570
12 Electric Circuits
∴
10 = i 1 + 30i
or
31i 1 + 30i 2 = 10
(3)
where we have used (1). Solving (2) and (3)
i 1 = −3.04 A, i 2 = 3.48 A
Power dissipated through E 1 is i 2 1 r 1 = (3.04) 2 × 1 = 9.24 W
Power dissipated through E 2 is i 2 2 r 2 = (3.48) 2 × 2 = 24.2 W
Power dissipated through R is i 2 R = (0.44) 2 × 30 = 5.8 W
(∵ i = i 1 + i 2 = −3.04 + 3.48 = 0.44 A)
12.44 Referring to Fig. 12.44
i 1 + i 2 = i
(junction theorem)
(1)
Traversing the loop B E 2 AE 1 B counterclockwise the loop theorem gives
ξ 2 − r 2 i 2 − (ξ 1 − r 1 i 1 ) = 0
∴
1 − 2i 1 − (2 − 1i 1 ) = 0
or
i 1 − 2i 2 = 1
( 2 )
For the loop B E 1 ARB
ξ 1 − i 1 r 1 − i R = 0
∴
2 − 1i 1 − 10i = 0
or
11i 1 + 10i 2 = 2
( 3 )
Fig. 12.44