 电阻R1上的电压为零，电感L上的电压不为零b. 电阻R1上的电压不为零，电感L上的电压为零c. 电阻R1上的电压和电感L上的电压均为零d. 电阻R1上的电压和电感L上的电压均不为零答案：c. 电阻R1上的电压和电感L上的电压均为零2. 填空题1) 欧姆定律描述了电路中的电压、电流和________之间的关系。
答案：电阻2) 串联电路中，总阻值等于各个电阻的阻值之________。
答案：和3) 在一个RLC电路中，当电流的频率与电路的谐振频率相同时，该电路呈________。
答案：共振3. 解答题1) 请简要解释电容器的作用和原理。
答：电容器是一种被用来存储电能的器件。
它由两块金属导体（即电极板）和介质（即电容介质）组成，电容介质可以是空气、塑料、陶瓷或电解质等。
当电容器接通电源后，电极板上会产生正负电荷，形成电场。
这时，电容器会储存能量，其中储存的能量量与电容器的电压和电容成正比。
2) 解释直流和交流电的区别。
答：直流电（DC）和交流电（AC）是两种常见的电流类型。
直流电的电流方向始终保持不变，电压稳定，例如电池供电的电路。
交流电的电流方向会定期变换，电压随时间变化，例如在家中使用的电源。
因此，直流电和交流电的主要区别在于电流方向和电压变化形式。
东南大学2010-2011计算机结构与逻辑设计考试A卷
东南大学考试卷（ A 卷）适用专业信息工程考试形式闭卷考试时间长度120分钟（开卷、半开卷请在此写明考试可带哪些资料）一填空题（20分）1、将二进制数10101.011 化成十进制数____21.375____________________；2、将十进制数21.7 化成二进制数(误差ε<2-6)_____10101.101100__________；3、写出二进制原码10101101对应的补码_____0 1010011__________________；4、将十进制数132表示为N B C D码___________000100110010 ______________________；5、写出等式B=+的对偶等式___________________________；A+ABA6、求逻辑函数DF+=的反函数F___________________________；∙CBA7、一个某计算机指令字长16位，能完成32条不同的指令，那么该指令字中操作码长度至少需要________位；此时地址长度为________位；可以访问__________________ * 16 bits的存储空间。
8、一个由256字×1位的SRAM芯片构成的存储器，其总容量为2048字×8位，该存储器总共有____________________ 片SRAM芯片构成共8 页第1页共 8 页 第 2 页二 作出逻辑函数]))[((A C B B A B A C F +⋅+⋅+=的卡诺图，将逻辑表达式化成最简与或表达式， 并用与非门实现之。
（10分）三分析如图由移位寄存器74195和双四选一74153构成的电路，写出状态方程、输出方程、状态转移真值表，画出其状态图。
（15分）共8 页第3页四已知4位二进制全加器74283的框图如下图所示，其完成的功能是将A3~A0与B3~B0以及CI进位输入相加起来，得到D3~D0和CO的输出。
电子电路基础期末试卷及参考答案
XX 中学XX 年XX 学期期末考试电子电路基础试卷（XX 班）（试卷满分100分 考试时间120分钟）班级 姓名 学号 得分一、填空题（每空1分，共20分）1.在时间上和数值上均作连续变化的电信号称为 模拟 信号；在时间上和数值上离散的信号叫做 数字 信号。
2.在正逻辑的约定下，“1”表示 高 电平，“0”表示 低 电平。
3.数字电路中，输入信号和输出信号之间的关系是 逻辑 关系，所以数字电路也称为 逻辑 电路。
4.在逻辑关系中，最基本的关系是 与逻辑 、 或逻辑 和 非逻辑 。
5.十进制数转换为八进制和十六进制时，应先转换成 二进 制。
6.最简与或表达式是指在表达式中 或项 最少，且 与项 也最少。
7.具有基本逻辑关系的电路称为 门电路 ，其中最基本的有 与门 、 或门 和非门。
8.两个与非门构成的基本RS 触发器的功能有 清零 、 置1 和 保持 。
9.组合逻辑电路的基本单元是 门电路 ，时序逻辑电路的基本单元是 触发器 。
二、选择题（每题2分，共30分）1.逻辑函数中的逻辑“与”和它对应的逻辑代数运算关系为（ B ）。
A 、逻辑加B 、逻辑乘C 、逻辑非 2.十进制数100对应的二进制数为（ C ）。
A 、1011110B 、1100010C 、1100100D 、11000100 3.和逻辑式AB 表示不同逻辑关系的逻辑式是（ B ）。
A 、B A + B 、B A ∙C 、B B A +∙D 、A B A + 4.数字电路中机器识别和常用的数制是（ A ）。
A 、二进制B 、八进制C 、十进制D 、十六进制 5.所谓机器码是指（ B ）。
A 、计算机内采用的十六进制码B 、符号位数码化了的二进制数码C 、带有正负号的二进制数码D 、八进制数 6.具有“有1出0、全0出1”功能的逻辑门是（ B ）。
A 、与非门B 、或非门C 、异或门D 、同或门 7.下列各型号中属于优先编译码器是（ C ）。
东南大学历年电路试卷2010~2011期中A(答案)
t
15 5e 10 tV
9
t 0
六、图示电路中，已知 U=8V。求： （1） R 为何值时它消耗的功率为最大？并求出此功率。 （2） 求 R=12Ω时电流 I 和 I1 的值
I1 ＋ U － U － ＋ U － 16Ω 16Ω ＋ U － 8Ω 16Ω ＋ U － 8Ω 16Ω ＋ U － 8Ω 16Ω ＋ 8Ω 16Ω I R
i1 4A R S 10i1
i L
九、叙述并证明戴维宁定理。
共5页 第5页
共5页
第1页
i1 2 i1 2 A 20 20i 20i 10 10 I i 1A 2 3 2 7i1 20i2 7 20 3i3 4 i3 1A I 1A I i1 i2 CCVS 的功率为 P 10I i2 10 11 10W
Ix －
2Ω
U
2Ω 1.5U
共5页 第2页
五、在 t＝0 时刻，图中开关闭合，求电流 i(t)及电容的电压 u(t)。
t=0
5W
i
+ 20V -
2A
5W
+
u
- 400pF
解：开关闭合前，电容断V
所以
u0 u0 10V
开关闭合无限长时间后，电容断路处理，则
共5页 第3页
解：反复利用电源的等效变换，可得到除 R 之外的电路其它部分的戴维宁等效电路为：
8Ω ＋ 8V －
所以当 R=8Ω时，其消耗的功率最大，此功率为 (2)当 R=12Ω时，易求 I=8/(12+8)=0.4A 反复倒退计算，得 I1=0.0125A 七、求图中电阻能获得的最大功率。
计算机电路基础期末考试试题及答案
计算机电路基础期末考试试题及答案一、单项选择题1、在下列电路元件中，能够实现电信号的放大的是（）A.电阻B.电感C.电容器D.三极管2、下列哪一项不是LC振荡电路的特性（）A.能在电路中产生振荡电流B.能实现信号的发射和接收C.能作为高Q值滤波器D.能放大交流信号3、以下哪种效应是由霍尔元件产生的？（）A.磁场效应B.热效应C.霍尔效应D.压电效应二、多项选择题1、下列哪些是数字电路的特点？（）A.处理模拟信号B.容易实现复杂控制C.高速度、高效率D.集成度高、功能强大2、在计算机电路中，常见的反馈类型有哪些？（）A.正反馈B.负反馈C.交流反馈D.直流反馈3、LC振荡电路的振荡频率由哪些因素决定？（）A.电容器的容量B.电感器的电感量C.激励源的频率D.电路的连接方式三、判断题1、计算机电路中的电信号只有数字信号。
（）2、LC振荡电路可以产生连续的振荡电流。
（）3、三极管可以放大交流信号，也可以放大直流信号。
（）4、在计算机电路中，数字信号和模拟信号是可以互相转换的。
（）5、在计算机电路中，正反馈和负反馈都是为了使电路正常工作。
（）6、在计算机电路中，集成运算放大器是一种常见的模拟电路元件。
（）7、在计算机电路中，二极管是一种常见的数字电路元件。
（）8、在计算机电路中，电阻是一种常见的模拟电路元件。
（）电路理论期末考试试题及答案一、单项选择题1、在下列电路元件中，能够产生持续电流的是（）。
A.电源B.导线C.电容器D.电感器2、下列电路元件中，不能用来存储电能的是（）。
A.电动机B.电容器D.电阻器3、下列电路元件中，具有记忆功能的是（）。
A.电动机B.变压器C.电容器D.电阻器4、在下列电路中，能够实现电压、电流和功率的测量的是（）。
A.电阻器B.电感器C.电容器D.三相交流电机5、在下列电路中，能够实现电能转换的是（）。
A.电阻器C.电容器D.三相交流电机二、多项选择题1、下列电路元件中，能够实现电能转换的是（）。
东南大学2011电工基础真题回忆版
11年的专业课我感觉比较简单，基本没有难题。
可能这与10年试卷较难有关系，个人觉得难度与09年持平，所以今年电气的分数线。
填空题10X6，计算题有13，17两种分值各六道。
填空题第一道为已知某支路电流为0，求电路中一个电阻的值，用戴维南即可。
第二道求等效电压，用等电位法以及加压求流法。
第三道利用电容的电流电压关系以及kcl，kvl
额，后面具体顺序记不清楚了。
有一道求含运放电路的时间常数，
还有一题用最大功率定理做。
还有考察电路串并联谐振以及向量法的，都为基础题。
还有求满足衰减振荡的二阶电路中某一参数的条件
最后一道填空题为对称三相电路，已知二瓦计法中2个功率表的读数，反求电路阻抗z
大题第一道为结点电压法计算复杂电阻网络、
第二道为含理想变压器的电路中求解最大功率问题
三道为非正玄电路求电路中电阻电容等参数，注意电路的谐振
四道为求一阶响应，是老题，貌似邱关源书后习题原题。
五道为拉氏变换求二阶电路响应
六道为小信号分析法。
总体计算难度都不大。
没有往年那种要死算的复数运算。
over。
电路基本分析电路基础期末考试试卷及答案
