131
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
及解析
答案
第一章 电路模型和电路定律
【题1】:由
V可得:
A:
:
V。
【题2】:D。
【题3】:300;-100。
【题4】:D。
【题5】:
EMBED Equation.2 ;
EMBED Equation.2 ;
EMBED Equation.2 ;
EMBED Equation.2 。
【题6】:3;-5;-8。
【题7】:D。
【题8】:
EMBED Equation.2
EMBED Equation.2 ;
;
;
;
;
。
【题9】:C。
【题10】:3;-3。
【题11】:-5;-13。
【题12】:4(吸收);25。
【题13】:0.4。
【题14】:
;
。
【题15】:
A;
A;
A;
A。
【题16】:
A;
V;X元件吸收的功率为
W。
【题17】:由图可得
V;流过
电阻的电流
A;由回路ADEBCA列KVL得
;又由节点D列KCL得
;由回路CDEC列KVL解得;
;代入上
式,得
V。
【题18】:
;故
;
;
⑴ KCL:
;
A;
V或
V;或
。
⑵ KCL:
;
A;
V。
第二章 电阻电路的等效变换
【题1】:[解答]
;
V;
A;
;吸收功率7.5W。
【题2】:[解答]
【题3】:[解答] C。
【题4】:[解答] 等效电路如图所示,
A。
【题5】:[解答] 等效电路如图所示,IL=0.5A。
【题6】:[解答]
【题7】:[解答]
I=0.6A;U1=-2A=-12V;U2=2I+2=32V
【题8】:[解答]由图可得U=4I-4。
【题9】:[解答]
⑴
V
4
⑵1 V电压源的功率为
W (吸收功率)
7
⑶1 A电流源的功率为
W (供出功率)
10
【题10】:[解答]A
第三章 电阻电路的一般分析方法
【题1】:
【题2】:
;
;
;
;
解得:
=-1.5 A,
=-0.5 A,
=1.5 A,
=-3.5 A。
【题3】:[解答]
;
A
【题4】:[解答]
;
A;
W
【题5】:[解答]答案不唯一,有多解。
【题6】:[解答]
设4A电流源两端电压为
,各网孔电流为
、
、
,参考方向如图所示
【题7】:[解答]
;解得
A;
V。
【题8】:[解答]
去掉
支路,设网孔电流如图所示
;解得
;
。
【题9】:[解答]
设15 A电流源两端电压为
EMBED Equation.2 ;解得
A;
V。
【题10】:[解答]
选节点
为参考点
;解得
。
【题11】:[解答]
【题12】:[解答]
-6.5W;供出功率。
【题13】:[解答]
用节点法
;
;令
;解得
。
【题14】:[解答]
⑴电路如图:
⑵解法一:将第2方程乘以2再减第3方程,可得
,即
与
公共支路电流为零。⑵解法二:电路为一平衡电桥,0.1 S公共支路中电流为零。
【题15】:该电路的一种可能的结构形式:(本答案不唯一,可有多解)
【题16】:[解答]
;
;
V;
V
【题17】:[解答]
选电压源负端为参考点,设其余各节点电压自左至右分别为
、
、
、
;解得
。
【题18】:[解答]
选电压源负端为参考点:
解得
V;
V
【题19】:[解答]
选节点
为参考点:
化简得
;解得
;故
V;
V。
第四章 电路定律
【题1】:用叠加定理求
;
=3V;
V;
=2V;
1(等效电路为:
【题2】:
;
;
【题3】:B
【题4】:D
【题5】:A
【题6】:B
【题7】:D
【题8】:
【题9】:
【题10】:
设
则
V,
V
V,
100 V
900 V,
1000 V
9000 V,
10000 V
可知此电路输出电压逐级衰减10倍
当
50 V时,则
5 V
0.5 V
0.05 V
0.005 V
【题11】:C
【题12】:4.5、2。
【题13】:
EMBED Equation.3
EMBED Equation.3 时能获得最大功率
第八章 相量法
【题1】:电流
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
A2读数为10A,电压表V2读数为100
V
【题2】:
,
=1.39S,L=0.543H
【题3】:L=1.2H
【题4】:I=1A
【题5】:
’
【题6】:
,上述关系也可从相量图得出
【题7】:
=
,
=
,
=
;
相量图:
【题8】:R=86.603Ω,L=0.159H,C=31.831
第九章 一般正弦稳态电路的分析
【题1】:C=64
【题2】:
,
,
【题3】:⑵、⑸正确,⑴、⑶、⑷不正确
【题4】:当
时,电流
值最大,即
,此时频率
]
【题5】:52.35∠-43.45oV
【题6】:
=
=7.07∠-8.2oA,
∠-161.6oVA,
EMBED Equation.3 ∠-108.4oVA,
∠-26.5oVA,]
【题7】:
【题8】:
【题9】:⑴PL=250W,⑵PL=310W,⑶PL=500W
【题10】: 当
时可获最大功率,且
【题11】: r=3.47Ω,C=53.2μF
【题12】: (1)25∠53.1oΩ (2) 25∠53.1oVA (3) 10
V
【题13】:
(t)=2.03 cos(t-82.96o)V
【题14】: r=1000Ω,
=j125V
【题15】: L=109.7 mH,r=6.27Ω
【题16】:
,
,(b)图为戴维南等效电路
【题17】:
=7.07∠-8.13oA
【题18】: 71
【题19】:
(产生)=600W,
(产生)=0Var;
(产生)=-100W,
(产生)=500Var
【题20】:
=2W,
=2Var,
=2
VA,
=2+j2VA
【题21】: L=0.02H,R=1Ω,Q=50
【题22】: 4.124A
【题23】:
【题24】: 电压表读数为200V,电流表读数为10
A
第十章 耦合电感和变压器电路分析
【题1】:
,(b)图为戴维南等效电路
【题2】: 0.64H
【题3】: 电压
60∠180oV,等效阻抗Zab=j9Ω,(b)图为戴维南等效电路
