[电路分析基础]第2版-习题参考答案-2014-tjh
《电路分析基础》各章习题参考答案
第1章 习题参考答案
1-1 (1) 50W;(2) 300 V、25V ,200V 、75 V; (3) R 2=12.5Ω,R 3=100Ω,R 4=37.5Ω 1-2 V A =8.5V,V m =6.5V,V B =0.5V,V C =−12V ,V D =−19V ,V p =−21.5V ,U AB =8V,U BC =12.5,
U DA =−27.5V
1-3 电源(产生功率):A 、B 元件;负载(吸收功率):C 、D 元件;电路满足功率平衡
条件。
1-4 (1) V A =100V,V B =99V,V C =97V,V D =7V,V E =5V,V F =1V,U AF =99V,U CE =92V,
U BE =94V,U BF =98V,U CA =−3 V;(2) V C =90V,V B =92V,V A =93V,V E =−2V ,V F =−6V ,V G =−7V ,U AF =99V,U CE =92V,U BE =94V,U BF =98V,U CA =−3 V
1-5 I ≈0.18A , 6度,2.7元
1-6 I =4A, I 1=11A,I 2=19A
1-7 (a) U =6V,(b) U =24 V,(c) R =5Ω,(d) I =23.5A
1-8 (1) i 6=−1A ; (2) u 4=10V,u 6=3 V; (3) P 1 =−2W 发出,P 2 =6W吸收,P 3 =16W吸收,
P 4 =−10W 发出,P 5 =−7W 发出, P 6 =−3W 发出
1-9 I =1A,U S =134V,R ≈7.8Ω
1-10 S 断开:U AB =−4.8V ,U AO =−12V ,U BO =−7.2V ;S 闭合:U AB =−12V ,U AO =−12V ,U BO =0V 1-11 支路3,节点2,网孔2,回路3
1-12 节点电流方程:(A) I1 +I3−I 6=0,(B)I 6−I 5−I 7=0,(C)I 5 +I4−I 3=0
回路电压方程:①I 6 R6+ US5 +I5 R5−U S3 +I3 R3=0,②−I 5 R5−U S5+ I7R 7−U S4 =0,③−I 3 R3+ U S3 + US4 + I1 R2+ I1 R1 =0
1-13 U AB =11V, I 2=0.5A,I 3=4.5A, R 3≈2.4Ω
1-14 V A =60V,V C =140V,V D =90V,U AC =−80V ,U AD =−30V ,U CD =50V
1-15 I 1=−2A ,I 2=3A,I 3=−5A ,I 4=7A,I 5=2A
第2章 习题参考答案
2-1 2.4 Ω,5 A
2-2 (1) 4 V,2 V,1 V;(2) 40 mA,20 mA,10 mA
2-3 1.5 Ω,2 A,1/3 A
2-4 6 Ω,36 Ω
2-5 2 A,1 A
2-6 1 A
2-7 2 A
2-8 1 A
2-9 I 1 = −1.4 A,I 2 = 1.6 A,I 3 = 0.2 A
2-10 I 1 = 0 A,I 2 = −3 A,P 1 = 0 W,P 2 = −18 W
2-11 I 1 = −1 mA,I 2 = −2 mA,E 3 = 10 V
2-12 I 1 = 6 A,I 2 = −3 A,I 3 = 3 A
2-13 I 1 =2 A,I 2 = 1A,I 3 = 1 A,I 4 =2 A,I 5 = 1 A
2-14 V a = 12 V ,I 1 = −1 A,I 2 = 2 A
2-15 V a = 6 V,I 1 = 1.5 A,I 2 = −1 A,I 3 = 0.5 A
2-16 V a = 15 V,I 1 = −1 A,I 2 = 2 A,I 3 = 3 A
2-17 I 1 = −1 A,I 2 = 2 A
2-18 I 1 = 1.5 A,I 2 = −1 A,I 3 = 0.5 A
2-19 I 1 = 0.8 A,I 2 = −0.75 A,I 3 = 2 A,I 4 = −2.75 A,I 5 = 1.55 A 2-20 I 3 = 0.5 A
2-21 U 0 = 2 V,R 0 = 4 Ω,I 0 = 0.1 A
2-22 I 5 = −1 A
2-23 (1) I 5 = 0 A,U ab = 0 V;(2) I 5 = 1 A,U ab = 11 V
