12.3 Solutions
559
Joule heat in R 1 , H 1 = i 2 R 1 = 1936 i 2
Joule heat in R 2 , H 2 = i 2 R 2 = 484 i 2
Therefore, 25 W bulb glows brighter.
12.17 By prob. (12.16), R 1 = 1936 and R 2 = 484 :
H 1 = i 1 V =
i V R 2
R 1 + R 2
H 2 = i 2 V =
i V R 1
R 1 + R 2
Therefore, 100 W bulb glows brighter.
12.18 Effective resistance in parallel is given by (Fig. 12.38)
1
R
=
1
4
+
1
6
+
1
12
=
1
2
∴ R = 2
Total resistance, R AB = 2 + 1 + 2 = 5
Current in the circuit
i =
V AB
R AB
=
120
5
= 24 A
V AC = i R AC = 24 × 2 = 48 V.
Fig. 12.38
12.19 i =
ξ
R + r
P = i
2 R =
ξ 2 R
(R + r ) 2
Maximum power delivered to R is obtained by setting
∂ P
∂ R
= 0. This gives
R = r :
∴
P max =
ξ 2 r
(r + r ) 2 =
ξ 2
4r