Important Multiple choice Question Of Transformer 2022

Ques.91. From an open circuit test on a transformer, the no-load power factor was determined. Out of the following choices which is the most likely value.

  1. 0.4
  2. 0.9
  3. Unity
  4. 0.8

Answer. 1. 0.4

Explanation

  • Open circuit test or no load test on a transformer is performed to determine ‘no-load loss (core loss)’ and ‘no-load current I0‘ in the transformer.

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  • At load, the primary current has two components: Shunt branch current & secondary current when referred to the primary side.

  • At no load, the secondary current becomes zero due to open-circuited at secondary.

  • Therefore, only the shunt branch component is left which is used to magnetize the transformer coils. The current flowing in coils leads to winding losses. Hence, the current supplied by the source at no load is shunt branch current which is known as No Load Current.

  • No load current has two components (a)Iron loss component (b) Magnetizing Component.
  • Iron loss component: It is the component of no-load current responsible for the resistive loss in the core.
  • Magnetizing Component: It is the component of no-load current responsible for hysteresis loss in the core
  • In this method, the secondary of the transformer is left open-circuited.
  • A wattmeter is connected to the primary.
  • An ammeter is connected in series with the primary winding. A voltmeter is optional since the applied voltage is the same as the voltmeter reading.
  • Rated voltage is applied at primary.
  • If the applied voltage is normal voltage then normal flux will be set up. Since iron loss is a function of applied voltage, normal iron loss will occur. Hence the iron loss is maximum at rated voltage. This maximum iron loss is measured using the wattmeter.
  • As the secondary side is open, the entire coil will be purely inductive in nature. So, the power will be lagging due to the inductive property of the circuit. So LPF (Low Power Factor) Wattmeter is used in open circuit test of transformer.
  • So that from the given options most suitable power factor is 0.4 because it is the lowest among all of the given power factors.

 

Ques.92. A 20 kVA, 2000/200 V, single-phase transformer has a leakage impedance of 8%. What voltage applied to the HV side will result in full-load current flow in the LV side, when the LV side is short-circuited?

  1. 64 V
  2. 160 V
  3. 86 V
  4. 132 V

Answer.2. 160 V

Explanation

Rated short circuit current: It is the rated current in the windings under short circuit conditions corresponding to the reduced applied voltage.

%Z of transformer = percentage rated voltage required to produce rated short circuit current.

Calculation:

Given that leakage impedance = 0.08 p.u

Voltage develop across secondary at the time of short circuit = percentage impedance = 0.08 × 1 = 0.08 pu

The base voltage on the secondary side = 200 V

Voltage develop across secondary at the of short circuit = 0.08 × 200 = 16 V

The voltage applied on the primary side (16 × 2000/200) = 160 V

 

Ques.93. A transformer on no-load has a core loss of 50 W, draws a current of 2 A, and has an induced EMF of 230 V. Determine the no-load power factor and core-loss current.

  1. Power factor = 0.208 lagging, core-loss current = 0.416 A
  2. Power factor = 0.108 lagging, core-loss current = 0.216 A
  3. Power factor = 0.108 lagging, core-loss current = 2 sin(cos-1 0.108) A
  4. Power factor = 0.208 lagging, core-loss current = 2 sin(cos-1 0.208) A

Answer.2. Power factor = 0.108 lagging, core-loss current = 0.216 A

Explanation:-

Formula:

W0 = wattmeter reading

V1 = voltmeter reading

I0 = Ammeter reading

cosϕo  = No load power factor.

∴ Iron loss of the transformer, Pi = W0

W0 = V1 I0 cosϕ0

No-load power factor = cosϕ0  = W0/V1 I0

Working component or core loss component,

IW = I0 cos ϕ0

Calculation:

Given:

W0 = 50 W

V1 = 230 V

I­0 = 2 A

W0 = V1 I0 cos ϕ

50 = 230 × 2 × cos ϕ0

cos ϕ0  = 0.108 lagging

Core loss component

IW = 2 × 0.108 = 0.216 A

 

Ques.94. A short circuit test is performed on a transformer with a certain impressed voltage at rated frequency. If the short circuit test is now performed with the same magnitude of impressed voltage, but at a frequency higher than the rated frequency then the magnitude of current:

  1. And power-factor will both increase
  2. Will decrease but the power factor will increase
  3. Will increase, but power factor will decrease
  4. And power factor will both decrease

Answer.4. And power factor will both decrease

Explanation:-

The short-circuit test on the transformer is performed to determine the equivalent resistance and leakage reactance either referred to the primary side or referred to the secondary side.

Short-circuit-test-transformer

Since the H.V. rated current is less, a short-circuit test is conducted preferably on H.V. side keeping L.V short-circuited.

ower Factor (cos ϕ) = R / Z

Where,

R is the equivalent resistance

Z is the equivalent impedance

Z = R + j 2π f L

f is the supply frequency

L is the equivalent inductance

 cos ϕ = R / (R + j 2π f L)

Power factor during the short circuit test is inversely proportional to supply frequency

Current (I) = V / Z

Where,

V is the supply voltage

⇒ I = V / (R + j 2π f L)

The magnitude of current during the short circuit test is inversely proportional to supply frequency

Therefore as the frequency of a transformer increases both power factor and current decrease.