电路基本分析电路基础期末考试试卷1A卷一、判断题每题2分1.在换路瞬间,如果电感电流不跃变,则电感电压将为零;2.当电感元件在某时刻t的电流it = 0时,电感元件两端的电压ut不一定为零；同样,当ut = 0时,it不一定为零;3.实际电源的两种模型,当其相互等效时,意味着两种模型中理想电压源和理想电流源对外提供的功率相同;4.某三层楼房的用电由三相对称电源供电,接成三相四线制系统,每层一相;当某层发生开路故障时,另二层电器一般不能正常工作;5.如图所示,当i1按图示方向流动且不断增大时,i2的实际方向如图所示;分享于上学吧资料分享课后答案二、填空题每题1分1.正弦量的三要素是指、、2.线性无源二端网络的阻抗与网络的结构、元件参数及有关;3.复功率~S的定义为~S ,它的实部表示功率,它的虚部表示功率,它的模表示 ,它的辐角表示 ;4.对称三相电路中,电源线电压为220V,负载作三角形联结,每相阻抗Z = 22,则线电流为；三相总有功功率为 ;5.由q u平面中一条曲线所确定的元件称为元件；由i平面中一条曲线所确定的元件称为元件;6.只具有单一节点对的电路的节点电压方程U = ,它称为定理;7. n个节点、b条支路的连通图,其树支数为 ,连支数为 ;8.已知某无源网络的导纳Y = 2 + j2S,则该网络的阻抗为 ,网络为性 ;9.选择u、i为关联参考方向时,50 电阻元件的u、i关系为 ,100F电容元件的u、i关系为 ,10mH电感元件的u、i关系为 ;10. RLC串联电路中,R = 50 ,L = 50 mH,C = 10F；电源电压US=100V,则谐振时电流I = ；品质因数Q = ;11.试为a、b两个图各选两种树,分别画在括号内;,12.图示电路中,N0为不含独立源的线性网络;当US= 3V、IS= 0时,U = 1V；当US= 1V,IS= 1A时,U = 0.5V;则当US = 0,IS= 2A时,U为 V;13.图示为RLC串联电路的频率特性曲线,其中感抗XL 的曲线为____________；容抗XC的曲线为____________；电抗X的曲线为___________；阻抗Z的曲线为___________;14. 含源二端电阻网络N S 如图所示,若测得开路电压U ab 为10V,短路电流I ab 为5A,则该网络的电压电流关系为U = ;15. 当一线圈含R 、L 与直流电压12V 接通时,其电流为2A,与正弦电压 V 100cos 212t u π=接通时,电流为1.2A,则线圈的电阻R = ,电感L = ;16. 对称三相负载作星形联结接至线电压为380V 的对称三相电压源,线电流为2A;若将负载改为三角形联结后接至线电压为220V 的对称三相电压源,线电流为 A;17. 设某电压的表达式为 ut = tt t 3t 3 V,在 t = 2 s 时,该电压值为 ;18. 某电压源的电压与电流为非关联参考方向,已知其电压为5V,电流为2A,则该电压源提供的功率为 W;19. 图示三相电路中,当开关S 闭合时,各电流表读数均为5A;当开关S 打开后,各电流表读数：A 1为 ,A 2为 ,A 3为 ;20. 电容C 经电阻R = 50充电,充电电源的直流电压U = 110V,接通电源后1.5 ms 时电流为 0.11A;则1充电时间常数 = ；2电容量C = ;三、计算题每题10分1. 电路如图所示,试用节点法求开路电压U OC ;2. 求图示二端网络的端口电压U OC ;3. 图示并联网络中,R 1 = 20 ,R = 6 ,L = 20 mH,C = 30 F;试求端口正弦电压有效值为220V 、频率为50Hz 时的端口电流,并求端口电压与电流的相位差;4. 试求图示电路中的电流I 1 、I 2、电压U 1和2 A 电流源的功率 ;5. 图示电路,已知L 1 = 0.1H,L 2 = 0.2H,M = 0.1H,R 1 = 5,R 2 = 10,C = 2F,试求顺接串联与反接串联两种情况下电路的谐振角频率0和品质因数Q ;6. 求图示电路的戴维宁等效电路;7. 试求图示电路在 = 1 rad/s 时的输入阻抗Z i ;电路基本分析 电路基础期末考试试卷1 A 卷答案一、判断题每题2分1.错误2.正确3.错误4.错误5.错误二、填空题每题1分1. 角频率或周期、频率；有效值或最大值；初相位2. 电源频率3. *I U；有功；无功；视在功率；阻抗角或功率因数角 4. 17.3A ；6600W5. 电容；电感6. ∑∑G GU S ；弥尔曼7. n -1；b -n -18. 0.25－j0.25；容9. u = 50 i ；t u i d d 104-=；t iu d d102-=10. 2A ；1.41411.；a ab b 12. 3113. 曲线A ；曲线B ；曲线D ；曲线C14. 10+2I15. 6；80 mH16. 3.4617. 2V18. -1019. 2.89A ；5A ；2.89A 20. 0.5 ms ；10 F三、计算题每题10分1. 选c 为参考节点,列节点电压方程：解得：V 38b =U2. U oc = -3 U oc +54所以：U oc =5V3. 设 220=U /0° V325 2.L j R U I =+=ω/－46.3°A=17.4－j18.3A= 29.74. I 1 =2+39A=5 AI 2 =24-2A= 0U 1 = 4-2V= 8 V电流源产生功率为：24-U 1-9W= 6 W5. 顺 接 串 联 时 L ' = L 1 + L 2 + 2M = 0.5 H故 rad/s 1000 10==C L ω反 接 串 联 时 L " = L 1 + L 2 2M = 0.1 H故 rad/s 2236"10==C L ω6. 求开路电压U oc ：解得： I 1=4A所以：U oc =6I 1=24V求等效电阻R o ：用外加电源法⎩⎨⎧-+=+=)(426621111I I I I I I U 解得：R o =Ω=5I U 所以戴维宁等效电路如下图：7. 将原电路去耦,并作出去耦电路的相量模型Z i = 0.098 + j0.804 = 0.81/83.05。
电工电子基础期末考试题及答案
电工电子基础期末考试题及答案一、选择题（每题2分，共20分）1. 电流的单位是：A. 伏特（V）B. 安培（A）C. 欧姆（Ω）D. 瓦特（W）答案：B2. 欧姆定律的公式是：A. V = I * RB. I = V / RC. R = V / ID. V = R * I答案：A3. 一个电路中，电阻R1为10Ω，R2为20Ω，串联后的总电阻为：A. 10ΩB. 20ΩC. 30ΩD. 40Ω答案：C4. 并联电路中，总电阻与各支路电阻的关系是：A. 总电阻等于各支路电阻之和B. 总电阻等于各支路电阻之积C. 总电阻小于任一支路电阻D. 总电阻大于任一支路电阻答案：C5. 电容器的单位是：A. 法拉（F）B. 安培（A）C. 欧姆（Ω）D. 伏特（V）答案：A6. 电感器的单位是：A. 法拉（F）B. 亨利（H）C. 安培（A）D. 欧姆（Ω）答案：B7. 交流电的有效值与峰值的关系是：A. 有效值等于峰值B. 有效值等于峰值的√2倍C. 峰值等于有效值的√2倍D. 峰值等于有效值的2倍答案：C8. 三相交流电的相位差是：A. 30°B. 45°C. 60°D. 90°答案：A9. 变压器的工作原理是：A. 电磁感应B. 静电感应C. 光电效应D. 热电效应答案：A10. 整流器的作用是将：A. 直流电转换为交流电B. 交流电转换为直流电C. 直流电转换为高频电D. 交流电转换为高频电答案：B二、填空题（每空2分，共20分）1. 电流通过导体时，导体内部的自由电子在电场力的作用下发生________。
答案：定向移动2. 欧姆定律表明，电流的大小与电压成正比，与电阻成________。
答案：反比3. 串联电路中，总电阻等于各部分电阻的________。
答案：和4. 电容器在交流电路中起到________的作用。
答案：滤波5. 电感器在交流电路中起到________的作用。
东南大学电子电路基础考试样卷
东南大学考试卷课程名称电子电路基础考试学期08- 09- 3 得分适用专业测控技术与仪器考试形式半开卷考试时间长度120分钟1、单项选择题（共20分，每小题2分）①在PN结外加正向电压时，扩散电流_______ 漂移电流，当PN结外加反向电压时,扩散电流___ 漂移电流。
A.小于，大于B.大于，小于C.大于，大于D•小于，小于②在三极管放大电路中，下列等式不正确的是（）。
A. |E=|B I cB. | cC. I CBO 二（1 「）1 CEOD.:: :=：③下面的电路符号代表（）管。
A.耗尽型PMOSB.耗尽型NMOSC.增强型PMOSD.增强型NMOS④放大电路如图所示，已知三极管的 E -50，则该电路中三极管的工作状态为（）°A.截止B.饱和C.放大D.无法确定⑤放大电路A、B的放大倍数相同，但输入电阻、输出电阻不同，用它们对同一个具有内阻的信号源电压进行放大，在负载开路条件下测得A的输出电压小，这说明A 的（）。
A.输入电阻大B.输入电阻小C.输出电阻大D.输出电阻小⑥把差分放大电路中的发射极公共电阻改为电流源可以（）A •增大差模输入电阻B •提高共模增益C .提高差模增益D •提高共模抑制比⑦ 乙类互补对称功率放大电路（）A •能放大电压信号，但不能放大电流信号B •既能放大电压信号，也能放大电流信号C •能放大电流信号，但不能放大电压信号D •既不能放大电压信号，也不能放大电流信号⑧ 为了减小放大电路从信号源索取的电流并增强带负载能力，应引入 （ ）负反馈。
A.电压串联B.电压并联C.电流串联D.电流并联⑨已知变压器二次电压为U 2=：：j 2U 2sin’t v ，负载电阻为R L ，则桥式整流电路流⑩关于三极管高频参数，下列说法中不准确的为（ ）。
A.许二 0B. f ：. = （4：o ）f 1 C. 2、（15分）放大电路如图所示，已知电容量足够大， Vcc=18V , R Bi =75k 「R B 2=20^ 1,R E2=1.8k 「R EI =200」R C =8.2k 「R L =6.2k 「, R s =600「三极管的 3 =100 r bb ' =200门,U BEQ =0.7V 。
电子电路基础 样卷(东南大学)
样卷课程名称 电子电路基础 考试学期 08-09-3得分适用专业 测控技术与仪器考试形式半开卷考试时间长度 120分钟1、单项选择题（共20分，每小题2分）①在PN 结外加正向电压时，扩散电流 漂移电流，当PN 结外加反向电压时，扩散电流 漂移电流。
A. 小于，大于B. 大于，小于C. 大于，大于D. 小于，小于②在三极管放大电路中，下列等式不正确的是（ ）。
A.C B E I I I +=B. B C I I β≈C. CEO CBO I I )1(β+=D. βααβ=+ ③下面的电路符号代表（ ）管。
A.耗尽型PMOSB.耗尽型NMOSC.增强型PMOSD.增强型NMOS④放大电路如图所示，已知三极管的05=β，则该电路中三极管的工作状态为（ ）。
A. 截止B. 饱和C. 放大D. 无法确定⑤放大电路A 、B 的放大倍数相同，但输入电阻、输出电阻不同，用它们对同一个具有内阻的信号源电压进行放大，在负载开路条件下测得A 的输出电压小，这说明A 的（ ）。
A. 输入电阻大B. 输入电阻小C. 输出电阻大D.输出电阻小⑥把差分放大电路中的发射极公共电阻改为电流源可以（ ）A ．增大差模输入电阻B ．提高共模增益C ．提高差模增益D ．提高共模抑制比 ⑦乙类互补对称功率放大电路（ ） A ．能放大电压信号，但不能放大电流信号 B ．既能放大电压信号，也能放大电流信号 C ．能放大电流信号，但不能放大电压信号D ．既不能放大电压信号，也不能放大电流信号⑧为了减小放大电路从信号源索取的电流并增强带负载能力，应引入（ ）负反馈。
A. 电压串联B. 电压并联C. 电流串联D. 电流并联 ⑨已知变压器二次电压为t U u ωsin 222=V ，负载电阻为R L ，则桥式整流电路流过每只二极管的平均电流为（ ）。
