SINGLE PHASE TRANSFORMERSCompiled and presented by Mr. D. Nedrick

*What is a transformerA transformer is a device used for changing voltage values in a circuit.

The change is brought about by using the magnetic effect of an alternating current.

*Induced e.m.f. (electromotive force) When a conductor cuts, or is cut by, magnetic lines of force, an e.m.f. (voltage) is induced into that conductor.

*Self-induced e.m.f The e.m.f. which is induced in a coil due to the changing flux in the coil cutting the conductors of the coil. This self-induced e.m.f. is in opposition to the voltage producing it. The self-induced e.m.f, tends to limit the current in the circuit: it chokes the current flowing in a coil when it is carrying an alternating, or changing, current. The greater the change in the current the greater the e.m.f.

*Self Induced EMF

*E.m.f. due to Mutual Inductance When a coil A carrying an alternating, or changing, current is placed beside another coil B, an e.m.f. will be induced into coil B. This e.m.f. induced into coil B, termed the e.m.f. of mutual inductance, is in opposition to the force producing it.

*Mutual InductanceNOTE: There is no direct electrical connection between the two coils, only a magnetic connection (i.e., a magnetic linkage).

*Double-wound TransformerConstruction: The construction of the double-wound transformer is as follows:

(1) primary winding (insulated copper conductors)(2) secondary winding (insulated copper conductors)(3) core (silicon steel laminations)(4) INSULATION

Construction contd*

*1. There are two coils, insulated from one another, and wound round a silicon steel core.2. The windings consist of insulated copper conductors, wound on a bobbin or former of insulating material.

*3. The Core of the TransformerThe core of the transformer is made up of layers of stampings, or laminations of silicon steel. These sheets of steel are insulated from one another with shellac, paper, or kaolin. Silicon steel is used because it retains very little magnetism and provides a low reluctance (magnetic 'resistance') path for the magnetic lines of force. There are two basic types of core: (a) the core type; and (b) the shell type. The vertical parts of the core are termed the limbs.

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*The Core4. The larger transformers are fitted into a welded sheet-steel case which is filled with oil for cooling purposes.

*Cycle of Operation There are two distinct circuits: (a) the primary winding (coil A) to which the incoming supply is connected, and (b) the secondary winding (coil B) across which the load is connected.

*How Does the Transformer Work?

*ContdThe cycle of operation is as follows:1. An alternating current supply (IA) flows in the primary winding A. 2. This alternating current produces an alternating magnetic field linking the primary and secondary windings.3. The variation of magnetic flux induces (a) an e.m.f. of self-inductance into the primary winding, and (b) an e.m.f. of mutual inductance into the secondary winding.

*Contd4. When a load is connected across the secondary winding B a current (IB) flows through it.5. The field due to this alternating secondary current (IB) has a demagnetizing effect on the primary winding and to neutralize this effect a greater current must flow in the primary winding.

*Transformation Ratio The ratio of the input voltage (the primary voltage VA) to the output voltage (VB) is determined by the ratio of the turns.

*ContdFor example: if the primary winding has 500 turns and the secondary winding has 1000 turns, the ratio will be two to one (2:1) step-up. Thus 100V applied across the primary will produce 200V across the secondary. The ratio is the same for voltage current and the number of turns hence:

*ContdTurns ratio = Vs/Vp; Ns/Np; Ip/Is therefore the formulaVs/Vp = Ns/Np = Ip/Is (the use of any two ratio constitute a formula)Where: Vs = Secondary Voltage Vp = Primary voltage Ns = Number of turns on the secondary Np = Number of turns on the primary Is = Current through the secondary coil Ip = Current through the primary coil

*ExampleA transformer has a step-down ratio of 20:1. Calculate the secondary voltage when the primary is supplied at 240V d.c. Trick you, transformer is used isolate dc circuits anyways work it a.c.NP = VPNs VS cross multiplyNP x VS = VP x NSVS = VP x NS NP

*Example SolutionVS = 240V x 1 20 = 12V How much current would flow if a 6 load is connected across the secondary winding?I = V/R = 12V/6 = 2A

*Solution 2Calculate the primary Current.NP/NS = IS/IP Cross multiplyNP x IP = IS x NSIP = IS x NS NP = 2A x 1t 20t = 0.1A

Exercise1. Calculate the secondary voltage in a double-wound transformer having a step- down ratio of 30:1. The primary voltage is 150V.2. A double-wound transformer is used to supply 50 V from the 250 V mains. The primary winding contains 1500 turns. Find (a) the number of secondary turns and (b) the secondary current when the primary current is 5 A.

