### Field Test

This is one of the methods of testing the d.c. series motors. Unlike shunt motors, the series motor can not be tested by the methods which area available for shunt motors as it is impossible to run the motor on no load. It may run at dangerously high speed on no load. In case of small series motors brake test may be employed.
The series motors are usually tested in pairs. The field test is applied to two similar series motors which are coupled mechanically. The connection diagram for the test is shown in the Fig. 1.
 Fig. 1 Field test

As shown in the Fig. 1 one machine is made to run as a motor while the other as a generator which is separately excited. The field of the two machines are connected in series so that both the machines are equally excited. This will make iron losses same for the two machines. The two machines are running at the same speed. The generator output is given to the variable resistance R.
The resistance R is changed until the current taken by motor reaches full load value. This will be indicated by ammeter A1. The other readings of different meters are then recorded.
Let      V = Supply voltage
I = Current taken by motor
V2 = Terminal p.d. of generator
Ra, Rse = Armature and series field resistance of each machine
Power taken from supply = VI
Output obtained from generator = V I
Total losses in both the machines, W = VI - V I
Armature copper and field losses, WCU  = ( Ra + 2 Rse ) I12   +   I22 Ra
Total stray losses = W - WCU

Since the two machines are equally excited and are running at same speed the stray loses are equally divided.

For Motor ;
Input to motor = VI
Total losses = Aramture Cu loss + Field Cu loss + Stray loss
= I12 ( Ra + Rse) + Ws
Output of motor = Input - Total losses = VI -  [ I12  ( Ra + Rse) + Ws ]

For Generator :
Efficiency of generator is of little importance because it is running under conditions of separate excitation. Still it can be found as follows.
Output of generator = VI
Field Cu loss =  I12  Rse
Armature Cu loss = I22  Ra
Total losses = Armature Cu loss + Field Cu loss + Stray loss
= I22  Ra +  I12  Rse + Ws
Input to generator = Output + Total losses = VI+ [ I22  Ra + I12  Rse + Ws ]

The important point to be noted is that this is not regenerative method though the two machines are mechanically coupled because the generator output is not fed back to the motor as in case of Hopkinson's test but it is wasted in load resistance.

See solved example