522 RHEOSTATS AND OTHER BALLAST [Cn. XIII 



As stated by Norris it is: "The electromotive force consumed in the 

 resistance of a conductor, is proportional to the current." P. 8. 



Using the terms now employed in place of electromotive force (voltage), 

 resistance (ohmage) , and current (amperage) , the law can be stated thus : 



(1) The voltage in a conductor is equal to the amperage multiplied by the 

 ohmage: V = A O. 



(2) The amperage is equal to the voltage divided by the ohmage: A = 



(3) The ohmage is equal to the voltage divided by the amperage: O = 



A 



As V = A O = i . From this form is derived the very simple dia- 



A O 



gram used practically in getting the formula for the value of any single quantity 

 if two are known. The formula for the unknown quantity is found thus : 



Cover the letter representing the unknown 

 V quantity, and the remaining letters will indicate 



the value of the unknown quantity. 



" Examples: 



FIG. 279. DIAGRAM OF i. If the voltage and amperage are known, 



OHM'S LAW FOR SOLV- wh at is the ohmage? 



ING PROBLEMS ( 7243). Cover the O and there remain V/A and this is 



V = Voltage equal to O, i. e., O = V/A. Suppose the volt- 



A _ . age is 1 10 and the amperage is 20, what is the 



O = Ohma e ohmage? Applying the formula, O = 1 10/20, 



2. If the voltage and the ohmage are known what is the amperage? Here 

 if A is covered there is left V/O, whence the amperage equals the voltage divided 

 by the ohmage. If the voltage is 220 and the ohmage is 5.5 as before, what is 

 the amperage? A = 220/5.5 =40 amperes. This example also illustrates the 

 fact that if the ohmage remains constant the amperage will increase in direct 

 proportion to the voltage. (See Dr. Nichols' definition above). 



3. If the amperage and ohmage are known what is the voltage? Here the 

 unknown quantity is represented by V. If this is covered there will be left 

 A O, whence V = A O. If the amperage is 15, and the ohmage 8 then the 

 voltage must be 15 x 8 = 120, i. e., V = 120 volts. 



As a further example suppose one wished to make a water-cooled rheostat 

 (fig. 283) and he had some wire which had an ohmage or resistance of 0.25 ohm 

 per meter, how much wire would be needed with a voltage of no and an 

 amperage of 15? Here voltage and amperage are known. From the formula 

 it is seen that ohmage equals voltage divided by amperage: whence 1 10/15 = 

 7.33 ohms, the total resistance required. 



Now as 55 is the voltage required by the arc with the direct current arc 



lamp, the lamp itself must offer a resistance of . ~ f r 3-66 ohms. 



A = 15 



As the total ohmage needed is 7.33, the rheostat must possess the difference 

 between 7.33 and 3.66 or 3.67 ohms. 



If each meter of the wire to be used offers a resistance of 0.25 ohm, it will 



require for 3.67 ohms, ^' ? = 14.68 meters of the wire for the rheostat. (For 



0.25 

 the wattage of the current see 660). 