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*Volts Per TurnVolts per turn in a transformer are calculated by dividing the voltage across the winding by the number of turns in the winding.

Volts per turn =

winding voltage no. of turns

*Example: A double-wound transformer has a 240V primary consisting of 2400 turns. Calculate the 'volts per turn'.

Volts per turn =

240V 2400t

0.1

winding voltage no. of turns

Volts per turn =

Volts per turn =

Exercise1. Determine the respective number of turns in each winding of a double-wound transformer, with a step-down ratio of 6000V to 250V if the volts per turn are 2.

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*Auto-transformerThis type of transformer uses the principle of self-induction. The primary and secondary are on the same winding and the lines of force from the primary cut the turns of the secondary and induce a voltage into them.

*LimitationsNOTE. This transformer may be used for step-up or step-down transformation, but its practical application is limited, because there is a direct electrical connection between the input supply and the output. The ratio is not greater than 2 to 1 unless special precautions are taken.

Dual Voltage Transformer*

*Losses in TransformersA transformer is a highly efficient piece of equipment, mainly because there are no moving parts. There are two sources of loss.1. Copper losses: This loss is a heat loss due to the current flowing through the copper of the windings. It is termed an I2R loss.2. Iron losses. These are of two types:(a) Losses due to eddy currents (b) Hysteresis losses

*Eddy current Eddy currents are alternating currents which are induced into the metal core of the transformer by the alternating field in the core.

*Eddy CurrentThis loss is minimized by using laminations. The laminations are insulated from one another to keep the eddy current paths separate, to reduce the e.m.f., and to increase the resistance per path.

*Hysterisis LossHysteresis losses are due to the energy used in the core during the changing cycle of magnetism. A certain amount of magnetism remains after the current has collapsed. This retained, or residual, magnetism must be neutralized and the energy used to neutralize it represents a loss. This loss is minimized by using a core in which the residual magnetism is small: silicon steel is the most common core material as it retains little magnetism and provides a low reluctance (magnetic resistance') to the lines of force.

*Calculation of Transformer Efficiency A transformer is the most efficient machine because it has no moving parts and has its efficiency usually about 90% to 98%. Efficiency, at any power factor, may be calculated as follows:Efficiency percent = Output Power x 100 Input Power

*Relationship between VA input and VA output: Assuming that the power factor is unity (i.e., kVA = kW) and that losses are negligible (power input = power output), thenVs x Is = Vp x Ip

*ExampleA single phase step down transformer having a ratio of 10:1 has a primary voltage of 6600V and a load of 13.2kVA. Since losses are negligible:VA (primary) x IA = VB (Secondary) x IBNP /NS = VP/VSVS = 6600V/10 = 660V

*Finding Secondary CurrentSince losses are negligible:Input = Output13200VA = 660V x II = 20A

ExerciseA 240V to 4800V transformer has a primary current of 95 A and a secondary current of 4A. What is its efficiency?*

*Types of Cooling The losses in a transformer are generated in the form of heat and it is essential that the temperature is kept below a certain level (generally about 80 C). This is achieved by one of the following methods1. Air cooling (AN). This method is used with small transformers. Convection currents in the air dissipate the heat.2. Oil immersion (ON). In this method the transformer winding is immersed in a tank containing oil.

*Oil ImmersionThe heat is transferred from the windings through the oil; the dissipation of the heat is assisted by increasing the surface area of the tank with cooling tubes.Variations of the above methods are also used, for example, forced air blast (AB) or the forced circulation of oil with a pump assisted by water cooling (OFW).

*RatingsA transformer is rated in kilovolt-amperes (i.e., apparent power). Since the heating of the windings, and hence temperature rise, depends on the current, true power (kW) is not used because the current through the windings will increase as the power factor decreases but the kilowatts will remain the same.

*ApplicationsThe main application of the transformer is in distribution systems. The effectiveness of the transformer in this application may be illustrated by the following examples:A 132,000V (132kV) distribution line has a total resistance of 1 and carries a current of 30A. Calculate the voltage drop in the line as a percentage of the total voltage. Since V = I x R V = 30A x 1 = 30V

*Example

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*Further Applications Welding (supplying a low voltage-high current output), supplies to rectifiers and radio equipment. The auto-transformer is often used for motor starting, and for discharge lamp circuits.

*Past paper questionsA double-wound power transformer has ten times as many primary windings as the secondary windings. The primary winding of the transformer is connected to a 120 V a.c. supply. The secondary winding of the transformer is connected to a load of 60 W. Assume the transformer losses to be negligible. Calculate the(i) secondary voltage(ii) current in the primary winding(iii) number of turns in the primary winding if there are 300 turns on the secondary winding.