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DC Machines Multiple Choice Questions on “Hopkinson’s Test”.
1. Hopkinson’s test of D.C. machines is conducted at _______________
a) No-load
b) Part load
c) Full-load
d) Overload
Answer: c
Clarification: Unlike Swinburne’s test Hopkinson’s test is carried out at loaded condition. Thus, we get stray load loss also, while finding out the efficiency. Hence, efficiency is not over-estimated like Swinburne’s test.
2. Hopkinson’s test requires ____________
a) One DC machine on which test is carried out
b) Two different DC machines
c) Two identical DC machines
d) Can be worked with one or two machines
Answer: c
Clarification: This is a regenerative test in which two identical dc shunt machines are coupled mechanically and tested simultaneously. One of the machines is used as a motor driving while the other one acts as a generator which supplies electric power to motor.
3. In Hopkinson’s test, two machines are connected in ______________
a) Series
b) Parallel
c) Can be connected in parallel or series
d) Two machines are not required
Answer: b
Clarification: The two machines are made parallel by means of switch S after checking that similar polarities of the machine are connected across the switch. Here, one machine is driving another machine.
4. What will happen if field current of generator in Hopkinson’s test is increased?
a) Current through motor armature will increase
b) Current through motor armature will decrease
c) Current through motor armature will remain constant
d) Motor armature current cannot be determined
Answer: a
Clarification: If field current through generator is increased, back emf of generator will increase thus it’ll become greater than back emf of a motor, so to compensate this effect armature current in generator will increase thus, motor armature current will also increase.
5. What will happen if field current of motor in Hopkinson’s test is decreased?
a) Current through motor armature will increase
b) Current through motor armature will decrease
c) Current through motor armature will remain constant
d) Motor armature current can’t be determined
Answer: a
Clarification: If field current through motor is decreased, speed of the motor will increase due to inverse proportionality, back emf of generator will increase thus it’ll become greater than back emf of a motor, so to compensate this effect armature current in generator will increase thus, motor armature current will also increase.
6. Hopkinson’s test is a regenerative test.
a) True
b) False
Answer: a
Clarification: Hopkinson’s test is a regenerative test, because the power drawn from the mains is only that needed to supply losses. The test is, therefore, economical for long duration test like a “heat run”.
7. For carrying out load test on Hopkinson’s test setup _________________
a) Actual load is needed
b) By changing field currents in two machines load can be changed
c) Can’t carry out
d) By changing the armature current test is carried out
Answer: b
Clarification: There is no need to arrange for actual load (loading resistors) which apart from the cost of energy consumed, would be prohibitive in size for large-size machines. By merely adjusting the field currents of the two machines, the load can be easily changed and a load test conducted over the complete load range in a short time.
8. Hopkinson’s test gives _______________
a) Combined iron losses of two machines which can be separated
b) Combined iron losses of two machines which can’t be separated
c) Doesn’t include iron losses
d) Depends on actual setup
Answer: b
Clarification: From Hopkinson’s test both machines are not loaded equally and this crucial in small machines. Thus, it is important to know the separate iron losses for given machines. But, test gives combined losses which are different for different machines as excitation differs.
9. Hopkinson’s test is suitable for ______________
a) Small machines only
b) Small and medium machines
c) All machines
d) Only large machines
Answer: d
Clarification: A large variation of field currents is required for small machines, the full-load set speed is usually higher than the rated speed and the speed varies with load. The full load in small machines cannot obtained by cutting out all the external resistances present in the generator field. Sufficient reduction in the motor field current is necessary to achieve full-load conditions resulting in speeds greater than the rated value.
10. Why field test is conducted even if Hopkinson’s test is present?
a) Instability of an operation
b) Possibility of run-away speed
c) Both instability and possibility of run-away speed
d) Field test is not conducted
Answer: c
Clarification: Regenerative test on two identical series motors is not feasible because of instability of such an operation and the possibility of run-away speed. Therefore, we have toconduct a loading test.
11. In field’s test generator field and motor field are connected in ______________
a) Series
b) Parallel
c) Alternatively, series and parallel
d) Not connected
Answer: a
Clarification: The generator field is connected in series with motor field circuit. The generator is thus separately excited and its excitation is identical to that of motor at all loads. This ensures that the iron-loss of both the machines are always equal.