2-24 I L = 2 A
2-25 I S =11 A,R 0 = 2 Ω
2-26 18 Ω,−2 Ω,12 Ω
2-27 U =5 V
2-28 I =1 A
2-29 U =5 V
2-30 I =1 A
2-31 10 V,180 Ω
2-32 U 0 = 9 V,R 0 = 6 Ω,U =15 V
第3章 习题参考答案
3-1 50Hz ,314rad/s,0.02s ,141V ,100V ,120°
3-2 200V ,141.4V
3-3 u =14.1sin (314t −60°) V
3-4 (1) ψu 1−ψu =120°;(2) ψ
1=−90°,ψ2=−210°,ψu 1−
ψu 2=120°(不变) 3-5 (1)
U 1=90︒V ,U 2=∠0︒V ;
(2) u 3ωt +45°)V ,u 4ωt +135°)V
3-6 (1) i 1=14.1 sin (ωt+72°)A ;(2) u 2=300 sin (ωt-60°)V 3-7 错误:(1) ,(3),(4),(5)
3-8 (1) R ;(2) L ;(3) C ;(4) R
3-9 i =2.82 sin (10t −30°) A,Q ≈40 var
3-10 u =44.9sin (314t −135°) V,Q =3.18 var
3-11 (1) I =20A;(2) P =4.4kW
3-12 (1)I ≈1.4A ,I ≈1.4∠-30︒A ;(3)Q ≈308 var,P =0W;(4) i ≈0.98 sin (628t −30°) A 3-13 (1)I =9.67A,I =9.67∠150︒A ,i =13.7 sin (314t +150°) A ;(3)Q =2127.4 var ,P =0W;
(4)I C =0A
3-14 (1)C =20.3μF ;(2) I L =0.25A ,I C =16A
第4章 习题参考答案
4-1 (a) Z =5∠36.87︒Ω,Y =0.2∠-36.87︒S ;
(b) Z =45︒Ω
,Y =45︒S 4-2 Y =(0.06-j0.08) S,R ≈16.67 Ω,X L =12.5 Ω,L ≈0.04 H 4-3 U R =60∠0︒V ,U L =80∠90︒V ,U S =100∠53.13︒V 4-4 I =20∠-36.87︒A
4-5
Z =45︒Ω,
I =1∠0︒A ,U R =100∠
0︒V ,U L =125∠90︒
V ,U C =25∠-90︒V 4-6 Y =45︒S ,U =0︒V ,I R 0︒A ,I L =-90︒A ,I C =90︒A 4-7
I =14︒5,A U S =100∠90︒V
4-8 (a) 30 V;(b) 2.24 A
4-9 (a) 10 V;(b) 10 A
4-10 (a) 10 V;(b) 10 V
4-11 U =14.1 V 4-12 U L1 =15 V,U C2 =8 V,U S =15.65 V
4-13 U
X1 =100 V,U 2 =600 V,X 1=10 Ω,
X 2=20 Ω,X 3=30 Ω 4-14
Z =45︒Ω,I =2∠-45︒A ,I 1=0︒A ,I 2=90︒A ,U ab =0V 4-15 (1)I =A ,Z RC =Ω,Z =Ω;(2)R =10Ω,X C =10Ω 4-16 P = 774.4 W,Q = 580.8 var,S = 968 V·A
4-17 I 1 = 5 A,I 2 = 4 A
4-18 I 1 = 1 A,I 2 = 2 A,I 26.565︒A ,=44.72∠-26.565︒V ⋅A 4-19 Z =10Ω,I =1∠90︒A ,U R2=135︒V ,P =10W 4-20 ω0 =5×106 rad/s,ρ = 1000 Ω,Q = 100,I = 2 mA,U R =20 mV,U L = U C = 2 V 4-21 ω0 =104 rad/s,ρ = 100 Ω,Q = 100,U = 10 V,I R = 1 mA,I L = I C = 100 mA 4-22 L 1 = 1 H,L 2 ≈ 0.33 H
第5章 习题参考答案
5-3 M = 35.5 mH
5-4 ω01 =1000 rad/s,ω02 =2236 rad/s
5-5 Z 1 = j31.4 Ω,Z 2 = j6.28 Ω
5-6 Z r = 3+7.5 Ω
5-7 M = 130 mH
5-8 I 245︒A
5-9 U 1 = 44.8 V
