The AutoTransformer

Objective:

• To study the voltage and current relationship of an autotransformer.
• To learn how to connect a standard transformer as an autotransformer.

Discussion:

There is a special type of transformer which physically has only one winding. Functionally, though, the one winding serves as both the primary and secondary. This type of transformer is called an autotransformer. When an autotransformer is used to step up the voltage, part of the single winding acts as the primary, and the entire winding acts as the secondary. When an autotransformer is used to step down the voltage, the entire winding acts as the primary, and part of the winding acts as the secondary.

Figure-1 and Figure-2 show autotransformers connected for both step-up and step-down operation.

The action of the autotransformer is basically the same as the standard two-winding transformer. Power is transferred from the primary to the secondary by the changing magnetic field, and the secondary in turn, regulates the current in the primary to set up the required condition of equal primary and secondary power. The amount of step-up or step-down in voltage depends on the turn’s ratio between the primary and secondary, with each winding considered as separate, even though some turns are common to both the primary and secondary.

Voltages and currents in the various windings can be found by two simple rules:

a) Primary apparent power (VA) equals Secondary apparent power (VA).

(VA)_P = (VA) _S---------------- (1)

E_PI_P = E_SI_S------------------ (2)

b) The primary (source) voltage and the secondary (load) voltage are directly proportional to the number of turns N.

These equations depend upon one important fact, that voltage E_{A to B} and E_{B to C} add in the same direction and do not oppose each other. We have assumed that the voltages are in phase.

The load current, of course, cannot exceed the current carrying capacity of the winding. Once this is known it is relatively easy to calculate the VA load which a particular autotransformer can supply.

A disadvantage of the autotransformer is the lack of isolation between the primary and secondary circuits, because the primary and secondary both use some of the same turns.

Equipment Required:

• A Single Phase Transformer
• Power Supply
• Resistive Load
• AC Ammeter
• AC Voltmeter
• Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Using a Single-Phase Transformer, Power Supply, Resistive Load, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure-3. Note that winding 5 to 6 is connected as the primary winding across the 220 V ac source. The centre tap of the winding, terminal-9 is connected to one side of the load and the 6 to 9 portion of the primary winding is connected as the secondary winding.

2. a. Place all of the Resistive Load switches in their open positions for zero loads current.

b. Turn on the power supply and adjust for exactly 220 V ac as indicated

by voltmeter E₁. (This is the rated voltage for winding 5 to 6).

c. Adjust the load resistance R_L to 4400 Ω.

d. Measure and record currents I₁, I₂ and the output voltage E₂.

I₁ = 0.01 A ac

I₂ = 0.02 A ac

E₂ = 105 V ac

      e. Return the voltage to zero and turn off the power supply.

3. a. Calculate the apparent power in the primary and secondary circuits.

        E₁ (220) × I₁ (0.01) = 2.2 (VA) _P

        E₂ (105) × I₂ (0.02) = 2.1(VA) _S

    b. The primary and secondary apparent powers are not equal. Because we can see from Figur-1; the primary winding turns is greater than the secondary winding turns. So the secondary winding apparent power is less than the primary winding apparent power. This is a step-down Autotransformer.

4. Connect the circuit shown in Figure-3. Notice that winding 6 to 9 is now connected as the primary winding across the 110 V ac source. The 5 and 6 winding is now connected as the secondary winding.

5. a. Place all of the Resistive Load switches in their open positions for zero load current.

   b. Turn on the power supply and adjust for exactly 110 V ac as indicated by voltmeter E₁.(This is the rated voltage for winding 6 to 9.

   c. Adjust the load resistance R_L to 2200 Ω.

   d. Measure and record currents I₁, I₂ and the output voltage E₂.

I₁ = 0.01 A ac

I₂ = 0.10 A ac

E₂ = 210 V ac

   e. Return the voltage to zero and turn off the power supply.

6. a. Calculate the apparent power in the primary and secondary circuits.

        E₁ (110) × I₁ (0.01) = 1.1 (VA) _P

        E₂ (210) × I₂ (0.10) = 21(VA) _S

    b. The primary and secondary apparent powers are not equal. Because we can see from Figur-2; the primary winding turns is less than the secondary winding turns. So the secondary winding apparent power is greater than the primary winding apparent power. This is a step-up Autotransformer.

Transformers in Parallel

Objective:

    · To learn how to connect transformers in parallel.

    · To determine the efficiency of parallel-connected transformers.

Discussion:

Transformers may be connected in parallel to furnish load currents greater than the rated current of each transformer. There are two precautions to be observed when connecting transformers in parallel.

1. The windings to be paralleled must have identical output voltage ratings.

2. The windings to be paralleled must have identical polarities.

Very large short-circuit currents can be developed if these rules are not followed. In fact, transformers, circuit breakers and associated circuitry can be severely damaged, or may even explode, if these short-circuit currents are large enough.

The efficiency of any machine or electrical device is given by the ratio of output power to input power. (Apparent power and reactive power are not used in calculating transformer efficiency). The equation for percent efficiency is:

Equipment Required:

• A Single Phase Transformer
• Resistive Load
• Power Supply
• AC Ammeter
• AC Voltmeter
• Wires

Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Using your Single-Phase Transformer, Power Supply, Single-Phase Wattmeter, Resistive Load, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure – 1. Note that the two transformers are connected in parallel. The primary windings (1 and 2) are connected together to the 220 V ac power source. The wattmeter will indicate the input power. Each secondary winding (3 to 4) is connected in parallel to the load R_L. Ammeters are inserted to measure load current I_L and transformer secondary currents I₁ and I₂.

Connect two resistance sections of the Variable resistance module in series to implement the resistive load R_L shown in Figure – 1. This is required to dissipate the large amount of power involved in this exercise.

2. