12.3 Solutions
571
where we have used (1). Application of Kirchhoff’s laws to the loop
B E 2 ARB does not yield anything extra. From (1), (2) and (3) we find
i 1 = 0.44 A, i 2 = −0.28 A and i = 0.16 A. The negative sign of i 2 shows
that its direction is opposite to that has been assumed.
12.45 i 1 + i 2 + i 3 = i = 5 A
(1)
As ρ = const and R = π d 2 /4 (Fig. 12.45)
R 1 : R 2 : R 3 =
l 1
d 2
1
:
l 2
d 2
2
:
l 3
d 2
3
∴
R 1 : R 2 : R 3 =
2
3 2 :
3
4 2 :
4
5 2 =
2
9
:
3
16
:
4
25
(2)
Since P.D across all the resistors is identical,
i 1 R 1 = i 2 R 2 = i 3 R 3
(3)
∴
i 2 = i 1
R 1
R 2
=
2
9
×
16
3
i 1 =
32
27
i 1
(4)
i 3 = i 1
R 1
R 3
=
2
9
×
25
4
=
25
18
i 1
(5)
∴
i 1 +
32
27
i 1 +
25
18
i 1 = i = 5
( 6 )
∴
i 1 = 1.4 A, i 2 = 1.66 A, i 3 = 1.94 A
where we have used (1), (4) and (5).
Fig. 12.45
12.46 The P.D across 8 and 2 resistors are equal, Fig. 12.46:
2i 2 = 8i 3
(1)
As P.D across 3 and 6 resistors are equal
3i 4 = 6i 1
(2)
As P.D across 4 and 6 resistors are equal
6i 6 = 4i 5
(3)