Kvl: u = 2× (I-0.5u)+ 10i, then U=6I.
KVL:3×(I-U/3)+U= 15. Therefore, I=5.
So: uoc = uab =10i =10× 5 = 50 (v).
Short circuit 15V voltage source. Voltage U0 is applied from ports A and B, and the inflow current is I0.
The 3 Ω resistor is connected in parallel with 6 Ω, and the voltage at both ends is equal to U, so their currents are U/3 and U/6 respectively. The current of 2Ω resistor obtained by KCL is: u/3+u/6 = 0.5u.
The resistance current of 10ω is U0/ 10=0. 1U0, KCl: I0+0.5U = 0. 1U0+0.5U ... So: I0=0. 1U0.
The equivalent resistance is: req = rab = u0/i0 =1/0.1=10 (ω).