12.3 Solutions
569
(b) Initially V = i G = 1 × 10 −3 × 80 = 0.08 V
N =
80
0.08
= 1000
R = (N − 1)G = (1000 − 1) × 80 = 79, 920
A resistance of 79,920 must be connected in series with the moving
coil galvanometer to give the required full scale deflection.
12.41 ξ = i(R + r )
(by Ohm’s law)
9 = i(120 + 15)
∴
i =
1
15
A
The voltmeter would read
V = ξ − ir = 9 −
1
15
× 15 = 8 V
12.42 S =
G
N − 1
i 1
i
=
1
N
=
1
100
or
N = 100
∴
G = (N − 1)S = (100 − 1) × 1 = 99
12.3.4 Kirchhoff’s Laws
12.43 By the junction theorem
i = i 1 + i 2
(1)
Applying the loop theorem to the loop B E 1 AE 2 B, traversing clockwise
ξ 2 − i 2 r 2 − ξ 1 + i 1 r 1 = 0
∴ 2i 2 − i 1 = 20 − 10 = 10
(2)
Applying the loop theorem to the loop B E 1 ARB
ξ 1 = i 1 r 1 + i R
569
(b) Initially V = i G = 1 × 10 −3 × 80 = 0.08 V
N =
80
0.08
= 1000
R = (N − 1)G = (1000 − 1) × 80 = 79, 920
A resistance of 79,920 must be connected in series with the moving
coil galvanometer to give the required full scale deflection.
12.41 ξ = i(R + r )
(by Ohm’s law)
9 = i(120 + 15)
∴
i =
1
15
A
The voltmeter would read
V = ξ − ir = 9 −
1
15
× 15 = 8 V
12.42 S =
G
N − 1
i 1
i
=
1
N
=
1
100
or
N = 100
∴
G = (N − 1)S = (100 − 1) × 1 = 99
12.3.4 Kirchhoff’s Laws
12.43 By the junction theorem
i = i 1 + i 2
(1)
Applying the loop theorem to the loop B E 1 AE 2 B, traversing clockwise
ξ 2 − i 2 r 2 − ξ 1 + i 1 r 1 = 0
∴ 2i 2 − i 1 = 20 − 10 = 10
(2)
Applying the loop theorem to the loop B E 1 ARB
ξ 1 = i 1 r 1 + i R