12.3 Solutions
565
ξ × 20
20 + r
=
150
l 0
ξ 0
ξ × 10
10 + r
=
120
l 0
ξ 0
Dividing the last two equations
2 (10 + r )
20 + r
=
5
4
whence r = 6.67 .
12.32 For the section ABC
i =
ξ 1
R + r
=
2.2
1 + 0.1
= 2 A
In the section BCD also i = 2 A.
The resistance of 26 cm wire =
26
50
× 1 = 0.52 .
Neglecting the internal resistance of the second cell
ξ 2 = 2 × 0.52 = 1.04 V
as no current flows through the galvanometer
12.33 (a) V AB = V AD ;
V BC = V DC
∴ i 1 P = i 2 R
(1)
i 1 Q = i 2 S
(2)
Dividing (1) by (2)
P
Q
=
R
S
(3)
(b) Assume that a non-zero current flows through the galvanometer of resistance G. Applying the junction theorem at A
i = i 1 + i 2
(4)
Applying the loop theorem to the loop ABDA and noting that there is no
emf in this loop
i g G + i 1 P − i 2 R = 0
( 5 )
565
ξ × 20
20 + r
=
150
l 0
ξ 0
ξ × 10
10 + r
=
120
l 0
ξ 0
Dividing the last two equations
2 (10 + r )
20 + r
=
5
4
whence r = 6.67 .
12.32 For the section ABC
i =
ξ 1
R + r
=
2.2
1 + 0.1
= 2 A
In the section BCD also i = 2 A.
The resistance of 26 cm wire =
26
50
× 1 = 0.52 .
Neglecting the internal resistance of the second cell
ξ 2 = 2 × 0.52 = 1.04 V
as no current flows through the galvanometer
12.33 (a) V AB = V AD ;
V BC = V DC
∴ i 1 P = i 2 R
(1)
i 1 Q = i 2 S
(2)
Dividing (1) by (2)
P
Q
=
R
S
(3)
(b) Assume that a non-zero current flows through the galvanometer of resistance G. Applying the junction theorem at A
i = i 1 + i 2
(4)
Applying the loop theorem to the loop ABDA and noting that there is no
emf in this loop
i g G + i 1 P − i 2 R = 0
( 5 )