20 THE ELECTRIC ARC 



than the potential difference between the terminals of the 

 arc plus the amount indicated by b/I. For example, if we 

 take the values of a and b for the arc 5 mm. in length, as 

 given by Mrs. Ayrton, we have E = 50 + 64/7. There- 

 fore, for 10 amperes the resistance in series with the arc 

 must be more than 0.64 ohm and the E.M.F. of the dynamo 

 must be more than 62.8 volts, although the potential 

 difference between the terminals of the arc is only 56.4 

 volts. For 5 amperes the resistance must be more than 

 2.48 ohms and the E.M.F. more than 75.6 volts. This is 

 quite in agreement with the well-known fact that the 

 smaller the current the greater the resistance that must be 

 in series with it. 



The energy used in the resistance is E"I. If we again 

 assume that E = a -f b/I, we have E' > E + b/I. From 

 this and the equation E" = E' E it can be shown that 

 E"I must be greater than b. The energy used in the arc 

 is al + b, so that the ratio of that lost in the resistance to 



that used in the arc is greater than , , , . This becomes 



al -f- 



smaller as I becomes larger. That is, the percentage of 

 energy necessarily wasted in the resistance, when a con- 

 stant voltage dynamo is used, is less the greater the cur- 

 rent. This, of course, does not state what is the best 

 current to use in practice, for there is still the question as 

 to the relation between the light efficiency of the arc and 

 the current, a question that will be considered in the 

 chapter on photometry. 



It should be noted that to have stable equilibrium it is 

 necessary that the stability be not destroyed either be- 

 cause of the sudden change in potential difference between 

 the electrodes when the current changes, or by the slower 



