Page 174 - Basic Electrical Engineering

Figure 2.76

At point E, this I = 2 A gets divided equally; 1 A going in branch EF and 1 A
to branch EC (CD being shorted). At point C, 1 A current will flow through

the short-circuited path provided between terminals A and B. Therefore, I SC
= 1 A.

Now, the resistance of the network viewed from the terminals AB when
the battery is short circuited is

(see also Fig. 2.76)

Norton’s equivalent circuit is shown in Fig. 2.77. Now, using the current
divider rule, the current through the load resistance is calculated as,