38 TIFK MRK<T-rn;REvr MOTOR < n. 11 



and hence the induced tension; the area DK'F'd' is 

 therefore the product of the current and the induced 

 tension, and therefore equal to the work done per second. 

 The induced tension is now greater than that of the line, 

 the difference, Tf'F', being what is called the 'effective' 

 tension. 



Example 13. -Suppose that we have a shunt-wound 

 dynamo directly connected to a steam engine running at 

 462 r.p.m., and giving 100 volts at the brushes on open 

 circuit. The induction factor is then 13. Let the dynamo 

 be acting as a generator, and charging accumulators. 

 Assume the speed and induction factor to be constant. 

 Suppose, also, that the torque required to turn the engine 

 and armature is 400 inch-pounds. When the dynamo 

 is delivering 100 amperes into the accumulators, the 

 torque due to the current is 1,300 inch-pounds and resists 

 the motion. The friction also opposes the motion, so that 

 we have 1,700 inch-pounds of torque opposing the motion. 

 Suppose, now, that the steam valve is suddenly closed. 

 If there is no automatic cut-out, the tension of the cells 

 immediately reverses the current in the armature, but the 

 direction of rotation remains the same, since the current 

 in the magnets is not reversed. The load on the dynamo 

 is now only that due to friction ; hence the current taken 

 from the cells will be 22 amperes. The speed is given by 

 Equation 13. If #=0-015, we see that the effect of 

 shutting off the steam will simply be to reduce the speed 

 of the dynamo by l - 5 r.p.m., and to reverse the current 

 from 100 amperes in one direction to 22 amperes in the 

 other. 



We are now dealing with a motor of constant induc- 

 tion factor and resistance, whose terminals are connected 



