PROBLEMS. 477 



The A.I.E.E., rule, see page 165, Art. 28, applied to this problem would give 

 43. 7 instead of the correct value 44.39. The rule is, however, nearly enough exact for 

 all practical purposes. Indeed tenths of degrees are wholly meaningless in connec- 

 tion with such uncertain quantities as rises of temperature of dynamo machines, so that 

 to the nearest degree the A.I.E.E. rule and the rigorous formula give the same result. 



125, The armature of a dynamo has a resistance between com- 

 mutator bars b and c (see problem 52) of 0.0678 ohm at room 

 temperature of 10 C. After the dynamo has been run for a long 

 time at full load it is shut down and the resistance of the arma- 

 ture between commutator bars is found to be 0.0806 ohm. (a) 

 Find the difference between room temperature and the running 

 temperature of the armature ; (b) what would this temperature 

 difference be if the room temperature were 25 C.? Ans. (a) 49 

 centigrade degrees ; (&) 5 3 centigrade degrees. 



126. The armature and all field windings ot a given 5oo-kilowatt 

 500- volt generator are connected together (machine not running) 

 and a circuit is made from one main of a 5OO-volt supply, through 

 a direct reading voltmeter to one brush of the given generator, 

 whence a very small current passes through the armature and 

 field insulation to the iron frame of the machine which is con- 

 nected to the other main of the 500- volt supply. The voltmeter 

 indicates 6.2 volts and its resistance is 55,000 ohms. Find the 

 insulation resistance of the given generator as connected. Ans. 

 4.38 megohms. 



NOTE. The voltmeter, arranged as specified in this problem, is used as an am- 

 meter, and the current flowing through it is equal to its reading in volts divided by its 

 resistance in ohms. 



CHAPTER VII. THE PRACTICAL OPERATION OF DYNAMOS. 



127. The voltage of an over-compounded generator rises from 

 115 volts at zero-load to 125 volts at full-load. When the 

 series field coil is not used the machine becomes a shunt gener- 

 ator and its voltage falls from 1 1 5 volts at zero-load to 90 volts 

 at full-load. What would the voltage of this generator be if it 

 were operated alone at full-load, with its series field coil in par- 



