there may be formed the two following equations:
| H E | |
| T = H + E, and S = | — + —, |
| 2 2 |
from which the elimination of H will give E = 2T - 4S.
The value of the eddy current loss thus found will be about 1½ per cent., and constant at all loads.
Having previously ascertained the power lost in both eddy currents and hysteresis, and knowing now the power lost in eddy currents alone, it is easy to find that lost in hysteresis by simply subtracting the latter known value from the former. The value of the hysteresis loss is therefore approximately 1½ per cent., and it is constant at different loads.
There yet remains to be determined the armature resistance loss and the field resistance loss. As for the calibrated motor, this may be disconnected from the dynamo, as it need not be used further in the test.
The armature resistance is the resistance of the armature winding of the dynamo, between the commutator bars upon which press the positive and negative brushes. Assume that the value of the armature resistance be known, call this value R ohms, together with that of the full load armature current, which is also known and which call I amperes, this is sufficient data for calculating the armature resistance loss at full load. It is evident that to force the full load current I through the armature resistance R will require a pressure of R volts, and that the watts lost in doing so will be the voltage multiplied by the current. The armature resistance is consequently
I R × I = I2R watts
or, expressed in horse power it is
I2R ÷ 746