(4) 109+254302V 1i2+o=501/s,1=2.5∠40A 25220m Find us(t). j500×20×10×2.5∠40° =1∠130°4 25 1=12+IL=1∠1300+25∠409=-0.643+j0.767+1915+607 1.272+j2374=2.69∠618° =10/1+25∠-30°+500×20×10-°Ir=35.34∠58.5° U、(t)==35342cos(500t+58.5°)ALI • ( ). 500 1/ , 2.5 40 . Find t If s I A s L   = =   • • 1 I • 2 I 25 500 20 10 2.5 40 3 2      = • − j I = + =   +   • • • I 1 I 2 I L 1 130 2.5 40 • − • • Vs = I +  −  + j   I L 3 10 1 25 30 500 20 10 = 35.3458.5  s (t) = = 35.34 2 cos(500t + 58.5) A = 1130 A = 1.272 + j2.374 = 2.6961.8 = −0.643 + j0.767 + 1.915 + j1.607 (4)