ELECTRIC OSCILLATIONS AND ELECTRIC WAVES. 243 



has a frequency of 10,000 cycles per second, then the wave- 

 length X will be 1 8. 6 miles. The voltage nodes (current anti- 

 nodes) of this standing wave train are at the points n, n, n, 



(0) alternator 



Fig. 184. 

 Voltage distribution over oscillating transmission line closed at far end. 



etc., and the voltage antinodes (current nodes) are at the points 

 a, a, a, etc. The voltage across the line at any point is an 

 alternating voltage of which the frequency is equal to the fre- 

 quency of the alternator, and the maximum value of the voltage 

 across the line at any point is represented by the ordinate at 

 that point of one of the sine curves in Figs. 181 to 184. Let E 

 be the maximum value of the voltage between wires at any voltage 

 antinode. Then the maximum voltage between wires at a point 



distant x from a voltage antinode is E sin -r The condition 



A 



which must be satisfied when the ultimate steady state of line 

 oscillation is reached is that the distribution of voltage over the 

 line as represented by the sine curves in Figs. 181 to 184 must 

 conform to or fit the alternator voltage at the generator end. 

 Therefore the ultimate state of line oscillation may be determined 

 as follows : Begin at the distant end of the line. Lay off a number 

 of quarter wave-lengths and determine the distance d which is 

 left over at the generator end of the line as represented in Figs. 

 1 8 1 to 1 84. Then the maximum voltage E at one of the voltage 



