急电工技术基础:3-13

2025-04-28 13:19:33
推荐回答(1个)
回答1:

U`=I2∠-30ºZ2∠acrtg2/3=4∠-30ºx√(3^2+2^2)∠33.7º=4x3.6=14.4∠3.7ºV
Iz3=U`/2=7.2∠3.7º
I=I2+Iz3=4[cos(-30º)+jsin(-30º)]+7.2[cos(3.7º)+jsin(3.7º)]
=3.464-j2+7.185+j0.465
=10.65-j1.535
=10.76∠-8.2ºA
Z`=U`/I=14.4∠3.7º/10.76∠-8.2º=1.34∠11.9º=1.31-j0.28
电路总阻抗Z=U/I=220∠0º/10.76∠-8.2º=20.45∠8.2º=20.24+j2.92
Z=Z1+Z`
Z1=Z-Z`=20.24+j2.92-1.31+j0.28=18.93+j3.2=19.2∠9.6º
P=UIcosφ=220x10.76cos(-8.2º)=2340.8W