 

Ques.95. Non-loading heat run test on transformers is performed by means of

  1. SC test
  2. OC test
  3. Sumpner’s test
  4. Core balance test

Answer.3. Sumpner’s test

Explanation

With the help of Sumpner’s test both the open-circuit and short-circuit tests can be performed simultaneously. The heat run test can also be performed by this test. The only need of Sumpner’s test is that another identical transformer is required.

Sumpner’s test: is a non-loading heat run test on transformers

While OC and SC tests on a transformer yield its equivalent circuit parameters, these cannot be used for the ‘heat run’ test wherein the purpose is to determine the steady temperature rise if the transformer was fully loaded continuously; this is so because under each of these tests the power loss to which the transformer is subjected is either the core-loss or copper-loss but not both. The way out of this impasse without conducting an actual loading test is the Sumpner’s test which can only be conducted simultaneously on two identical transformers.

 

Ques.96. Which of the following is a type test for transformers?

 

  1. Temperature rise test
  2. Lightning impulse test
  3. Partial discharge test
  4. All of the above

Answer.2. Lightning impulse test

Explanation:-

Type tests of the transformer include:

  • Transformer winding resistance measurement
  • Transformer ratio test.
  • Design test
  • Production test
  • Applied voltage test
  • Transformer vector group test.
  • Measurement of impedance voltage/short circuit impedance (principal tap) and load loss (Short circuit test).
  • Measurement of no-load loss and current (Open circuit test).
  • Measurement of insulation resistance.
  • Lightning impulse test
  • Tests on on-load tap-changer.
  • Vacuum tests on tanks and radiators.

Lightning Impulse Test

Distribution lines are routinely disturbed by voltage surges caused by lightning strokes and switching transients. A standard impulse wave with a peak equal to the BIL (basic impulse insulation level) of the primary system is applied to verify that each transformer will withstand these surges when in service.

 

Ques.97. A 2000/200 V, 50 Hz single-phase transformer has an exciting current of 0.5 A and a core-loss of 600 W. When H.V side is energized by the rate voltage and frequency, the magnetizing current is

  1. 0.1 A
  2. 0.2 A
  3. 0.3 A
  4. 0.4 A

Answer.4. 0.4 A

Explanation:-

In open-circuit test,

No-load current I0 = 0.5 A

Core loss = 600 W

VI= 600

I= 600/2000 = 0.3A

Magnetising current $I_{M}=\sqrt{(0.5)^{2}+(0.3)^{2}}= 0.4A$

 

Ques.98. While performing a short circuit test on a single-phase 110/220 V, 50 Hz transformer with LV side shorted, wattmeter reading is found to be 20 W. If the same test is performed on the transformer with the HV side shorted, the wattmeter reading will be?

  1. 5 W
  2. 10 W
  3. 20 W
  4. 40 W

Answer.3. 20 W

Explanation:-

Short-Circuit Test

The short-circuit test on the transformer is performed to determine the equivalent resistance and leakage reactance either referred to the primary side or referred to the secondary side.

Short-circuit-test-transformer

  • Since the H.V. rated current is less, a short-circuit test is conducted preferably on H.V. side keeping L.V short-circuited.
  • For conducting a short-circuit test only a small percentage (of the order 5% to 10%) of the rated voltage is sufficient to cause the rated current to flow through the windings.
  • As the voltage is less, the core flux density is very much less compared to the normal value, so the iron loss which depends on flux density (voltage) can be neglected. The input is therefore equal to the total copper loss.
  • So, whatever power flows into the transformer thus gets dissipated, in the primary and secondary windings as well as the core.
  • Now if the same test is performed with the HV side short-circuited then the wattmeter reads the same reading.

Therefore, the wattmeter reads 20 W only.

 

Ques.99.  Which of the following transformer type does not have primary winding?

  1. Window type
  2. Shell type
  3. Core Type
  4. Air core type

Answer.1. Window type

Explanation

Window type Transformer

The window-type current transformer contains no primary winding. The current transformer has an insulated hole through the core and secondary winding. The circuit conductor is inserted through the window of the current transformer, and thus this conductor then becomes the primary of the current transformer.

 

Ques.100. In an auto-transformer, the number of turns in the primary winding is 210 and in the secondary winding is 140. If the input current is 60 A, the current in output and in common winding are respectively.

  1. 40A, 20A
  2. 40A, 100 A
  3. 90A, 30 A
  4. 90A, 150A

Answer.3. 90A, 30 A

Explanation

The auto-transformer ratio is given as

I1/I2 = N2/N1

60/I2 = 140/210

I2 = 90A

Current in common Winding Ic

Ic = (N2 – N1)I1/N2

=  60 x (210 – 140)/140

= 30 A

 


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