Calculation of Efficiency by Hopkinson's Test
Let, V = supply voltage of the machines
Then, Motor input = V(I1 + I2) I1 = The current from the generator I2 = The current from the
external source And, Generator output = VI1..................(1) Let, both machines are operating at
the same efficiency 'η'. Then, Output of motor = η x input = η x V(I1 + I2) Input to generator =
Output of the motor = η X V(I1 + I2) Output of generator = η x input = η x [η x V(I1 + I2)] = η2
V(I1 + I2)..................(2) From equation 1 an 2 we get, VI1 = η2 V(I1 + I2) or I1 = η2 (I1 + I2)
Now, in case of motor, armature copper loss in the motor = (I1 + I2 I4) Ra. Ra is the armature resistance of both motor and generator. I4 is the shunt field current of
the motor. Shunt field copper loss in the motor will be = VI4
Next, in case of generator armature copper loss in generator = (I1 + I3)2Ra I3 is the shunt field
current of the generator. Shunt field copper loss in the generator = VI3 Now, Power drawn from
the external supply = VI2 Therefore, the stray losses in both machines will be W = VI2 - (I1 + I2 I4)2 Ra + VI4 + (I1 + I3)2 Ra + VI3 Let us assume that the stray losses will be same for both the
machines. Then, Stray loss / machine = W/2
Efficiency of Generator
Total losses in the generator, WG = (I1 + I3)2 Ra + VI3 + W/2 Generator output = VI1 Then,
efficiency of the generator,
Efficiency of Motor
Total losses in the motor, WM = (I1 + I2 - I4)2 Ra + VI4 + W/2 Motor input = V(I1 + I2) Then,
efficiency of the motor,
Advantages of Hopkinson's Test
The merits of this test are…
This test requires very small power compared to full-load power of the motor-generator
coupled system. That is why it is economical. Large machines can be tested at rated load
without much power consumption.
Temperature rise and commutation can be observed and maintained in the limit because
this test is done under full load condition.
Change in iron loss due to flux distortion can be taken into account due to the advantage
of its full load condition.
Efficiency at different loads can be determined.
Disadvantages of Hopkinson's Test
The demerits of this test are
It is difficult to find two identical machines needed for Hopkinson's test.
Both machines cannot be loaded equally all the time.
It is not possible to get separate iron losses for the two machines though they are different
because of their excitations.