Question 1: Show from first principles the relationship between turns ratio (N1N 2 ), voltage ratio (

V 1V 2 ) and current ratio ( I 2

I 1).

Answer: In an ideal transformer, the rate of change of flux is the same for both primary and

secondary and thus E1N 1

= E2N 2 i.e. the induced emf per turn is constant. If no losses E1=V1 and

E2= V2. Therefore V 1N 1

=V 2N 2

∨V 1V 2

=N 1N 2 ……Equation 1

In an ideal transformer, losses are neglected therefore efficiency is 100%. This means input power= output power or V1I1=V2I2. Therefore

V 1V 2

= I 2I 1 …….Equation 2

Combining both equations 1 and 2: V 1V 2

= N 1N 2

= I 2I 1

Question 2: Identify with aid of diagrams the core construction of the two types of double-wound transformers and explain their differences.

Answer:

Core Type Double Shell Type Double Wound Transformer Wound Transformer

The shell type core has 3 limbs, the centre limb being a greater cross-sectional area than the outer limbs. Both primary and secondary coils are wound over the centre limb, no coils being wound on the outer limbs. The core type has only 2 limbs; primary and secondary coils being wound on each limb.

Question 3: Explain what is meant by “fringing”

Answer: The absence of a centre limb in the core type transformer will result in a distortion between the magnetic fields produced by both windings. This effect is known as fringing.

Question 4: Explain how the iron (core) loss is developed in a transformer.

Answer: Iron losses are caused when small leakage of current flow from the coil of the transformer onto the core of the transformer. It is a combination of hysteresis and eddy current losses.Hysteresis losses are energy losses associated with the continued magnetisation and demagnetisation of the core.Eddy current losses are energy losses due to circulating currents resulting from voltages induced into the core by the winding.

Question 5: Explain how the Cu. (I2R) loss is developed in a transformer.

Answer: Copper loss or I2R loss is caused due to the resistance of the primary and secondary winding. When electrical energy is converted into heat energy in the windings, this loss is developed during the energy conversion also due to the coils resistance, the power loss in both windings, [(I1

2R1) + (I22R2)].

Question 6: State the condition for maximum efficiency in a transformer.

Answer: The condition for maximum efficiency of a transformer is when the variable, Copper loss, is equal to the constant, Iron loss.

Question 7: Sketch and explain the equivalent circuit of a transformer:(a) On load(b) On no-load

Answer: a)

Equivalent Circuit of an ON LOAD Transformer

R and X are omitted since the no load current, Io is normally about 3-5% of the full load primary current.

b)

Equivalent Circuit of an NO LOAD Transformer

R1 and R2 represent the resistances of the primary and secondary winding and X1 and X2 represent the reactance of the primary and secondary windings, due to leakage flux. The core

V1

losses due to hysteresis and eddy current are allowed for by resistance, R which takes a current Ic, the core loss component of the primary current. Reactance, X takes the magnetising component, Im.

Question 8: Define voltage regulation of a transformer.

Answer: When the secondary of a transformer is loaded, the secondary terminal voltage, V2 falls. As the power factor decreases, this voltage drop increases. This is called voltage regulation of a transformer. It is usually expressed as a percentage or per unit of the secondary no load voltage, E2.

Voltage Regulation = (E2−V 2E2

)×100

Question 9: a) Sketch and explain the sandwich type construction for 1∅ transformer.

b) State its main advantage over the concentric type

Answer:

a)

Sandwich Type Winding for a 1ф Transformer

The sandwich type transformer has its primary and secondary winding lapping on each other with little or no iron core separating the windings.

b) The main advantage of the sandwich type winding is that there is less losses hence the transformer operates at a higher frequency

Question 10: Sketch and explain the following:a) Open circuit testb) Short circuit test

State the purpose of each of the tests mentioned above.

Answer:

a)

OPEN CIRCUIT TEST

The ratio of the voltmeter readings, V 1V 2 , gives the ratio of the number of turns. Ammeter, A

gives the no load current, and its reading is a check magnetic quality of the ferro-magnetic core and joints. The wattmeter, W, reading indicates the Iron Loss.

b)

SHORT CIRCUIT TEST

The short circuit test is used to determine the copper loss of a transformer where the reading obtained from the wattmeter, W will indicate the Copper Loss.

Question 11: Identify two methods to reduce the iron loss in a transformer.

Answer: Two methods to reduce the iron loss in the transformer are:1. The core of the transformer is laminated

2. Vanish is applied to the coreQuestion 12: State the relationship between load, power factor and efficiency of a transformer.

Answer: Efficiency of a transformer will depend on the operating power factor as well as the load since the efficiency varies directly proportional to the load and the power.

Question 13: a) Explain how iron loss varies with load. b) Explain how copper loss varies with load

Answer: a) The iron loss remains constant despite changes in the load.

b) The copper loss varies in squared proportion with respect to changes in the load.

Question 14: Sketch phasor diagram for the following as it relates to transformers:(a) No-load(b) On-load

Answer: a) Phasor diagram for a no load transformer

b) The phasor diagram for an on load transformer

Question 15: Sketch and explain the distribution of currents in an auto-transformer.

Answer:

Auto- transformer Circuit

The circuit above shows the secondary being part of the primary, the current in the secondary being: (I2-I1).Question 16: Identify (3) advantages and (1) disadvantage of the auto-transformer compared to the double-wound transformer.

Answer: Advantages: It is cheaper since less copper is used Less volume, hence less weight A higher efficiency, resulting from lower I2R losses A continuously variable output voltage is achievable if a sliding contact is used A smaller percentage voltage regulation.

Disadvantage: The primary and secondary windings are not electrically separate; hence if an open circuit occurs in the secondary winding, the full primary voltage appears across the secondary.

Question 17: Explain (1) advantage of the shell type core construction compared to the core type.

Answer: one advantage of the shell type core construction compared to the core type is the flux leakage is less in the shell type than of the core type.

Question 18: State two suitable uses of an auto-transformer.

Answer: Auto-transformers are used for reducing the voltage when starting induction motors and for interconnecting systems that are operating at approximately the same voltage.

Question 19: Identify the relationship between the secondary voltage of a transformer between no-load and full-load.

Answer: the relationship is on no load the secondary voltage is E2 and it is larger than the secondary voltage on full load which is V2. On full load V2<E2.

Question 20: Explain why an auto-transformer with a high voltage transformation ratio is not suitable for certain applications.

Answer: Since the primary and secondary windings electrically connected; hence if an open circuit occurs in the secondary winding, the full primary voltage appears across the secondary.