Hopkinson Test

The Hopkinson test is a useful method for testing the efficiency of a DC machine. It requires two identical machines, one acting as a generator and the other as a motor. The generator supplies mechanical power to the motor, which then drives the generator. This setup is why the Hopkinson test is also called the back-to-back or regenerative test.
If there were no losses, no external power supply would be needed. However, due to generator output voltage drop, an extra voltage source is required to supply the correct input voltage to the motor. The external power compensates for internal losses in the motor-generator set. This is why the Hopkinson test is also known as the regenerative or heat run test.

Connection Diagram of Hopkinson’s Test

In the Hopkinson test, a motor and a generator, both identical, are coupled together. The machine starts as a motor, and the shunt field resistance is adjusted so the motor can run at its rated speed.

The generator voltage is now made equal to the supply voltage by adjusting the shunt field resistance connected across the generator. This equality of these two voltages of generator and supply is indicated by the voltmeter as it gives a zero reading at this point connected across the switch. The machine can run at rated speed and at desired load by varying the field currents of the motor and the generator.

Calculation of Efficiency by Hopkinson’s Test

Let, V = supply voltage of the machines.
Then,
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 =

From equation 1 an 2 we get,


Now, in case of motor, armature copper loss in the motor = .
R_a 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 =
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

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,
Generator output = VI₁
Then, efficiency of the generator,

Efficiency of Motor

Total losses in the motor,

Then, efficiency of the motor,

Advantages of Hopkinson’s Test

The merits of this test are…

1. 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.
2. Temperature rise and commutation can be observed and maintained in the limit because this test is done under full load condition.
3. Change in iron loss due to flux distortion can be taken into account due to the advantage of its full load condition.
4. Efficiency at different loads can be determined.

Disadvantages of Hopkinson’s Test

The demerits of this test are

1. It is difficult to find two identical machines needed for Hopkinson’s test.
2. Both machines cannot be loaded equally all the time.
3. It is not possible to get separate iron losses for the two machines though they are different because of their excitations.
4. It is difficult to operate the machines at rated speed because field currents vary widely.