564
12 Electric Circuits
12.29 The current flows from positive to negative terminal inside the battery. The
potential difference between the two terminals of the battery would be
V = ξ + ir = 2 + 5 × 0.1 = 2.5 V
12.30 (a) The effective resistance in the circuit (Fig. 12.40) from 3 in series with
5 and 2 in parallel
R 0 = R +
R 1 R 2
R 1 + R 2
= 3 +
2 × 5
2 + 5
= 4.43
i =
ξ
R 0
=
6
4.43
= 1.35 A
(b) i 1 =
i R 2
R 1 + R 2
=
1.35 × 2
5 + 2
= 0.386
P 1 = i 1
2 R 1 = (0.386)
2
× 5 = 0.74 W
i 2 = i − i 1 = 1.35 − 0.386 = 0.964 A
P 2 = i 2
2 R 2 = (0.964)
2
× 2 = 1.86 W
P = i
2 R = (1.35)
2
× 3 = 5.47 W
(c) Total power dissipated by the resistances
= P 1 + P 2 + P = 0.74 + 1.86 + 5.47 = 8.07 W 8.1 W
Power supplied by the battery = ξi = 6 × 1.35 = 8.1 W
Fig. 12.40
12.3.3 Instruments
12.31 P.D across AB is
ξ R
R + r
=
l
l 0
ξ 0
12 Electric Circuits
12.29 The current flows from positive to negative terminal inside the battery. The
potential difference between the two terminals of the battery would be
V = ξ + ir = 2 + 5 × 0.1 = 2.5 V
12.30 (a) The effective resistance in the circuit (Fig. 12.40) from 3 in series with
5 and 2 in parallel
R 0 = R +
R 1 R 2
R 1 + R 2
= 3 +
2 × 5
2 + 5
= 4.43
i =
ξ
R 0
=
6
4.43
= 1.35 A
(b) i 1 =
i R 2
R 1 + R 2
=
1.35 × 2
5 + 2
= 0.386
P 1 = i 1
2 R 1 = (0.386)
2
× 5 = 0.74 W
i 2 = i − i 1 = 1.35 − 0.386 = 0.964 A
P 2 = i 2
2 R 2 = (0.964)
2
× 2 = 1.86 W
P = i
2 R = (1.35)
2
× 3 = 5.47 W
(c) Total power dissipated by the resistances
= P 1 + P 2 + P = 0.74 + 1.86 + 5.47 = 8.07 W 8.1 W
Power supplied by the battery = ξi = 6 × 1.35 = 8.1 W
Fig. 12.40
12.3.3 Instruments
12.31 P.D across AB is
ξ R
R + r
=
l
l 0
ξ 0