Learning Objectives:

• Understand the differences between two types of losses in a transformer.
• Calculate efficiency of a transformer by evaluating iron and copper losees.

• No load losses or core losses

  • No load losses remains the same irrespective of the load connected to the transformer.

  • It is the power consumed to sustain the magnetic field in the transformer’s core. It is of two types - hysteresis loss and eddy current loss

  • Hysteresis loss is the energy lost by reversing the magnetizing field in the core as the AC changes direction in every cycle.

  • Eddy current loss is a result of induced currents circulating in the core.

  • Hysteresis loss is minimized by using steel of high silicon content for the core Eddy current loss is minimized by using very thin laminations polished with varnish.

• Load losses or copper losses

  • They are associated with load current flow in the transformer windings

  • Copper loss is power lost in the primary and secondary windings of a transformer due to the Ohmic resistance of the windings.

Solved Example: 9978-01

Hysteresis and eddy current losses of single-phase transformer working on 200 V, 50 Hz is $P_h$ and $P_c$ respectively. The percentage decrease in $P_h$ and $P_c$, when the transformer operates on 160 V, 40 Hz supply, will respectively be: (ISRO Scientist Electrical 2019)

A. 32, 36

B. 25, 50

C. 20, 36

D. 40, 80

Correct Answer: C

Solved Example: 9978-02

A single phase transformer when supplied with 220 V, 50 Hz has eddy current loss of 50 W. If the transformer is connected to a voltage of 330 V, 50 Hz the eddy current loss will be: (ISRO (VSSC) Technical Assistant Electrical 2017)

A. 168.75 W

B. 112.5 W

C. 75 W

D. 50 W

Correct Answer: B

Solved Example: 9978-03

In a single-phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are: (GATE EE 2021)

A. 1600 W and 900 W

B. 900 W and 1600 W

C. 600 W and 250 W

D. 250 W and 600 W

Correct Answer: B

Solved Example: 9978-04

When are eddy-current losses in a transformer reduced? (ISRO Scientist Electrical 2013)

A. If laminations are thick

B. If number of turns in primary winding is reduced

C. If the number of turns in secondary winding is reduced

D. If laminations are thin

Correct Answer: D

Solved Example: 9978-05

Generally the no-load losses of an electrical machine is represented in its equivalent circuit by a: (SSC JE EE Mar 2017 Evening)

A. Parallel resistance with a low value

B. Series resistance with a low value

C. Parallel resistance with a high value

D. Series resistance with a high value

Correct Answer: C

Learning Objectives:

• Describe principle of working of an ideal transformer.
• Analyze operation of such a transformer.

Transformer is a static device. It transfer electrical energy from one part of the electrical or electronic circuit to other part of circuit without changing the frequency. It works on the Michael Faraday’s law of Electromagnetic Mutual Induction.

Principle:

When the current is provided to the primary winding it behaves as electromagnet due to this the EMF is induced in the secondary winding as it comes in the area having magnetic field lines due to primary Winding. \[V_s = N_s \dfrac{d\phi}{dt}\] where,
\(V_s\) = Secondary Winding Voltage
\(N_s\) = Secondary Winding Turns

Construction

• Laminated Core: Cores are designed not to have current in it. But still there is a leakages current Or the eddy current present in it. To minimize these current the cores are laminated.

• Winding: Made up of copper or aluminum coated with very thin layer of insulation.

• Tank: The main function of conservator tank of transformer is to provide adequate space for expansion of oil inside the transformer. It is usually Cylindrical or cubical.

• Isolation: It is provided by using Synthetic Oil between Core and Tank.

• Breather: When the temperature changes occur in Transformer insulating oil, the oil expands or contracts and there an exchange of air also occurs when transformer is fully loaded. When transformer gets cooled, the oil level goes down and air gets absorbed within.

\[\mathrm{Turns\ Ratio} = \dfrac{\mathrm{No.\ windings\ in\ Secondary}}{\mathrm{No.\ Windings\ in\ Primary}} = \dfrac{\mathrm{Voltage\ in\ Secondary}}{\mathrm{Voltage\ in\ Primary}} = \dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\]

Solved Example: 9167-01

A single-phase 111-V, 50-Hz supply is connected to a coil with 200 turns of a coil-core assembly. Find the magnitude of maximum flux in the core. (SSC JE EE March 2021 Morning)

A. 10 mWb

B. 2.5 mWb

C. 1 mWb

D. 25 mWb

Correct Answer: B

Learning Objectives:

• Determine the reflected impedance and draw an equivalent circuit based on it.

• In a transformer, the load 'seen' from the voltage source is different from the actual load.
• The effective resistance as seen by the primary source will depend upon the turns ratio of the primary and secondary sides of the transformer.
• The effective impedance as seen by the voltage source is called reflected impedance.
• Once the reflected impedance is evaluated, the ciruit can be considered like a simple resistive series circuit.
\[R_p = R \left(\dfrac{N_p}{N_s}\right)^2\]

Solved Example: 9169-01

Which is to be short circuited on performing short circuit test on a transformer? (SSC JE EE Jan 2018 Evening)

A. Low Voltage Side

B. High Voltage Side

C. Primary side

D. Secondary side

Correct Answer: A