Задача по электротехнике, дорешайте пожалуйста

Неправильно: R234=R23*R4/(R23+R4)=23*2/25=1.84 Ом. R1234=1.84*1/2.84=0.65 Ом

R23 = 23;
1/R1234 = 1/1 + 1/23 + 1/2 = 1 + 0.043 + 0.5 = 1.543; R1234 = 0.648(Om);
Rобщ = 0.648 + 20 = 20.648(Om);
Iобщ = 12/20.648 = 0.581(А); I5 = Iобщ = 0.581(А);
U5 = 20 * 0.581 = 11.623(V);
U12345 = 12 - 11.623 = 0.377(V);

I1 = 0.377(A);
I23 = 0.377/23 = 0.016(A);
U2 = 8 * 0.016 = 0.13(V);
U3 = 15 * 0.016 = 0.24(V);
I4 = 0.377/2 = 0.188(A);

P1 = 0.377 * 0.377 = 0.142(W);
P2 = 0.016 * 0.13 = 0.002(W);
P3 = 0.016 * 0.24 = 0.004(W);
P4 = 0.377 * 0.188 = 0.071(W);
P5 = 0.581 * 11.623 = 6.753(W);

0.142+0.002+0.004+0.071+6.753 = 6.972(W) - контроль.
Pобщ = 0.581 * 12 = 6.972;

Порнография какая-то, вообще-то.

Решение: