SI a Us1(6=4V R1292 L u 6V R2 8Q +10 解 b L L-=l1+1-le+ ab l531+Ri-l2+R2 I=Mabls t u 2(6-4+10)V 1.2A R1+R2 (2+8)2 两个电压源的吸收功率分别为 ps1=l51i=4V×12A=48W Ps2 l,i=-10V×1.2A=-12Wu u u u u u R i u R i ab = S1 + 1 − S2 + 2 = S1 + 1 − S2 + 2 1.2A (2 8) (6 4 10)V 1 2 ab S1 S2 = +  − + = + − + = R R u u u i 10V 1.2A 12W 4V 1.2A 4.8W S2 S2 S1 S1 = − = −  = − = =  = p u i p u i 两个电压源的吸收功率分别为 解： 6V uS1(t)=4V uS2(t)=10V 2 8