A. L R U 29.0 B. L R U 2C. L R U 245.0 D. LR U 22 ⑩关于三极管高频参数，下列说法中不准确的为（ ）。
东南大学《电路基础》期末试卷B及答案
《电路基础》试卷B专业：姓名：共五道计算题，每题20分一、如图所示电路，a,b,c,d表示4个节点。
12R=Ω,26R=Ω,36R=Ω,求电压bU二、已知电路如图，已知1236,4,12R R R =Ω=Ω=Ω,4L R =Ω,利用戴维南定理，求该电阻上的电流i ，并画出中间过程等效电路图。
三、如图所示电路，已知()i t A =,4r =Ω， 2.4L H =，0.025C F =求电压u四、如图所示电路, 122,1R R =Ω=Ω,1,3L C X j X j =Ω=-Ω,.100()S U V =∠ 求:(1)各元件吸收的功率;(2)电源供给的功率.五、如图所示的对称三相负载，已知线电压380l U V =,负载阻抗68Z j =+,求各项负载电流和负载总功率。
试卷B 答案一、 解：由图可见，ab 间的电压6ab ad bd a b b U U U U U U =-=-=-所以 1622ab bU U I -== bc 间的电压(3)3bc bd cd b c b b U U U U U U U =-=-=--=+所以 2366bc b U U I +== 366bd b U UI ==对于节点b,根据KCL 有123I I I =+,将1I ，2I ，3I 代入上式，得 63266b b bU U U -+=+ 可解得 b U ＝3V二、解：利用戴维南定理(1)求等效电阻0R130213R R R R R R ⨯=++6124612⨯=++ 8=Ω(2)求端口开路电压OC U1212612120.54612612OC U U U ⨯⎛⎫=+=⨯+⨯+ ⎪++⎝⎭8412V =+=利用戴维南得原理图：所以 12184OC O L U i A R R ===++三、解：取电流i 的相量为：.20()I A =∠令未知电压相量为.0U .由于5/rad s ω=，根据各元件值可得 5 2.412()j L j j ω=⨯=Ω118()50.025j j C j ω==-Ω⨯ 可得各元件电压分别为：..42080()r U r I V ==⨯∠=∠ ..12202490()L U j L I j V ω==⨯∠=∠..0018201690()C U I j V j Cω==-⨯∠=∠- 由KVL,得 ....008024901690r L C U U U U =++=∠+∠+∠-0824168845()j j j V =+-=+= 将电压相量变换为正弦函数形式，得00()45)16cos(545)()u t t t V =+=+四、 解 (1)支路1的阻抗为121()Z j =+Ω,故电流..011100 4.7426.57()21S U I A Z j ∠===∠-+ 1R 吸收的有功功率和L 吸收的无功功率分别为1211214020varR L L P I R W Q I X ====支路2的阻抗为213()Z j =-Ω,故电流..022100 3.1671.57()13S U I A Z j ∠===∠-2R 吸收的有功功率和C 吸收的无功功率分别为22222210()30varR C C P I R W Q I X ===-=-(2)电路的总电流...0012 4.4726.57 3.1671.5751()I I I j A =+=∠-+∠=+ 电源供给的复功率: ..*(100)(51)5010()S S U I j j j VA ==+-=- 五、解 与线电压l U 相对应的相电压220P U V === 由于三相电路对称，因而中线两端电压'.0n n U =，根据置换定理，将'n n 短路。
精选-东南大学电子学院《电路基础》期末考试样卷及答案
Solve the following problems. (100 points)1、( 6 points)Find U ab in the circuit in Figure 1.Figure 1a b2、( 8 points)Find u o and i o in the circuit in Figure 2.u oFigure 2-3、( 6 points)In the circuit of Figure 3, readings of voltmeter ○V1，○V2and○V3 are 10V, 18V and 6V, respectively. Please determinate the reading of the voltmeter ○V.aFigure 34、( 8 points)The resonant or tuner circuit of a radio is portrayed in Figure 4, where u s1 represents a broadcast signal, given that R=10Ω，L=200μH, Us1rms=1.5mV，f1=1008kHz. If the circuit is resonant with signalus1, please determinate: (1) the value of C; (2) the quality factor Q of the circuit; (3) the current I rms; (4) the voltage U c rms.Figure 4u5、( 8 points) A balanced three-phase circuit is shown in Figure 5. Calculate the phase currents and voltages in the delta-connected load Z, if V U A ︒∠=•0220, Z =(24+j15)Ω，Z L =(1+j1)Ω.Figure 5•6、( 8 points) Find they parameters for the circuit in Figure 6.••Figure 67、( 15 points) For the circuit in figure 7, find the rms value of the current i and the average power absorbed by the circuit, given that Ω10=L ω,Ω51=R ,Ω202=R ,Ω20/1=C ω,V )302cos(50cos 10080)(︒+++=t t t u s ωω.Figure 72u s8、( 12 points) In the circuit of Figure 8, determinate the value of Z that will absorb the maximum power and calculate the value of the maximum power.Figure 8V ︒∠050ω=103•rms9、( 14 points) If the switch in Figure 9 has been closed for a long time and is opened at t = 0, find u c , i L , u k for t ≥ 0.20Figure 9u c +-10、( 15 points) The switch in Figure 10 has been in position a for a long time. At t = 0, it moves to position b. Find i 2(t) for t ≥ 0 and sketch it. (Attention: Please mark the key data in your graph.)Figure 1010ΩSolve the following problems. (100 points)1、( 6 points) Find U ab in the circuit in Figure 1.Figure 1a bSolution:U ab = 9V2、( 8 points) Find u o and i o in the circuit in Figure 2.u o Figure 2-Solution:V 4.2=o u , 0.256mA -=o i3、( 6 points) In the circuit of Figure 3, readings of voltmeter ○V1，○V2 and ○V3 are 10V, 18V and 6V, respectively. Please determinate the reading of the voltmeter ○V .aFigure 3Solution:The reading of the voltmeter ○V is 10V .4、( 8 points) The resonant or tuner circuit of a radio is portrayed in Figure 4, where u s1 represents a broadcast signal, given that R =10Ω，L =200μH , U s1rms =1.5mV ，f 1=1008kHz. If the circuit is resonant with signal u s1, please determinate: (1) the value of C ; (2) the quality factor Q of the circuit; (3) the current I rms ; (4) the voltage U c rms .Figure 4uSolution:pF 5.12)1(≈C 6.126)2(≈Q μA 150)3(=rms I mV 190)4(=Crms U5、( 8 points) A balanced three-phase circuit is shown in Figure 5. Calculate the phase currents and voltages in the delta-connected load Z, if V U A ︒∠=•0220, Z =(24+j15)Ω，Z L =(1+j1)Ω.Figure 5•Solution:A 3.69-11.74''︒∠=•B A I , V 31.28332''︒∠=•B A U6、( 8 points) Find the y parameters for the circuit in Figure 6.