# What is Hopkinson’s Test : Circuit Diagram & Its Working

The DC machines like motor and generator are used in different electrical applications. The main function of the generator is to convert the power from mechanical to electrical whereas the motor is used to convert the power from electrical to mechanical. Therefore, the input power of the dc generator is in electrical form whereas the output is in mechanical form. Similarly, the input power of the motor is in electrical form whereas the output is in mechanical form. But in practice, the power conversion of a DC machine cannot be done completely due to the power loss so that the efficiency of the machine can be reduced. It can be defined as the ratio of o/p power and i/p power. So the DC machine’s efficiency can be tested with the help of a Hopkinson’s test.

## What is Hopkinson’s Test?

Definition: A full load test which is used to test the efficiency of a DC machine is known as Hopkinson’s test. An alternate name of this test is back to back, heat run, and regenerative test. This test uses two machines which are connected electrically and mechanically with each other. From these machines, one acts as a motor whereas another works as a generator. The generator provides the mechanical power to the electric motor whereas the motor is used to drive the generator.

Therefore, the o/p of one machine is used as an input to another machine. Whenever these machines run on the condition of full load, then the input supply can be equivalent to the whole losses of the machines. If there is no loss within any machine, there is no need for external power supply. However, if the o/p voltage of the generator is dropped then we need an additional voltage source to provide proper i/p voltage to the motor. Therefore, the power which is drawn from the exterior supply can be used to conquer the inside losses of the machines.

### Circuit Diagram of the Hopkinson’s Test

The circuit diagram of the Hopkinson’s test is shown below. The circuit can be built with a motor as well as a generator together with a switch. Whenever the motor is started then the shunt filed resistance of this motor can be adjusted so that it runs at its rated speed.

Now, the voltage of the generator can be made identical to the voltage supply through regulating the shunt field resistance which is allied across the generator. This equality of generator’s two voltages & its supply can be specified with the help of the voltmeter because it provides a zero reading across the ‘S’ switch. The machine works at rated speed as well as the desired load through changing the motor’s field currents as well as the generator.

### Calculation of the Efficiency of the Machine by Hopkinson’s Test

Let the voltage supply of the machine is ‘V’, then the input of the motor can be derived by the following equation.

The input of the motor = V (I1+I2)

I1 = Generator’s current

I2 = External source current

The o/p of the generator is VI1…….(1)

If the machines work at the same efficiency that is ‘η’

The motor’s o/p is η x i/p = η V (I1+I2)

The input of the generator is the output of the motor then, η V (I1+I2)

The o/p of the generator is the input of the motor then, η [η x V (I1+I2)] = η2 V (I1+I2) ….(2)

From the above two equations, we can get

VI1= η2 V (I1+I2) then I1= η2 (I1+I2) = η√I1/(I1+I2)

The armature copper loss within the motor can be derived by (I1+I2-I4)2Ra

Where,

‘Ra’ =Armature resistance of the machine

‘I4’ = Motor’s shunt field current

Shunt field copper loss within the motor is ‘VI4’

The armature copper loss within the generator can be derived by (I1+I3)2Ra

I3 = Shunt field current

Shunt field copper loss within the motor is ‘VI3’

Power supply drawn from the exterior supply is’ VI2’

So, the stray losses within the machines will be

W =VI2-(I1+I2-I4)2Ra +VI4+(I1+I3)2 Ra+VI3

The stray losses for the machines are similar so W/2 = stray loss/machine

#### The Efficiency of the Motor

The losses in the motor can be derived by the following equation

WM = (I1+I2-I4)2Ra + VI4+W/2

The input of the motor = V(I1+I2)

Then the motor efficiency can be derived by ηM = output/input = (input-losses)/input

= (V (I1+I2) -WM)/ V(I1+I2)

#### The Efficiency of the Generator

The losses in the generator can be derived by the following equation

WG = (I1+I3)2Ra + VI3+W/2

O/p of the generator =VI1

Then the generator efficiency can be derived by ηG = output/input = output/(output + losses)

= VI1/(VI1+ WG)

The advantages of Hopkinson’s test are

• Hopkinson’s test uses very less power
• It is economical
• This test can be done under full-load conditions so that a rise in temperature & commutation can be examined.
• Iron loss change because of flux distortion is taken into account because of the full-load condition.
• Efficiency can be determined at dissimilar loads.

### The Disadvantage of Hopkinson’s Test

The disadvantages of Hopkinson’s test are

• It is complicated to discover two equal machines required for this test.
• The two machines which are used in this test cannot be loaded evenly constantly.
• It is impossible to acquire separate iron losses used for the machines due to their excitations.
• It is tricky to control the machines at the required speed due to the change in field currents extensively.

### FAQs

1). Why field test is conducted even if the Hopkinson test is present?

This test on two equal series motors is not possible due to the instability of operation as well as the run-away speed

2). What is the purpose of the retardation test?

The retardation test is used to discover the efficiency of a stable speed dc machine. In this technique, we discover the losses of the machine-like mechanical & iron.

3). Why generator efficiency is more than motor?

Because the windings are thicker, low resistance & low copper losses

4). What are the various types of losses?

They are iron, windage, and friction

5). What is the polarity test?

The polarity test is used to know the direction of current in an electrical circuit

Thus, this is all about an overview of Hopkinson’s Test. It is one kind of technique for efficiency testing of a DC machine by connecting with each other. It is also known as a full load test. Here is a question for you, what are the applications of the Hopkinson’s test?