1. The total core loss can be termed as ____________
a) Eddy current loss
b) Hysteresis loss
c) Copper loss
d) Magnetic loss
Explanation: The total core loss is due to iron core or any core material used. As iron loss is proportional to magnetic flux density or flux, these are also called as magnetic loss. The total core loss or magnetic loss in any given transformer totally consists of eddy current loss and hysteresis loss.
2. 2 KVA, 230 V, 50 Hz single phase transformer has an eddy current loss of 40 watts. The eddy current loss when the transformer is excited by a dc source of same voltage will be ___________
a) Equal to 40W
b) Less than 40W
c) More than 40W
d) Zero watts
Explanation: Eddy current loss is directly proportional to the frequency^2. So, for DC current frequency is equal to 0 Hz. Thus, eddy current losses being directly proportional to square of frequency they’ll be equal to 0
3.Which of the following is the correct formula for Bm?
a) Bm= (Yo2-Gi2)(1/2)
b) Bm= (Yo2+Gi2)(1/2)
c) Bm= (Yo2-Gi2)(2)
d) Bm= (Yo2+Gi2)(1/2)
Explanation: We get the value of Y0 from the no-load current and voltage reading as, Io/V1. Similalry we get the value of Gi from output power and voltage reading as, Po/V1. It then follows that, Bm= (Yo2-Gi2)(0.5).
4. How shunt branch component Y0 is calculated?
a) I0/V1
b) V1/I0
c) P0/V12
d) Cannot be determined
Explanation: Shunt branch admittance is defined as inverse of shunt branch impedance. As we know, impedance can be calculated by the simple ohm’s law; admittance is equal to the inverse of the impedance
5. While conducting short-circuit test on a transformer which side is short circuited?
a) High voltage side
b) Low voltage side
c) Primary side
d) Secondary side
Explanation: It’s a common practice to conduct SC test from HV side, while keeping LV side short circuited. Thus, short circuited current is made to flow from shorted low voltage terminals i.e. LV side.
6. During short circuit test why iron losses are negligible?
a) The current on secondary side is negligible
b) The voltage on secondary side does not vary
c) The voltage applied on primary side is low
d) Full-load current is not supplied to the transformer
Explanation: Very small amount of voltage is given to the transformer primary thus the magnetic losses which are dependent on magnetic flux density will get minimum value, hence iron losses are negligible.
7. Short circuit test on transformers is conducted to determine ______
a) Core losses
b) Copper losses
c) Hysteresis losses
d) Eddy current losses
Explanation: Short circuit test is used to determine the copper losses taking place in the transformer under operation, while open circuit test gives us the value of core losses taking place in the transformer.
8. When a short circuit test on a transformer is performed at 25 V, 50 Hz, the drawn current is I1. If the test is performed by 25 V and 25 Hz and power drawn current is I2, then
a) I1 > I2
b) I1 < I2
c) I1 = I2
d) Can’t be defined
Explanation: Current by ohm’s law is equal to voltage divided by impedance. So, I=V/Z. Here Z is inductive load, thus Z= 2πfL. So as the frequency decreases the impedance also decreases and ultimately it reduces the denominator term causing increase in current.
9. Why SC test is not conducted on LV side?
a) Difficult to arrange low voltage supply
b) Difficult to arrange high current supply
c) Difficult to arrange low voltage and high current supply to the LV
d) SC test on LV does not give correct results
Explanation: If rated voltages and power is considered we need only 5% of rated voltage to be applied at on HV side, while by calculations current requirement is also less. For the same test on LV side though voltage required is less compare to HV side, current required is very high.
10. SC test gives ______________
a) Series parameters of equivalent circuit
b) Parallel parameters of equivalent circuit
c) Both parameters of equivalent circuit
d) Neither series nor parallel parameter of equivalent circuit
Explanation: Short circuit test gives the copper losses; these losses are taken into consideration by series parameters of the equivalent circuit. While, Open circuit test gives us iron losses; which are shown by parallel components of equivalent circuit