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tiL12214_appe_643-698.indd Page 661 09/10/10 8:37 AM user-f463
tiL12214_appe_643-698.indd Page 661 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

12 V
_ _
V P I P
(b) I s = ? V P I P = V s I s ∴ I s = = 16.0 Ω
V s
12 V _
(120 V)(5.0 A)
__ = _

I s = 16.0 V _
360 V
A
_ _
120 × 5.0 V·A

= V _ _

A

360 V = 0.75 ×
1 V
= 1.7 A
= 0.75 A
(c) P s = ? P s = I s V s
The current through each bulb is 0.75 amp. The voltage drop

J
s (
)
C _ _ across each lamp can be calculated from Ohm’s law, and V = IR =

= (1.7 ) 360 (0.75 A)(8.0 V/A) = 6.0 V. As a check, you know this is correct
C
J
C _ _ because the voltage drop equals the voltage rise and 6.0 V + 6.0 V =
= 1.7 × 360 × 12.0 V. The power of each glowing bulb can now be calculated

s C
from P = IV and P = (0.75 A)(6.0 V) = 4.5 W. Two lamps
_ J
= 612 connected in a series circuit, as compared to one lamp in a simple
s
circuit, thus decreases the current by one-half and decreases the
= 610 W voltage by one-half, resulting in one-fourth the power for each lamp.
6.17. List the known and unknown quantities:
CHAPTER 7
resistance (R) 8.0

voltage (V) 12 V 7.1. The relationship between the speed of light in a transparent
8
power (P) ? W material (v), the speed of light in a vacuum (c = 3.00 × 10 m/s),
current (I) ? A and the index of refraction (n) is n = c/v. According to
Table 7.1, the index of refraction for water is n = 1.33 and for

For one bulb: The resistance (R) is given as 8.0 ohms (which is ice is n = 1.31.
8.0 volt/amp, by defi nition). The voltage source (V) is c _ c _

8
12 V, and the power of the glowing bulb is to be determined. (a) c = 3.00 × 10 m/s n = ∴ v =
v
n
Power can be calculated from P = IV. This equation requires n = 1.33 __

8
3.00 × 10 m/s

the current (I), which is not provided but can be calculated v = 1.33
v = ?
from Ohm’s law, V = IR.
8
= 2.26 × 10 m/s
V _
V = IR ∴ I = (b) c = 3.00 × 10 m/s

8
R
_ n = 1.31
12 V
=
V _
8.0 v = ?

A 3.00 × 10 m/s
8
__
v =
12 _ V _ A _ 1.31

= ×

8.0 1 V 8
= 2.29 × 10 m/s
= 1.5 A
8
7.2. d = 1.50 × 10 km
Now that we have the current, we can find the power: 11

= 1.50 × 10 m
8
P = IV c = 3.00 × 10 m/s
s ( )
J
C _ _ t = ?

= (1.5 ) 12
C
d _
d _

v = ∴ t =

J
C _ _ t v
= 1.5 × 12 ×

s C __
11
1.50 × 10 m
_ J t = 8 m _

= 18 3.00 × 10

s s
11
= 18 W _ s _
1.50 × 10

= ×
m

m
3.00 × 10 8
For two 8.0 ohm bulbs in series: Since the bulbs are connected
_
2 m·s
one after another, the current encounters resistance from each in = 5.00 × 10

turn, and the total resistance is equal to the sum of the m
2
_
individual resistances, or R total = R 1 + R 2 . Th erefore, = 5.00 × 10 s

s _
60.0

R total = 8.0 + 8.0 = 16.0 min
min
5.00 × 10

We can now find the current flowing through the entire = _ 2 _

s
×

circuit, which will be the same as the current fl owing through 60.0 s
each of the resistances. = 8.33 min
_
V

I =
R total
E-19 APPENDIX E Solutions for Group A Parallel Exercises 661