哈工大电路答案第14章(2)

?s?1.5?0.25? Y?????0.750.3125?

答案14.4

解 (a)：按网孔列写KVL方程得

(1)?(R?2R)I1?2RI2?U1 ?2RI?(R?2R)I?U?3U (2)221?1将式(1)代入式(2)整理得

?U1?3RI1?2RI2 ?U??7RI?3RI12?2所以

2R??3R Z?????7R?3R?(b) 消去互感后电路如图14.4(e)所示，列KVL方程如下

I1(s)I2(s)1/sC?U2(s)s(L2?M)?sM??s(L1?M)RU1(s)图14.4(e)

?U1(s)??R?s?L1?L2?2M??I1(s)?s?L2?M?I2(s)? ?U(s)?s(L?M)I(s)?(sL?1)I(s)22122?sC?所以

?R?s(L!?L2?2M)s(L2?M)?Z(s)??? s(L?M)sL?1/sC22??(c) 将?联接的三个阻抗转换成Y形联接，如图14.4(f)所示，由此电路可直

接写出Z参数

1??j?j?图14.4(f)

?1?jj?? Z????j0?(d)

?u1?Rbi1??u2 (1) ??u2?RCiC?RC(i2??i1) (2)将式(2)代入式(1) 整理得

?u1??Rb???RC?i1??RCi2 ?u???Ri?RiC1C2?2所以

?R???RCZ??b???RC

?RC? ?RC?答案14.5

解 (a)：列写节点电压方程

1?11(?)U?(1)?510110U2?I1 ?111??U1?(?)U2?I2 (2)510?10由式 (2)得

U1?3U2?10(?I2) 代入式(1)整理得

(3)

I1?0.8U2?3(?I2) 由式(3)和(4)得

(4)

?310??A??? 0.8S3??(b):根据A参数定义可得

A11?u1u2i1u2?i2?0u1u1??5u3?u40.4u1?0.6u1i1i1???0.1S

u3?u440?0.5i1?60?0.5i1?u1??240Ω0.5u1/60?0.5u1/40i1??5

0.4i1?0.6i1

A21??i2?0A12?u1?i2i1?i2

u2?0A22??u2?0所以

??5?240?? A????5???0.1S(c) 按网孔列写KVL方程得

???1?j2?I??jI? ?U (1)112 ???(2)?U2?jI1???j3?j1?I2 由式(2)得

???jU??2(?I?) (3) I122代入式(1)化简得：

??(2?j)U??(?2?j5)(?I?) (4) U122由式(4)和(3)得：

?2?j?(2?j5)??A??? ?jS?2??(d)解：根据基尔霍夫定律和理想变压器方程得

??R1i1?nu2 (1) u1?R1i1?u1?/R2?i1??nu2/R2?(?i2)/n (2) i1?u1将式(2) 代入式(1)整理得

u1?(1?R1/R2)nu2?(R1/n)(?i2) (3) 由式(3)和(1)得

??1?R1R2?nA???nR2R1n? ?1n?答案14.6

解：当开关断开时

U?U1?1A ,I2?0 , U2?3V U1?5V，I1?34Ω由传输参数方程得：

?5?A11?3 ? ?1?A?321?当开关接通时

U1?4V,I1??A11?53 ?A?13?21?I2?U21?A 6Ω3U3?U1?1A，U2?2V,4Ω由参数方程又得

51?4??2??A12??A12?233 ?? ?11A?1?22?1??2??A223?3所以

A???5/32???1/3S1??

答案14.7

解：(a) 列写网孔电流方程如下：

??U1?40?I1?I2?(1)?U2?40I

1?100I2 (2)由式(2)得

I2??0.4I1?0.01U2?H21I1?H22U2代入式(1)整理得

U1?24I1?0.4U2?H11I1?H12U2 由式(3)和(4)得

H???24?0.4???0.40.01S?? (b)解：列写节点电压方程如下 ??(1?1)?U?1?2U2?I1?RR1R ???1R?2U12?RU2?I2整理得

??U1?0.5RI11?U2???I2?? RU2所以

H???0.5R1??0?1/R?? (c) 根据KVL和理想变压器方程得

??u1?R1i1?nu?2 (1)?i2??ni1 (2) ??u?2?u2?R2i2 (3)将式(3)及式(2)代入式(1)整理得

??u1??R1?n2R2?i1?nu2?i2??ni 1所以

(3) (4) ?R1?n2R2H????nn?? 0?(d) 列写回路电流方程如下：

?u1?15i1?10i2 ???u?10i?10i?4i?10i?10i?4i?i?6i?6i12x121212?2整理得

10?u?5i?u21?16

?1?i2??i1?u26?所以

?5?5/3? H?????11/6S?

答案14.8

解：(1)由阻抗参数方程得

(1)?u1?4i1?3i2 ?(2)?u2?3i1?5i2 由式(2)得 代入式(1)得

(3)

i2??0.6i1?0.2u2

u1?4i1?1.8i1?0.6u2?2.2i1?0.6u2 (4)

由式(3)和(4)得

?2.2?0.6? H?????0.60.2S?(2) 若i1?10A，u2?20V，由式(3)和(4)解得

u1?(2.2?10?0.6?20)V?34V i2?(?0.6?10?0.2?20)A??2A 功率

p?u1i1?u2i2?34?10?20?(?2)?300W

注释：二端口消耗的功率等于两个端口消耗功率之和。

答案14.9

解：(a)中阻抗矩阵为对称矩阵，且矩阵中元素均为实数，故可用由电阻组成的T形电路来等效。如图(a)所示。其中