Question

For the circuit shown in Fig.calculate the current in the 6 Ω resistance by using Norton’s theorem.

Answer 1

As explained in , we will replace the 6Ω resistance by a short-circuit as shown in Fig.. Now, we have to find the current passing through the short-circuited terminals A and B. For this purpose we will use the mesh analysis by assuming mesh currents I₁ and I₂.

From mesh (i), we get 

3 − 4 I₁ − 4 (I₁ − I₂) + 5 = 0 or 2 I₁ − I₂ = 2 ...(i) 

From mesh (ii), we get 

− 2 I₂ − 4 − 5 − 4 (I₂ − I₁) = 0 or 4I₁ − 6I₂ = 9 ...(ii) 

From (i) and (ii) above, we get I₂ = − 5/4 

The negative sign shows that the actual direction of flow of I₂ is opposite to that shown in Fig. . Hence, I_sh = I_N = I₂ = − 5/4 A i.e. current flows from point B to A. 

After the terminals A and B are open-circuited and the three batteries are replaced by short circuits (since their internal resistances are zero), the internal resistance of the circuit, as viewed from these terminals’ is 

R_i = R_N = 2 + 4 || 4 = 4 Ω 

The Norton’s equivalent circuit consists of a constant current source of 5/4 A in parallel with a resistance of 4 Ω as shown in Fig. . When 6 Ω resistance is connected across the equivalent circuit, current through it can be found by the current-divider rule (Art). 

Current through 6 Ω resistor = 5/4 x 4/10 = 0.5 from B to A.