For finding open-circuit voltage Voc or Vth across terminals A and B, we must first find current I2 flowing through branch CD. Using Maxwell’s loop current method we have from Fig..
− 2 I1 − 4 (I1 − I2) + 8 = 0 or 3 I1 − 2 I2 = 4
Also − 2 I2 − 2 I2 − 4 − 4 (I2 −I1) = 0
or I1 − 2 I2 = 1
From these two equations, we get I2 = 0.25 A
As we go from point D to C, voltage rise = 4 + 2 × 0.25 = 4.5 V
Hence, VCD = 4.5 or VAB = Vth = 4.5 V.
Also, it may be noted that point A is positive with respect to point B.
In Fig. 2.138 (b), both batteries have been removed.
By applying laws of series and parallel combination of resistances, we get Ri = Rth = 5/4Ω = 1.25Ω.
(i) When RL = 2 Ω ; I = 4.5/(2 + 1.25) = 1.38A
(ii) When RL = 1 Ω ; I = 4.5 (1 + 1.25) = 2.0A