••Figure 6Solution:[]S y ⎥⎦⎤⎢⎣⎡=0.5 40.02 2.07、( 15 points) For the circuit in figure 7, find the rms value of thecurrent i and the average power absorbed by the circuit, given that Ω10=L ω,Ω51=R ,Ω202=R ,Ω20/1=C ω,V)302cos(50cos 10080)(︒+++=t t t u s ωω.Figure 72u sSolution:A 17.6357.2714.42.3222222120=++=++=I I I I rmsP =P 0+P 1+P 2=256+333.28+50=639.28W8、( 12 points) In the circuit of Figure 8, determinate the value of Zthat will absorb the maximum power and calculate the value of the maximum power.Figure 8V ︒∠050ω=103•rmsSolution:If Ω135jZ Z eq+==*, Z will absorb the maximum average power. The maximum average power is, W 7.2491356.1941412max=⋅=⋅=*eq OC R U P9、( 14 points) If the switch in Figure 9 has been closed for a long timeand is opened at t = 0, find u c , i L , u k for t ≥ 0.20Figure 9u c +-Solution:0 A,5.15.1)5.10(5.14040>-=-+=--t e e i t t L ， 0 V,60120)12060(12020/20/>-=-+=--t e e u t t C 0 ,V 6030902020/40>+--=-=--t e e u i u t t C L K10、( 15 points) The switch in Figure 10 has been in position a for a long time. At t = 0, it moves to position b. Find i 2(t) for t ≥ 0 and sketch it. (Attention: Please mark the key data in your graph.)Figure 1010ΩSolution: A 25.125.132t t e e i ---= 打发第三方（注：专业文档是经验性极强的领域，无法思考和涵盖全面，素材和资料部分来自网络，供参考。
《电路分析基础》期末考试试卷
《电路分析基础》期末考试试卷非统考科目一、判断题（每题5分共100分）1．负载在额定功率下的工作状态叫满载。
[判断题] *对(正确答案)错2．电流方向不随时间而变化的电流叫稳恒电流。
[判断题] *对错(正确答案)3．如果把一个6V的电源正极接地，则其负极为-6V。
[判断题] *对(正确答案)错4．电路中某点的电位值与参考点选择的无关。
[判断题] *对错(正确答案)5．电路中某两点间电压的正负，就是指这两点电位的相对高低。
[判断题] *对(正确答案)6．导体的长度增大一倍，则其电阻也增大一倍。
[判断题] *对(正确答案)错7．电源电动势的大小由电源本身性质决定，与外电路无关。
[判断题] *对(正确答案)错8．公式可以写成，因此可以说导体的电阻与它两端的电压成正比，与通过它的电流成反比。
[判断题] *对错(正确答案)9．在电源电压一定的情况下，负载大就是指大负载。
[判断题] *对错(正确答案)10．电源电动势等于内外电压之和。
[判断题] *对(正确答案)错11．当电源的内电阻为零时，电源电动势的大小就等于电源端电压。
[判断题] *对(正确答案)12．短路状态下，短路电流很大，电源的端电压也很大。
[判断题] *对错(正确答案)13．开路状态下，电路的电流为零，电源的端电压也为零。
[判断题] *对错(正确答案)14．电池存放久了，用万用表测量电压为1.4V，但接上灯泡却不发光，这主要是电源电动势变小了。
[判断题] *对错(正确答案)15．全电路电路中，若负载电阻变大，端电压将下降。
[判断题] *对错(正确答案)16．一般来说，负载电阻减小，则电路输出的功率增加，电源的负担加重。
[判断题] *对(正确答案)错17．通过电阻上的电流增大到原来的2倍时，它所消耗的功率也增大到原来的2倍。
[判断题] *对错(正确答案)18．把“110 V/50 W”的灯泡接在220 V电压上时，功率还是50 W。
东南大学数字电路期末试卷
数字电路期末试卷一一、一、 设计一个模18计数器（共40分）分） 要求:1．设计电路,写出设计过程并将逻辑图画在答题纸；(15分) 2．用单脉冲或秒脉冲验证实验结果；(由老师检查)(15分) 3．用示波器或者逻辑分析仪观察并记录时钟与个位的低两位信号用示波器或者逻辑分析仪观察并记录时钟与个位的低两位信号（（Q1、Q0）波形。
（10分）分）二、二、 设计一个具有自启动功能的序列信号发生器1011 （共60分）分） 要求:1．设计出电路图,写出设计过程并将逻辑图画在答题纸上；(20分) 2．根据设计搭试电路；（15分）分）3．用指示灯验证电路的正确性,并检查该电路是否具有自启动功能；(15分) 4．用示波器或者逻辑分析仪观察波形,并将测试结果画在答题纸上。
(由老师检查)(10分) 一、一、 设计一个模18计数器（共40分）分） 要求:1．设计电路，写出设计过程并将逻辑图画在答题纸；(15分) 评分标准：原理图完全正确15分；分；若其中低位或者高位单独正确给若其中低位或者高位单独正确给5分；如果两个单独均正确但级联错误给10分；接地不画扣2分。
2．用单脉冲或秒脉冲验证实验结果.(由老师检查)(15分) 3．记录结果．记录结果 （10分）分）评分标准:：相位对齐6分（每个输出端信号3分），画满一个周期3分，方波边沿画出1分。
分。
二、1．评分标准:原理图正确20分，输入没有使能端扣3分，接地不画扣2分。
分。
2．根据设计搭试电路；（15分）分）3．用指示灯验证电路的正确性,并检查该电路是否具有自启动功能；(15分) 评分标准:实验操作,仪器使用5分,指示灯验证和自启动功能检查15分 4．用示波器或者逻辑分析仪观察波形,并将测试结果画在答题纸上.(由老师检查)(10分) 评分标准:波形观察记录，相位对齐6分，至少画满一个周期（3分），且画出边沿（1分）10分。
南信大模拟电路期末试题
复习题二注意：1、先把书上知识点和做过的作业理解会做，然后再复习这部分内容。
2、期中考试前发的样卷里有很多第十章直流稳压电源相关的题目。
1、放大电路中的静态分量是指__________，动态分量是指____。
A 、直流电源所提供的电压、电流 ，B 、电压、电流中不随输入信号变化的部分，C 、电压、电流中随输入信号变化的部分 ，D 、正弦交流输入、输出信号2、NPN 和PNP 型晶体管的区别取决于____A 、半导体材料硅和锗的不同，B 、掺杂元素的不同 ，C 、掺杂浓度的不同，D 、P 区和N 区的位置不同。
N 沟道和P 沟道场效应管的区别在于____ A 、衬底材料前者为硅，后者为锗，B 、衬底材料前者为 N 型，后者为 P 型，C 、导电沟道中载流子前者为 电子，后者为空穴。
3、晶体管工作在放大区时，发射结为____，流过发射结的主要是____；集电结为____，流过集电结的主要是____。
A 、正向偏置，B 、反向偏置，C 、零偏置，D 、扩散电流，E 、漂移电流4、某硅二极管在正向电压V 6.0D =U 时，正向电流mA 10D =I 。
当D U 增加10％（即增大到0.66V ）时，则D I 约为____。
A 、10mA ，B 、11mA ，C 、20mA ，D 、100mA5、已知某二极管在温度为25℃时的伏安特性如图中实线所示，在温度为 T 1时的伏安特性如图中虚线所示。
1．在25℃时，该二极管的死区阈值电压、反向击穿电压和反向电流各是多大？ 2．温度T 1是大于25℃还是小于25℃？6、两个放大倍数相同、输入和输出电阻不同的放大电路A 和B ，对同一个内阻等于零的信号源电压进行放大，并接有相同大小的负载电阻，测得A 的输出电压比B 的大。
这说明A 的________________。
输出电阻比B 的________________。
A 、输入电阻， B 、输出电阻 C 、大， D 、小7、有两个放大倍数相同、输入和输出电阻不同的放大电路A 和B ，对同一个具有内阻的信号源电压进行放大，在负载开路的条件下测得A 的输出电压比B 的小，这说明A 的________________。
东南大学电工电子学试卷
共 45 页 第 1 页东 南 大 学 考 试 卷（A 卷）课程名称电工学 考试学期 11-12-2得分适用专业 机械工程及自动化 考试形式 闭卷 考试时间长度120分钟1)Birds routinely land and relax on power lines whichcarry tens of thousands of volts of electricity. Explain why these birds do not get electrocuted. (6 pts)The resistance of the air is too large for current to flow from the line to the bird to the ground.Alternately, the resistance is too large for current to divert from the wire, into the bird, and then back into the wire. It is not because of wire insulation. Birds could land on uninsulated wires just fine.6 pts for right answer. 