Question

Obtain current I_o in Fig. below using Norton’s theorem.

Answer 1

Our first objective is to find the Norton equivalent at terminals a-b. Z_Nis found in the same way as Z_Th. We set the sources to zero as shownin Fig. (a). As evident from the figure, the (8 − j2) and (10 + j4)impedances are short-circuited, so that

Z_N= 5Ω To get IN, we short-circuit terminals a-b as in Fig. (b) and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,

-j40 + (18 +j2) I₁ - (8 – j2) I₂ - (10 + j4) I₃ =0 …(1)

For the supermesh,

(13- j2)I₂ + (10 + j4) I₃ – (18 + j2) I₁ = 0 …(2)

At node a, due to the current source between meshes 2 and 3,

I₃= I₂+ 3 …(3)

Adding Eqs. (1) and (2) gives

−j40 + 5I₂= 0 ⇒I₂= j8

From Eq.(3),

I₃= I₂+ 3 = 3 + j8

The Norton current is

I_N= I₃= (3 + j8) A

Figure (c) shows the Norton equivalent circuit along with the impedanceat terminals a-b. By current division