5-10 M 12 = 20 mH,I 1 = 4 A
5-11 U 2 = 220 V,I 1 = 4 A
5-12 n = 1.9
5-13 N 2 = 254匝,N 3 = 72匝
5-14 n = 10,P 2 = 31.25 mW
第6章 习题参考答案
6-1 (1) A相灯泡电压为零,B 、C 相各位为220V
6-3 I L = I p = 4.4 A,U p = 220 V,U L = 380 V,P = 2.3 kW 6-4 (2) I p = 7.62 A,I L = 13.2 A
6-5 A 、C 相各为2.2A ,B 相为3.8A
6-6 U L = 404 V
6-7 U A 'N '=202∠-47︒V
6-8 cos φ = 0.961,Q = 5.75 kvar
6-9 Z =33.4∠28.4︒Ω
6-10 (1) I p = 11.26 A,Z = 19.53∠42.3° Ω; (2) I p = I l = 11.26 A,P = 5.5 kW 6-11 U l = 391 V
6-12
i A =ωt -53.13︒
) A
i B =ωt -173.13︒
) A
i C =ωt +66.87︒) A
6-13 U V = 160 V
6-14 (1)
负载以三角形方式接入三相电源
(2) I AB
=
15︒A ,I BC =
-135︒A ,I CA =105︒A
I A =-45︒A ,I B =165︒A ,I C =75︒A
6-15 L = 110 mH,C = 91.9 mF
第7章 习题参考答案
7-1 P = 240 W,Q = 360 var
7-2 P = 10.84 W
7-3 (1)i (t ) =4.7sin(ωt
+100︒) +3sin3ωt A
(2) I ≈3.94 A ,U ≈58.84 V,P ≈93.02 W
7-4 u 2(t ) =ωMU m
z sin(ωt +ωL π-arctan 1)V ,z =2R
7-5 直流电源中有交流,交流电源中无直流
7-6 U 1=54.3 V,R = 1 Ω,L = 11.4 mH;约为8%,(L ’ = 12.33 mH) 7-7 使总阻抗或总导纳为实数(虚部为0)的条件为R 1=R 2=R X 7-8 C 1=9.39μF ,C 2=75.13μF
7-9 L 1 = 1 H,L 2 = 66.7 mH
7-10 C 1 = 10 μF,C 2 = 1.25 μF
第8章 习题参考答案
8-6 i L (0+)=1.5mA ,u L (0+)=−15V
8-7 i 1(0+)=4A ,i 2(0+)=1A ,u L (0+)=2V ,i 1(∞) =3A ,i 2(∞) =0,u L (∞) =0 8-8 i 1(0+)=75mA ,i 2(0+)=75mA ,i 3(0+)=0,u L1(0+)=0,u L2(0+)=2.25V
8-9 i C (t ) =2e -1⨯10t A
8-10 u L (t ) =6e -4t V
8-11 u C (t ) =10(1-e -1⨯10t )V ,i C (t ) =5e -1⨯10t A
*8-12 u C (t ) =115e -500t sin(866+60︒) V 8-13 u L (t ) =12e -10t V ,i L (t ) =2(1-e -10t )A 8-14 u R (t ) =-U S e -1t R 2C 666V ,u R (3τ) =-U S e -3V 8-15 (1) τ=0.1s ,(2) u C (t ) =10e -10t V ,(3) t =0.1s 8-16 u C (t ) =10-9e -10t V
8-17 i L (t ) =5e -10t A
8-18 (a)f (t ) =1(t -t 0) -1(t -2t 0) ;
(b)f (t ) =1(t ) -1(t -t 0) -[1(t -t 0) -1(t -2t 0)]=1(t ) -2⨯1(t -t 0) +1(t -2t 0) 8-19 u C (t ) =[5(1-e -0.5t )1(t ) -5(1-e -0.5(t -1) )1(t -1)]V 8-20 u o 为三角波,峰值为±0.05V *8-21 临界阻尼R
-6t 5,欠阻尼R
,过阻尼R
-t -1
6*8-22 i L (t ) =[(1-e )1(t ) +(1-e
)1(t -1) -2(1-e -t -26)1(t -2)]