1 pt for something vaguely correct.2) Suppose your car battery is dead, and you need to charge it using another car’s battery. You have a pair of cables 电缆which you can use to connect the terminals终端of the batteries. In order to charge your car battery, should you connect the same terminals (positive of one battery to the positive of the other, and the same with the negative) or the opposite terminals (positive of one battery to the negative of the other battery)? Why did you choose this configuration配置；结构? (6 pts)First note that real batteries have internal内部的resistance, so the universe领域will not explode爆炸if you directly connect two car batteries.Next, if the opposite terminals are connected, then P=VI and common sense tells us that both batteries will be supplying power, and the internal resistances will be consuming消耗power, so clearly no power is being delivered 递送by either battery. If the same terminals are connected,共45 页第 2 页then we can see that if one battery is of a higher voltage than the other, it will provide power to theother battery. Key to this realization实现is to know that battery voltages drop下降；终止as the battery is depleted耗尽的.6 pts for right answer. 2 pts if you try to say there is no current if two batteries are connected + to+ because voltage sources are perfectly balanced.3) A standard procedure for testing the internal resistance of a battery is the “dual双重的pulse脉冲” test. We first attach an ideal 5 mA current source between the terminals of the battery, so that current flows in the usual direction (positive to negative), and measure the voltage across the battery terminals. We then remove the 5 mA source, and attach an ideal 505 mA current source instead,and again measure the battery terminals. (6 pts)共45 页第 3 页i.Suppose that we find a 1.485V voltage with the 5 mA source, and a 1.385 with the 505 mA source, what is the internal resistance?Easy way to solve these is to subtract减去；扣掉 2nd from first, giving, or .3 pts for right answer. 2 pts if sign error or algebra 代数学mistake.ii. Assuming the battery is perfectly linear 线型的(i.e. accurately modeled 模式化；被效仿by a Thevenin 戴维宁equivalent), is it possible to find the voltage provided by the battery with no load attachedusing the data above? If so, what is it? If not, why not?共45 页第 4 页Plug塞住；用插头将与电源接通 in to one of our equations above, e.g., giving , or finally. If you did it with the current source the other way, you should still have gotten 1.486V.3 pts for right answer.Find the Thevenin or Norton equivalent circuit model.(12 pts ).共45 页第 5 页(6 pts )(6 pts )Find the Voltage v and the currents i1 and i2 for the circuit shown below. (10 pts )共45 页第 6 页SolutionLabel标注 the meshes；网状物:Since i3=2 it is not a variable.Writing KVL around the first mesh:24*(i2-i3) + 12*i1 = 0Writing KVL around the second mesh:12*i2 + 6*(i2 –i3) = 0Substituting代替，取代 for i3 and rearranging 36*i1 = 4818*i2 = 12共45 页第7 页Namelyi1 = 4/3 Ai2 = 2/3A7. Find the current i for the circuit shown below (10 pts )SolutionZeroing the 1A source gives共45 页第8 页By the current division principle, the contribution of the 2A source to i1 is2*(-1/30) / (1/30+1/5) = -2/7A.Zeroing the 2A source givesBy the current division principle, the contribution of the 1A source to i1 is1*(1/15)/(1/15+1/20) = 4/7A.Hence i1 = 4/7-2/7 = 2/7A.共45 页第9 页Find the Thevenin and Norton equivalent circuits across terminals a and b for the following circuitWith an open circuit across terminals a and b the current through the 20 resistor must be 0.5ix and the voltage across it is then 10ix. ByKVL we also get30V = 5ix + 10ixSo ix = 2A and we calculateVTH = 10ix = 20VTo find RTH we zero the independent独立的 source to get the circuit in Figure 4.We wish to determine what resistance this circuit isequivalent 等价的to across共45 页第10 页the terminals a and b. One way to do this is to imagine connecting anindependent voltage source of voltage VS across the terminals and findingthe current drawn from this voltage source. We therefore consider thecircuit in Figure 5.KVL gives−5Ωix = VS → ix = −VS/5Ω A (1)KVL also gives20Ωi1 = VS → i1 =VS/20 Ω (2)KCL givesi = i1 + 0.5ix − ix = i1 − 0.5ix (3)共45 页第11 页Plugging (1) and (2) into (3) yieldsi =VS/20Ω+VS/10Ω=3VS/20ΩEventually,RTH =VS/i=20/3 ΩThe Thevenin equivalent circuit is depicted描述； in Figure 6.From this we get the Norton equivalent circuit in Figure 7 where we usedWe could have also determined IN by applying a short circuit across theterminals a and b and computing 计算；处理the current through the short circuit短路.This involves analyzing the circuit in Figure 8. Here wehave by KVL共45 页第12 页5ix = 30V → ix = 6AThis gives, by KCLi = ix − 0.5ix = 0.5ix = 3Awhich matches the earlier calculation计算；估计.Suppose that v1(t) = 80 cos (ωt) and v2(t) = 60 sin (ωt). Use phasors相量 to reduce the sum vs(t) =v1(t) + v2(t) to a single 单一的term of the form Vm cos (ωt +θ). Draw a phasor diagram, showing V1, V2, and Vs. State the phase relationships between each pair of these phasors. (16共45 页第13 页pts )The phasor corresponding tov1(t) = 80 cos (ωt)isV1 = 80∠0 = 80Equivalently, we get forv2(t) = 60 sin (ωt) + 60 cos (ωt −π/2)the phasorV2 = 60∠−π/2 = −j60To find the sum using phasors we add the complex numbers which yieldsV = V1 + V2 = 80 − j60 = 100e−j arctan 3/4This corresponds符合，一致 to the time signalvs(t) = 100 cos (ωt − jarctan3/4)The phasor diagram is depicted描述 in Figure 2.Note that V1 leads V2 by π/2 and leads Vs by arctan反正切 ( 4/3 ).