192 Mr. T. Gray on the Size of Conductors 
that is to say, 
r{l+|r 8 log5}=50 (12) 
Assuming, what are probably not far from the true values for 
gutta percha, e and k to be 4^00 anc * 2000 respectively, we 
get 
T'= - (13) 
l + Wog e ^ 
Again, since the amount of heat generated by the current 
must be equal to the quantity radiated from the surface of the 
covering, 
c 2 S(l+50a) _27rr 2 
A J ~4000* i; 
by equation (13), 
2irr 2 50 
4000 * 
77T 2 
4o{l + l'15r,log M ^} 
Take, for example, r x = l and r 2 =l*4. Then 
2030 ir x 1-4 
(14) 
ttx4-2x10 7 40{ 1 + 1-15 x 1-4 log 1-4}' 
. /^ 1-4 x 4-2 x W~ 
V 40 x 2030 {1 + 1-15x1-4 log 1-4 J. 
= ttx22-3. 
That is to say, a rod 1 centimetre in radius and covered with 
gutta percha to a thickness of 4 millimetres requires a current 
223 amperes per square centimetre to raise its temperature 
50° C. 
If we suppose r x = l and ?' 2 = 5, we find c = 7rx 21, or a cur- 
rent of 210 amperes would heat the conductor 50° C. In- 
creased thickness, therefore, in the covering does not greatly 
alter the capacity of the conductor. 
Regulation. 
It is of great importance, especially in electric-lighting cir- 
cuits, that the E.M.F. at which the electricity is supplied 
