ABSTRACT
R in parallel with r2 gives an equivalent resistance of
(5 × 2)/(5 + 2) = 10/ 7 = 1.429 as shown i n t he equival ent net work of Figure 32.3. From Fi gure 32.3,
current I1 = E1 (r1 + 1.429) =
8 2.429
= 3.294 A
From Fi gure 32.2,
current I2 = (
r2 R + r2
) (I1) =
( 2
5 + 2 )
(3.294)
= 0.941 A
and current I3 = ( 5
5 + 2 )
(3.294) = 2.353 A
(iii) Redraw the original network with source E1 removed and replaced by r1 only, as shown in Fi gure 32.4.
