# Hopkinson’s test

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## Hopkinson’s Test

Hopkinson’s test is another useful method of testing the efficiency of a dc machine. Hopkinson’s test is also called back-to-back test or regenerative test. It is a full load test and it requires two identical machines which are coupled to each other. One of these two machines is operated as a generator to supply the mechanical power to the motor and the other is operated as a motor to drive the generator. For this process of back to back driving the motor and the generator,

Two identical DC shunt machines are mechanically coupled and electrically connected in parallel across the DC supply. By adjusting the field excitation of the machine, one is run as a motor and the other as a generator.

The electric power from the generator and electrical power from the DC supply are fed to the motor. The electric power given to the motor is converted into mechanical power, the rest going to the various motor losses. This mechanical power is given to the generator. The electrical power of the generator is given to the motor except that which is wasted as generator losses.

The electrical power taken from the DC supply is the sum of motor and generator losses. this power can be measured by digital multi-meter . Since the power input from the DC supply is equal to the power required to supply the losses of the two machines. this test can be carried out with a small amount of power.

By adjusting the field strengths of the machines, any load can be put on the machines. Therefore, we can measure the total loss of the machines at load.

### Circuit Diagram of the Hopkinson’s Test

When the voltage of the generator is equal and the same polarity as the of the busbar supply voltage, the main switch S is closed, and the generator is connected to the busbar. Thus, both the machines are now in parallel across the supply. Under this condition, when the machines are running parallel, the generator are said to float. This means that the generator is neither taking any current nor giving any current to the supply.

Now with the help of a field rheostat, any required load can be thrown on the machines by adjusting the excitation of the machines with the help of field rheostats.

Armature copper losses in the generator = I2² rg

Armature circuit copper losses in the motor = (I1 + I2)² rm

Total power drawn by the armature circuit of the motor = V x I1 watts

Let the sum of iron losses and mechanical losses of each machine be Wc, then

V x I1 = 2Wc + I2 ² rg + (I1 + I2)² rm

Thus, Wc = ½ [ V x I1 – I2 ² rg – (I1 + I2) ² rm ]

#### Efficiency of Motor

Shunt field copper losses of the motor = V x I3

Hence, total losses of the motor = Wc + (I1 + I2) ² rm + V x I3

Total power input to the motor, P1 = V x (I1 + I2 + I3)

1.  Power drawn from the supply is low.
2. Both the DC machines are operating under loaded conditions, can such stray load losses are taken into account.
3. The temperature rise and the commutation conditions can be checked under rated load conditions.
4. Efficiency at different loads can be determined.