Hopkinson’s Test

[vc_row][vc_column width=”2/3″][vc_column_text]

  • Hopkinson’s Test is also known as Regenerative Test, Back to Back test and Heat Run Test.
  • This as an elegant method of testing DC machines.
  • Here it will be shown that while power drawn from the supply only corresponds to no load losses of the machines, the armature physically carries any amount of current (which can be controlled with ease).
  • Such a scenario can be created using two similar mechanically coupled shunt machines.
  • Electrically these two machines are eventually connected in parallel and controlled in such a way that one machine acts as a generator and the other as motor.
  • In other words two similar machines are required to carry out this testing which is not a bad proposition for manufacturer as large numbers of similar machines are manufactured.

Figure 1.Hopkinson’s Test: Machines before Paralleling


  1. Connect the two similar (same rating) coupled machines as shown in figure 1.
  2. With switch S opened, the first machine is run as a shunt motor at rated speed.

It may be noted that the second machine is operating as a separately excited generator because its field winding is excited and it is driven by the first machine.

Now the question is what will be the reading of the voltmeter connected across the opened switch S?

  • The reading may be
    1. either close to twice supply voltage or
    2. Small voltage.
  • In fact the voltmeter practically reads the difference of the induced voltages in the armature of the machines. The upper armature terminal of the generator may have either + ve or negative polarity. If it happens to be +ve, then voltmeter reading will be small otherwise it will be almost double the supply voltage.
  • Since the goal is to connect the two machines in parallel, we must first ensure voltmeter reading is small.
  1. In case we find voltmeter reading is high, we should switch off the supply, reverse the armature connection of the generator and start afresh.
  2. Now voltmeter is found to read small although time is still not ripe enough to close S for paralleling the machines. Any attempt to close the switch may result into large circulating current as the armature resistances are small.
  3. Now by adjusting the field current Ifg of the generator the voltmeter reading may be adjusted to zero (Eg ≈ Eb) and S is now closed.
  4. Both the machines are now connected in parallel as shown in figure 2.

Figure 2.Hopkinson’s Test: Machines Paralleled

Loading the machines

  • After the machines are successfully connected in parallel, we go for loading the machines i.e., increasing the armature currents. Just after paralleling the ammeter reading A will be close to zero as Eg ≈ Eb.
  • Now if Ifg is increased (by decreasing Rfg), then Eg becomes greater than Eb and both Iag and Iam increase,
  • Thus by increasing field current of generator (alternatively decreasing field current of motor) one can make Eg > Eb so as to make the second machine act as generator and first machine as motor.
  • In practice, it is also required to control the field current of the motor Ifm to maintain speed constant at rated value.
  • The interesting point to be noted here is that Iag and Iam do not reflect in the supply side line.
  • Thus current drawn from supply remains small (corresponding to losses of both the machines).
  • The loading is sustained by the output power of the generator running the motor and vice versa.
  • The machines can be loaded to full load current without the need of any loading arrangement.

Calculation of efficiency

Let field currents of the machines be are so adjusted that the second machine is acting as generator with armature current Iag and the first machine is acting as motor with armature current Iam as shown in figure 2.

Also let us assume the current drawn from the supply be I1. Total power drawn from supply is VI1 which goes to supply all the losses (namely Cu losses in armature & field and rotational losses) of both the machines, now:

Since speed of both the machines are same, it is reasonable to assume the rotational losses of both the machines are equal; which is strictly not correct as the field current of the generator will be a bit more than the field current of the motor, Thus,

Once Prot is estimated for each machine we can proceed to calculate the efficiency of the machines as follows,

Efficiency of the motor

As pointed out earlier, for efficiency calculation of motor, first calculate the input power and then subtract the losses to get the output mechanical power as shown below,

Efficiency of the generator

For generator start with output power of the generator and then add the losses to get the input mechanical power and hence efficiency as shown below,

Condition for maximum efficiency

We have seen that in a transformer,

Maximum efficiency occurs when copper loss = core loss,

where, copper loss is the variable loss and is a function of loading while the core loss is practically constant independent of degree of loading. This condition can be stated in a different way:

Maximum efficiency occurs when the variable loss is equal to the constant loss of the transformer.

Maximum efficiency for motor mode

Let us consider a loaded shunt motor as shown in figure 3. The various currents along with their directions are also shown in the figure.

Figure 3.Machine Operate as a Motor

We assume that field current If remains constant during change of loading.


Now the armature copper loss can be approximated to as  as Ia ≈ IL .This is because the order of field current may be 3 to 5% of the rated current. Except for very lightly loaded motor, this assumption is reasonably fair. Therefore replacing Ia by If in the above expression for efficiency m , we get,

Thus, we get a simplified expression for motor efficiency m in terms of the variable current (which depends on degree of loading) IL, current drawn from the supply.

So to find out the condition for maximum efficiency, we have to differentiate m with respect to IL and set it to zero as shown below.

Maximum efficiency for Generation mode

Similar derivation is given below for finding the condition for maximum efficiency in generator mode by referring to figure 4.

Figure 4. Machine operates as a Generator

We assume that field current If remains constant during change of loading.


Thus, we get a simplified expression for motor efficiency g in terms of the variable current

( which depends on degree of loading) IL, current delivered to the load.

So to find out the condition for maximum efficiency, we have to differentiate g with respect to IL and set it to zero as shown below.

Thus, maximum efficiency both for motoring and generating are same in case of shunt machines. To state, we can say at that

Armature current maximum efficiency will occur which will make variable loss = constant loss.

Eventually this leads to the expression for armature current for maximum efficiency as

Advantages of Hopkinson’s Test

The main advantages of using Hopkinson’s test are as follows:-

  1. This method is very economical.
  2. The temperature rise and the commutation conditions can be checked under rated load conditions.
  3. Stray losses are considered, as both the machines are operated under rated load conditions.
  4. Large machines can be tested at rated load without consuming much power from the supply.
  5. Efficiency at different loads can be determined.

Disadvantage of Hopkinson’s Test

  1. The main disadvantage of this method is the necessity of two practically identical machines for performing the Hopkinson’s test. Hence, this test is suitable for large DC machines.

[/vc_column_text][/vc_column][vc_column width=”1/3″][/vc_column][/vc_row][vc_row][vc_column width=”2/3″][vc_column_text]AUTHORS
1.Bunty B. Bommera
2.Dakshata U. Kamble[/vc_column_text][/vc_column][vc_column width=”1/3″][/vc_column][/vc_row]

Leave a Reply