例:求电路中u1(t)、i(t),并求各电源的发出功率。其中 (t)=5c0s(10)A l()=10cos(51-90°)7 190.1H 解:1、电流源单独作用: (1) (1) 5∠45 0 R 1Q 0.2F )_;(1) =-s -5cos 1Ot P=12.5W P.=50W 40=52c0s(10+45°)=1 2、电压源单独作用:3、时域叠力=52010+459 (2) =11+ (2) 0 =0 (2) i=i0+i2)=10√2cos(5t-45°)A 12)==10∠-45° l=10+ (2) ()?=A =10√2cos(5t-45 =5√2cos(10t+45°)+10cos(5t-909 i=i)+ l2)=ly=10cos(5t-909) -5cos10t+10√2cos(5t-45°)A例：求电路中u1(t)、i1(t)，并求各电源的发出功率。其中 u (t) 10cos(5t 90 )V. s = −  i (t) 5cos(10t)A. s = + u - i 解：1、电流源单独作用： =   • 5 45 (1) U1 0 (1) i 1 = i i t s 5cos10 (1) (1) = − = − (1) (1) 1 u = 5 2 cos(10t + 45) = u 2、电压源单独作用： 0 (2) 1 = • U = = −  • • 10 45 (2) (2) I 1 I 0 (2) u1 = 10cos(5 90 ) (2) u = u = t −  s 10 2 cos(5 45 ) (2) 1 (2) i = i = t −  3、时域叠加： u u u 5 2 cos(10t 45 )V (2) 1 (1) 1 1 = + = +  i i i 10 2 cos(5t 45 )A (2) 1 (1) 1 1 = + = −  (1) (2) u = u + u = 5 2 cos(10t + 45) +10cos(5t −90)V (1) (2) i = i + i = −5cos10t +10 2 cos(5t − 45)A Pi =12.5W Pu = 50W