In the original picture a), R3 can be ignored as an open circuit = ideal voltage source U 1= 10v, and R2 can be ignored as a short circuit = ideal current source is=2A because the current on the branch is constant, and figure b) can be obtained after finishing; Figure b) The left branch U 1 series R 1 can be equivalently converted into an ideal current source i 1=(U 1/R 1), and figure c) can be obtained after conversion;

I 1 =10/=10a, I1and is are added in the same direction = 10+2= 12A, 12A.

The current on (1) R I =12xr1/(r1+r) =12x1/2 = 6a.

(2) The original picture a) is=i+i[R 1], and is and I are known, I [r1] = is-I = 2-6 =-4a; R3 = u1=/terminal voltage of kloc-0/0v, that is, its current I [R3] =10/R3 =10/5 = 2A from top to bottom, I [R3] = I [u1]+.

The terminal voltage of the ideal current source UIS = IR+ISR2 = (6x1)+(2x2) =10v.

3) Us transmission power = US X I [u1] =10x6 = 60w, is transmission power = IS X UIS = 2x10 = 20w;

Power consumption of R 1 =i[R 1]? X R 1= 16w, R2 power consumption =2x2x2=8w, R3 power consumption =2x2x5=20w, and R power consumption = 6x6x61= 36w; Issue = consumption =80w, and the power is balanced.