【题4】:
=0.354∠8.13oV
【题5】:
=
=1.77∠-25.64o(A);
=
=-1.77 ∠-25.64o(A)
;
=
-
=3.54∠-25.64o(A)
【题6】:
=0
【题7】: n=2,
=5∠0o(mA) ,
=10∠0o(mA)
【题8】: L1=L2=183.75 mH,M=130.5 mH
【题9】:
【题10】: 设ω=100rad/s)[Z12= j1(Ω),Le =10(mH)]
【题11】: L1 [R1+jω(L1+L2-2M12)
+ jω(M12-M13+M23-L2)
=
]
L2 jω(M12-M13+M23-L2)
+[ jω(L2+L3-2 M23 )-
]
=0 }
【题12】:1.59∠-12.720 A
第十一章 三相电路
【题1】:220 220 190
【题2】:15 0 15
【题3】:D
【题4】
A
A
电流表A1读数为43.8A
//
A 电流表A读数为70.4A
【题5】:300V
【题6】:
对负载Z1
A
则相电流
A
Z1负载端电压
V
对星接负载Z2 线电压
相电压
EMBED Equation.2 A
=480W
【题7】:
相量图如下:
得
A
A
IC=2A
【题8】:D
【题9】:C
第十二章 非正弦周期电流电路
【题1】:串联电路应谐振于
故 L =
;
并联电路应谐振于
故
。
【题2】:
作用
A
作用
A
A
A P=(12+22)
W
【题3】:
作用时 Z=1
EMBED Equation.2
A
A
作用时
A
A
A
【题4】:
V
L、C对二次谐波谐振
方程
得
V
V
V 10
【题5】:A
【题6】:B
【题7】:D
【题8】:10 1
【题9】:C
【题10】:A
电路Ⅱ
第六章 一阶电路
题1:(t=0+时刻的等效电路)2.5A;1.5V
题2:(t=0+时刻的等效电路)25
;10
题3:0;2 A;0;2 A
题4:2.5 A; 7.5 V; 1.25 A
题5:(c)
题.6:
A;0。
题7:(c)
题8:(b)
题9:(
)
题10:(b)
题11:
题12:30; 1.5; 50; 48。
题13:
题14:
;
;
EMBED Equation.2 ;
,
;
;
EMBED Equation.2 ;
EMBED Equation.2
。
题15:5; 40; 0.5; 20。
题16:
题17:
;
;
。
题18:
题19:(c)
题20:
;
;
;
;
;
;
EMBED Equation.2 ;
题21:
时;
;
;
;
;得
,
;
,
;
时;
;
;
;
得
,
;或
,
;
题22:
;
;
;得
,
题23:
;
;
;
;
;
;得
,
;
,
。
题24:
第十三章 拉普拉斯变换答案
【题1】:
;
【题2】:c
【题3】:d
【题4】:d
【题5】:c
【题6】:A提示:可用比较系数法
,
【题7】:
EMBED Equation.2
【题8】:c
【题9】:d
【题10】:
【题11】:作
域模型,选用节点法,设节点电压
(电容电压),和节点电压
(受控源两端电压),可得:
;
;解得
;
【题12】:
;
;复频域模型如图
节点方程
得
EMBED Equation.2
,
【题13】:
EMBED Equation.2
EMBED Equation.2
【题14】:
【题15】:
EMBED Equation.2
EMBED Equation.2
或
【题16】:
【题17】:
【题18】:
作
域模型,选用网孔法
解得:
【题19】:
复频域模型如图
节点方程:
得
EMBED Equation.2
,
第十五章 电路方程的矩阵形式答案
题1
(画错一条(包括方向错误)扣2分,错4条以上则无分)
题2:(C)
题3:(D)
题4:(C)
题5:(C)
题6:(A)
题7:
EMBED PBrush
题8:
题9:
题10:
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
题11:
题12:
题13:
EMBED Equation.2
EMBED Equation.2
题14:
第十六章 二端口网络答案
3、典型习题
【题1】:(B)
【题2】:(B)
【题3】:(A)
【题4】:输出端开路时的转移阻抗; 输入端开路时的输出阻抗。
【题5】:
【题6】:
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
EMBED Equation.2
【题7】:
,得
EMBED Equation.2
,得
EMBED Equation.2
,得
EMBED Equation.2 ;
,得
EMBED Equation.2
【题8】:(B)
【题9】:(B)
【题10】:
【题11】:
【题12】:
【题13】:(D)
【题14】:(A)
【题15】:
;h11=
= 4
;h12=
=
;h21=
=1 ;
h22=
=
S
【题16】:S断开时 5
10(3h11(250h12=0。005
100 5
10(3h21(250h22=0;
S闭合时 5
10(3h11(125h12=0。005
100 5
10(3h21(125h22=
;
解得 [H]=
【题17】: (B)
【题18】: (C)
【题19】:由U1、I1、U2、I2的参考方向;
;
;
;
S;
【题20】:(C)
【题21】:(C)
【题22】:
;
解得
电源所提供的即网络N消耗的功率为
=
【题23】:1.断开R,置电压源为零值
由Y参数方程 I2
U2;可求得 Rab
2.开路电压Uab由下图求得
7
由Y参数方程:I2
可得 Uab=U2
V,则 Pmax
W
【题24】:
(设
参考方向指向2)
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
【题25】:(C)
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