共45 页第14 页Find an expression 表达式for v(t) of the form Vm cos (ωt +θ) when v(t) = v1(t) + v2(t) + v3(t) + v4(t) withv1(t) = 20 sin (ωt)v2(t) = 20 cos (ωt +π/6)v3(t) = 20 sin (ωt +π/3)v4(t) = −10 cos (ωt)Use phasors. (6 pts )Solution:The phasors corresponding 相应的to each component 成分；组件；[电子] 元件signal are given in Table表格 1. The complex number equivalent复数等效to each phasor is given as well.共45 页第15 页Adding all complex numbers giveswhich corresponds to the phasorwhich can be transformed变形 into the time signalFind the complex impedance阻抗 in polar form 点斜式极坐标形式极形式of the network shown below for ω = 10001/s , ω= 20001/s ,and ω = 40001/s . (6 pts )共45 页第16 页(5 pts)Both capacitors电容器 and inductors电感器 can be used to store energy for use later. Capacitors, though, are far more popular for energy storage than inductors, because they are easier to keep charged带电荷的带电的. In lab, for example, you can easily connect up a capacitor to a battery, disconnect it from the source, carry it around in your pocket to grab lunch午餐, go back to lab, and then find that it's still holding a charge even after a long time. Explain why this is relatively difficult with inductors.共45 页第17 页(5 pts) If you've ever had the batteries on a flash li ght闪光灯；手电筒 run out,you know that as soon as the ligh t starts dimming调光；变暗,your battery will very soon be dead.This is because a battery supplies its maximum vo ltage峰点电压 until right before it is depleted耗尽的.On t he axis 坐标轴on the left,I have plotted标绘的 the brightn ess亮度辉度 of a battery powered flashlight over the lif e of the battery.We can also power a flashlight using共45 页第18 页a capacitor.Assuming the brightness of a bulb 电灯泡is p roportional比例的 to the voltage across it,on the right axis,draw (qualitatively定性地) the brightness vs. Time for a capacitor powered flashlight.Draw so that bright ness is 1 at time=0,and has dropped to at least 0.1 by time=1.We just want the shape形状,not necessarily the exact values.c) (5 pts) A student is not satisfied with the amount of共45 页第19 页energy stored in his one-capacitor circuit. He wants to store more energy, so puts 10 capacitors in series串联 and charges them with a single voltage source. Is this a good idea? Why or why not?Find the Thevenin equivalent 戴维南定理of the circuit电路below at the two terminals on the far right:共45 页第20 页e) Find the Thevenin equivalent of the circuit below between terminals and [Think about the various algorithms算法 we’ve used to find Thevenin equivalents.If you get stuck被困住了,complete the rest of these tricky 棘手的复杂的problem and come back to this one and maybe some inspiration will reach you]:共45 页第21 页共45 页第22 页For the circuits below, please find expressions for the specified voltage表达式指定的电压over the indicated time 表示时间ranges in terms of the circuit parameters电路参数. Plot 绘图 the waveform 波形图on the provided axes轴线, and clearly identify. 确定 the key parameters参数 in your graph图表. (3A) Consider the circuit of Fig. 4. The switch is open for t <0, closed for 0≤t < t1, and open for t ≥t1, where t1 = 3RC. Find and plot the voltage v C (t).共45 页第23 页(3B) Consider the circuit of Figure 5, in which > -1.共45 页第24 页The switch is open for t <0, and closed for t 0. vC(0-) = V0. Find and plot the voltage v1(t).共45 页第25 页The circuit shown in Figure 4 has been at rest with theswitch open for a long time. At t = O the switch is closed.共45 页第26 页Sketch 素描；略图；梗概the current ic集成电路 through the capacitor for t > O on the axes below. Using the circuit parameters R1, R2, C, and Vo,indicate on your sketch (i) the initial value初始值 ic(Ot), (ii) the final value ic(oo), and (iii) the time constant时间常数.共45 页第27 页Problem 1F: 5 pointsThe capacitor in the circuit in Figure 6 has an initial初始最初的voltage V O on its terminals at t =0−when a step of voltage Vu(t) is applied at t = 0. Find an expression for the voltage across the capacitor v C for t> 0.共45 页第28 页Consider the current divider circuit分流电路 shown below:共45 页第29 页1a. How many branches分支 are there in the circuit?1b. How many nodes节点 are there in the circuit?2. Derive an expression派生一个表达式 for the output current 输出电流(Iout, shown above) as a function of theinput current (Iin) and the two resistors (R1 and R2). Use any technique you’d like.3. How should the resistors be chosen such that twice as much current flows throughR1 compared to R2?4. Using the equation 方程等式you obtained in #2, show what happens to Iout输出电流 if R2 is replacedwith a short circuit (i.e. R2 = 0). Repeat if R2 is replaced with an open circuit.