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in Physics by (63.5k points)

Use Thevenin’s theorem to find the current in a resistance load connected between the terminals A and B of the network shown in Fig. if the load is (a) 2 Ω (b) 1 Ω

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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 I= 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 R= 2 Ω ; I = 4.5/(2 + 1.25) = 1.38A 

(ii) When RL = 1 Ω ; I = 4.5 (1 + 1.25) = 2.0A

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