192 TIIK DIRECT-CURRENT MOTOR cil. IX 



car is started in 1GO seconds ; the total energy taken from 

 the line is then 972 x 10 4 foot-pounds ; in the first case 

 the car is started in 107 seconds, and the energy is 

 753 x 10 4 foot-pounds. 



CASE III. Motors started in series and run in 

 parallel. This is known as the ' Series-Parallel ' method. 

 Since the motors are to run at full speed in parallel they 

 will have the same induction factor as in Case I. ; hence 

 Jlf=95-8. 



STEP I. Motors in series. Starting rheostat of 1*7 

 ohms in the circuit. Frictional current 29 '6 amperes. 

 Current available for acceleration 170-4 amperes. The 

 initial acceleration is then I'OOo f.p.s. per second. 

 The speed of the motors when the rheostat is all out is 

 106'4 r.p.m. The time on Step I. is 1 5 - 3 seconds, and the 

 distance travelled is 1 1 8 feet. 



STEP II. Motors in parallel. Rheostat again in 

 circuit. On switching over to parallel the current from 

 the line must not exceed 200 amperes. If X is the 

 resistance in series with each motor at the beginning of 

 Step II. we have, since the speed is the same as it was 



250 - 200 x -4- 



immediately before switching over, 



M 



500 100 xX Q . 



- or A = 3-3 ohms. As the internal resist- 

 M 



ance is 0'4 ohm, we shall need a rheostat of 2-9 ohms in 

 series with each motor. The current per motor will then 

 be 100 amperes. The speed of the motors when this 

 resistance is all out is 288 r.p.m., as in Case I. The 

 acceleration here is 0'415 f.p.s. per second, the time 

 occupied is 63 seconds, and the distance travelled 1,789 feet. 

 STEP III. Rheostat all out. Acceleration diminish 



