请教以下电工题目

更新时间：2021-10-13 09:59:32 92次访问

LC并联阻抗  X = XL * Xc/( XL + Xc )

= j10 * (-j20 )/( j10 - j20 ) = - j10 * j20/(-j10) = j20 Ω = 20∠90°；

总阻抗 Z = 10 + j20 = 10√( 1^2 + 2^2 ) ∠arctan(2/1) = 10√5∠63.435° Ω；

总电流 i = u/Z = 20∠0°/(10√5∠63.435°) = [ 20/(10√5) ] ∠( 0° - 63.435° ) = 2/√5∠-63.435° A；

Uab = iX = 2/√5∠-63.435° * 20∠90° = 8√5∠26.565° v；

Ubc = iR = 10 * 2/√5∠-63.435° = 4√5∠-63.435° v；