Page 16 - Schaum's Outline of Theory and Problems of Electric Circuits

5
INTRODUCTION
CHAP. 1]
1.3 A certain circuit element has a current i ¼ 2:5 sin !t (mA), where ! is the angular frequency in
rad/s, and a voltage difference v ¼ 45 sin !t (V) between terminals. Find the average power P avg
and the energy W T transferred in one period of the sine function.

Energy is the time-integral of instantaneous power:
ð 2 =! ð 2 =!
2
W T ¼ vi dt ¼ 112:5 sin !tdt ¼ 112:5 ðmJÞ
0 0 !
The average power is then

W T
P avg ¼ ¼ 56:25 mW
2 =!
Note that P avg is independent of !.

1.4 The unit of energy commonly used by electric utility companies is the kilowatt-hour (kWh).
(a) How many joules are in 1 kWh? (b) A color television set rated at 75 W is operated from
7:00 p.m. to 11:30 p.m. What total energy does this represent in kilowatt-hours and in mega-
joules?
(a) 1 kWh ¼ð1000 J=sÞð3600 s=hÞ¼ 3:6MJ
(b) ð75:0WÞð4:5hÞ¼ 337:5Wh ¼ 0:3375 kWh
ð0:3375 kWhÞð3:6MJ=kWhÞ¼ 1:215 MJ

1.5 An AWG #12 copper wire, a size in common use in residential wiring, contains approximately
2:77 10 23 free electrons per meter length, assuming one free conduction electron per atom.
What percentage of these electrons will pass a fixed cross section if the conductor carries a
constant current of 25.0 A?
25:0C=s 20
1:602 10 19 C=electron ¼ 1:56 10 electron=s
ð1:56 10 20 electrons=sÞð60 s=minÞ¼ 9:36 10 21 electrons=min
9:36 10 21
ð100Þ¼ 3:38%
2:77 10 23

1.6 How many electrons pass a fixed point in a 100-watt light bulb in 1 hour if the applied constant
voltage is 120 V?
100 W ¼ð120 VÞ IðAÞ I ¼ 5=6A
ð5=6C=sÞð3600 s=hÞ 22
¼ 1:87 10 electrons per hour
1:602 10 19 C=electron

1.7 A typical 12 V auto battery is rated according to ampere-hours. A 70-A h battery, for example, at
a discharge rate of 3.5 A has a life of 20 h. (a) Assuming the voltage remains constant, obtain the
energy and power delivered in a complete discharge of the preceding batttery. (b) Repeat for a
discharge rate of 7.0 A.

(a) ð3:5AÞð12 VÞ¼ 42:0 W (or J/s)
ð42:0J=sÞð3600 s=hÞð20 hÞ¼ 3:02 MJ
(b) ð7:0AÞð12 VÞ¼ 84:0W
ð84:0J=sÞð3600 s=hÞð10 hÞ¼ 3:02 MJ