求各支路的电流
is = 3√2∠-60° A，e = 15√2∠-30° = 15√2[ cos(-30°) + jsin(-30°) ] = 15√6/2 - j15√2/2 v；
XL = j( 314 * 0.016) = j5 = 5∠90° ；，XC = -j10^6/( 314 * 637 ) = -j5 = 5∠-90° ；
5、用戴维宁定理计算
(1)、断开电流源支路；
RL并联阻抗 Z1 = RXL/( R+XL) = 2 * j5/( 2 + j5) = j10( 2 - j5 )/( 2 + j5)( 2 - j5) = ( 50 + j20 )/29
= (1/29)√( 50 ^2 + 20^2 )∠arctan(20/50) = 10√29/29∠21.8°
Z2 = Z1 + XC = ( 50 + j20 )/29 - j5 = ( 50 - j125 )/29
= (1/29)√( 50 ^2 + 125^2 )∠arctan(-125/50) = 25√29/29∠-68.2°；
由分压公式，等效电势 e1 = UZ1 = e * Z1/Z2 = e * (10√29/29∠21.8°)/(25√29/29∠-68.2°)
= (15√2∠-30°) * [ (2/5)∠90°) ] = 6√2∠60° = 6√2( cos60° + jsin60° ) = 3√2 + j3√6 v；
等效阻抗 Z = [ XL//XC ]//R2 = [ (j5)//(-j5) ]//R2 = R2 = 2 Ω；
(2)、接入电流源支路；
UZ = isZ = (3√2∠-60°) * 2 = 6√2∠-60° = 6√2[ cos(-60°) + jsin(-60°) ] = 3√2 - j3√6 v；
UR2 = UL = 电流源支路两端电压 = e1 + UZ = 3√2 + j3√6 v + 3√2 - j3√6 v = 6√2 v；
ie = ( UR2 - e )/XC = ( 6√2 - 15√6/2 + j15√2/2 )/(-j5) = -[ 3√2 + j√2( 1.5√3 - 0.6 ) ]
= -√26∠33.7° = -√26sin( 314t + 33.7° ) A；
iL = UL/XL = 6√2/(5∠90°) = 1.2√2∠-90° = 1.2√2sin( 314t - 90° ) A；
iR2 = UR2/R2 = 6√2/2 = 3√2sin(314t) A；
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