Page 296 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 24] ELECTRIC CURRENT 281

Here R = 5 and R = 9 − 5 = 4 . Since R = αR T ,
R 4
◦
T = = = 222 C
◦
αR (0.0036/ C)(5 )
The temperature of the furnace is T + T = 20 C + 222 C = 242 C.
◦
◦
◦
CIRCULAR MIL
In engineering practice the unit of area of a round conductor is often the circular mil,or cmil. The mil is a unit
of length equal to 0.001 in., which is 1 in. A circular mil is a unit of area equal to the area of a circle whose
1000
diameter is 1 mil, as in Fig. 24-2. The area A cm in circular mils of a circle whose diameter in mils is D m is equal
2
to D :
m
A cm = D 2
m
The advantage of using the circular mil as a unit of area is that it avoids multiplication and division by π. When
the length of a conductor is given in feet and its area in circular mils, the unit of resistivity is the ohm-cmil per
foot.

Fig. 24-2

SOLVED PROBLEM 24.13
The specific resistance of the copper used in electric wires is 10.4 ·cmil/ft. Find the resistance of 1500 ft
of copper wire whose diameter is 0.080 in.

Since 0.080 in. = 80 mils,
2
2
A cm = D = (80) cmil = 6400 cmil
m
and
ρl (10.4 ·cmil/ft)(1500 ft)
R = = = 2.44
A 6400 cmil
SOLVED PROBLEM 24.14
The specific resistance of Nichrome is 600 ·cmil/ft. How long should a Nichrome wire 20 mils in
diameter be for it to have a resistance of 5 ?
The cross-sectional area of the wire is
2
2
A cm = D = (20) cmil = 400 cmil
m
Next we solve for l and substitute ρ = 600 ·cmil/ft, R = 5 , and A = 400 cmil to find the value of the length l:
RA (5 )(400 cmil)
l = = = 3.33 ft
ρ 600 ·cmil/ft