(think about what R2 should be replaced with here) Now suppose the following values are used: Iin = 3 A, R1= 5 , R2 = 20 , and we wish to find the Thevenin共45 页第30 页equivalent circuit across the two indicated terminals below:5. Find Voc and Isc across the two terminals indicated on the circuit.6. Draw the Thevenin equivalent circuit for the two terminals indicated above.7. Suppose a load is now attached across the two terminals of the Theveninequivalent circuit:a. Write an equation for I in terms of V.b. Sketch the IV characterstics伏安特性. Use I as the“y-axis Y轴” and V as the“x-axis”. Label标注all relevant points相关点.共45 页第31 页8. Suppose the load has an IV characteristic given as follows:a. What are the values of I and V if the load is attached to the Thevenin equivalent circuit ] 等效电路?b. Looking at the IV characteristic, what is the load? (describe the load as much as you can)For problems 9 through 12, consider the following circuit:9. Using KVL and mesh analysis, write down the equations needed to solve the circuit. Convert转变the resistor voltages into currents.10. The dependent source requires one additional equat共45 页第32 页ion附加方程to be generated产生. Write down that equation by looking at the 3 resistor.11. Solve the equations simultaneously同时地to determine the two mesh currents网孔电流.12. To check your answer, confirm确认that power is conserved 守恒in the circuit. Find the power dissipated浪费的in each resistor. Find the power generated生成的(or dissipated in each source). Be clear in your answers whether power is being generated or dissipated in each circuit element. For problems 1–7, consider the following circuit:1. Find the transfer转移function, i.e. Vout输出电压()/Vin(), in terms of R, L, and .2. What is the magnitude大小of the transfer function传递函数, i.e. | H() |, in dB?3a. What is | H() | equal to (in dB) for very low共45 页第33 页frequencies频率?3b. What happens to | H() | at high frequencies? What kind of a filter滤波器do we have here?4. Derive the equation for the break frequency推导出方程为截止频率. The break frequency is definedas the frequency where | H() | = –3.0103 dB.5. Sketch the magnitude Bode波特图plot for this filter滤波器. Label标注all critical points andslopes. Show how you obtained the value for any non-zero slopes零斜坡.6. Suppose R = 1 , L = 2 H, and Vin = 3 cos(5t + 30o). Find Vout(t).7. Draw a phasor diagram, indicating the voltage and current phasors of all thecircuit elements电路参量.For problems 8–12, consider the following circuit:共45 页第34 页8a. Find the current that appears on the output side ofthe transformer变压器.8b. What is the voltage that appears across the capacitor. (phasor form is acceptable,assume an angular frequency角频率of 1 rad/s)9. Find the voltage that appears on each side of the transformer.10. Find the voltage that appears across the current source.11. Replace the transformer and input circuitry with an equivalent current sourceand inductor. In other words, draw a circuit which will have a current source,an inductor, and a 3 F capacitor all in series. Label the values for the current source and the inductor.Check your answer to #11 by comparing the amount of power共45 页第35 页generated by the currentsource in each circuit.12. Find the amount of power generated by the current source in the new circuit.Compare this with the amount of power generated by the current source in theoriginal circuit原电路 (they should be the same)For problems 4–7, consider the following circuit:4. Find the transfer function, i.e. Vout ()/Vin (), in terms of R, L, and (or f).5. What is the magnitude of the transfer function, i.e. | H() |?共45 页第36 页6. What is | H() | equal to for very low frequencies? For very high frequencies?What happens at resonance谐振? (i.e. = o =LC1) What kind of a filter滤波器do we have here?7. Sketch the magnitude Bode plot for this filter. Label all critical points andslopes斜率. Use R = 2 , L =41H, and C =81F. Show how you obtained the valuefor any non-zero slopes.In order to measure the voltage across a resistor in a circuit, a voltmeter电压表is used.Although we assume an ideal voltmeter in most cases, a real voltmeter has a finite有限的resistance (R3) which can have an effect on your circuit. Consider the followingmeasurement on a voltage divider分配共45 页第37 页Vin, R1, and R2 are just the standard elements of a voltage divider.1a. How many nodes are there in the circuit?1b. How many meshes are there in the circuit?2. Write a KCL equation at node A. Convert the currents into voltages.3. Solve the equation for Vout, so that you have an equation for Vout based onVin, R1, R2, and R3.4. Show that as R3 approaches infinity无限大(i.e. R3 ), you get the standardequation for a voltage divider.For problems 5 through 11, consider the following circuit:共45 页第38 页Find the Norton equivalent circuit, as follows:5. Short circuit the two terminals and redraw the circuit without the 2 resistor.Explain why the 2 resistor can be dropped from the circuit.