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since it is not given. The current can be obtained from the Motor: (0.20 hp)(746 W/hp) = 150 W

relationship of Ohm’s law:
150 J/s
P _ _
= 1.3 A
I = =

V _ V 120 J/C

I =
R Three 100 W bulbs: 3 × 100 W = 300 W
_ 300 J/s
3.00 V
P _
= V _ I = = _

= 2.5 A

15.0 V 120 J/C
A
600 J/s
P _ _

= 5.0 A
= 0.200 A Iron I = =
V 120 J/C
P = IV
The sum of the currents is 8.0 A + 1.3 A + 2.5 A +

= (0.200 C/s)(3.00 J/C)
5.0 A = 16.8 A, so the total current is greater than 15.0 amp
= 0.600 W and the circuit breaker will trip.
__
(watts)(time)(rate)

6.10. cost = 6.14. (a) V P = 1,200 V
_
W
1,000
kW N P = 1 loop
(
$0.10
)
(1,200 W)(0.25 h) _ N s = 200 loops
___ V s = ?
kWh
= _
W
V P _ _
1,000 = ∴ V s = _
V P N s
V s

kW
N P N s N P
__ h _ _ (1,200 V)(200 loop)
(1,200)(0.25)(0.10) W _ _
kW
$

= × × × __

1,000 1 1 kWh W V s = 1 loop
= $0.03 (3 cents)
= 240,000 V
V P I P
6.11. The relationship between power (P), current (I), and voltage _

(b) I P = 40 A V P I P = V s I s ∴ I s =
(V) will provide a solution. Since the relationship considers V s
power in watts, the first step is to convert horsepower to watts. __

1,200 V × 40 A
I s = ? I s =
One horsepower is equivalent to 746 watts, so 240,000 V
1,200 × 40 _
(746 W/hp)(2.00 hp) = 1,492 W _ V·A

= =
P _ 240,000 V

P = IV ∴ I =
V = 0.2 A
_ J V P _ _ N P _ _

V s
1,492 6.15. (a) V s = 12 V = ∴ =
V P

_

s
= J N P N s N s V s

_

12.0 I s = 0.5 A
10
120 V
C N P _ _ _

V P = 120 V = =
_ _ _
1,492 J C N s 12 V 1

= ×

12.0 s J 10 primary to 1 secondary
C _ _
V s I s
= 124.3 (b) I P = ? V P I P = V s I s ∴ I P =
s V P
(12 V)(0.5 A)
= 124 A __
I P =

120 V
6.12. (a) The rate of using energy is joule/s, or the watt. Since _ _

12 × 0.5 V·A
1.00 hp = 746 W, =
120 V
inside motor: (746 W/hp)(1/3 hp) = 249 W = 0.05 A
outside motor: (746 W/hp)(1/3 hp) = 249 W (c) P s = ? P s = I s V s
compressor motor: (746 W/hp)(3.70 hp) = 2,760 W = (0.5 A)(12 V)
J
249 W + 249 W + 2,760 W = 3,258 W C _ _

= 0.5 × 12 ×

s
C
(
)
$0.10
(3,258 W)(1.00 h) _ _ J

kWh
___ = 6
s
(b) _
= $0.33 per hour
W
1,000 = 6 W
kW
V P _ V s _ _
V P N s

(c) ($0.33/h)(12 h/day)(30 day/mo) = $118.80 6.16. (a) V P = 120 V = ∴ V s =
N P N s N P
N P = 50 loops

6.13. The solution is to find how much current each device draws 120 V × 150 loops
__

and then to see if the total current is less or greater than the N s = 150 loops V s = 50 loops
breaker rating. I P = 5.0 A
V·loops
120 × 150 _
_

V _ _ V s = ? = 50 loops
120 V

Toaster: I = = = 8.0 A
R 15 V/A
= 360 V
660 APPENDIX E Solutions for Group A Parallel Exercises E-18