放弃了电路6. Use node analysis to find Isc. (You will need to come up with one additionalequation to eliminate消除ix and get a numerical数字的answer) 7. Now consider the circuit when the two terminals are open circuited. Using meshanalysis, find ix under these conditions.8. Using your results from #7 (or using node analysis, which will take more time),find Voc, the voltage across the two terminals in an open circuit condition.共45 页第39 页9. Confirm that power is conserved in the circuit under open-circuit conditions.Find the power dissipated in each resistor. Find the power generated (ordissipated in each source). Be clear in your answers whether power is beinggenerated or dissipated in each circuit element.10a. Draw the Norton equivalent circuit. Label values for all circuit elements.10b. Draw the Thevenin equivalent circuit. (Note: you already have everything youneed)11a. Assume that a circuit element (that draws some current I and has somevoltage drop V) is attached to the terminals of the Norton equivalent circuit.Using KCL, write an equation for I in terms of V.11b. Graph the IV relationship (i.e. the load line). Use I as the “y-axis” and V as the共45 页第40 页“x-axis”. Label all relevant points.For problems 1–5, consider the following circuit:Use Vin = 100 v, R1 = 1000 , R2 = 250 , R3 = 1000 , and C = 1 F. At t = 0, theswitch is moved down, disconnecting R1 and connecting R2 to the circuit. We want to find the voltage v(t), and current i(t), across the capacitor as a function of time. 1a. Redraw the circuit, showing how the circuit looks just before the switch ismoved. Assume a long time has already elapsed before the switch is moved.1b. What is the continuity连续性condition for this circuit共45 页第41 页at t = 0?2. Using the circuit generated in #1a, find v(0–) and i(0–)3a. Find v(). Explain how you get this result.3b. Write a KCL equation for the circuit after the switch as been moved at t = 0.Convert the currents into voltages. (you should end up with a first-orderdifferential equation for v(t))4a. What form should the solution for v(t) take?4b. Substitute替代this solution into the differential equation generated in #3b. Use theresults of the substitution, along with the results for #1b and #2 to find all theconstants defined for v(t). (Clearly show your work) 5. Sketch curves of i(t) and v(t) as a function of time (t ). Indicate allcritical and asymptotic values.For problems 6–10, consider the following second-order共45 页第42 页circuit:The switch is opened at t = 0. The goal is to find the current through the inductor, i(t).6a. Assume that the switch has been closed for a long period of time. Find i(0–).6b. What should i(0+) be? Explain.7a. Find i(). Explain how you get this result. Draw the corresponding相应的circuitat this steady state condition.7b. Since this is a second-order circuit, another initial 初始/boundary condition is needed.Find that other value. Show your work.8. Derive导出the second-order differential equation for this circuit. Start by writing aKVL equation for the circuit after the switch has been opened. Then convert共45 页第43 页voltages in each passive element into currents.9a. What form does the solution for i(t) take?9b. Substitute your solution into the differential equation. Solve the correspondingcharacteristic equation.相应的特征方程10. Finish determining i(t) by using the results in #6b and #7b to find any constantsyou have defined in your solution for i(t).For problems 1–5, consider the following circuit:Use Vs = 5 v, R1 = 5 , R2 = 10 , and L = 1 H. After a long time has elapsed, theswitch is opened at t = 0. We want to find the voltage v(t), and current i(t), across共45 页第44 页R2 as a function of time.1a. Redraw the circuit, showing how the circuit looks just before the switch isopened.1b. Using this result, find v(0-).2a. What is the continuity/boundary condition for this circuit at t = 0?2b. Find the numerical value of the boundary condition.3. Write a KVL equation for the circuit after the switch as been opened at t = 0.(note: do not use the circuit you drew in #1a). Convert the voltages into currents.4. Solve the differential equation generated in #3 for i(t). Clearly show how youobtained all of the numerical values in i(t).5. Sketch curves of i(t) and v(t) as a function of time (t ). Indicate allcritical and asymptotic渐近的values.共45 页第45 页。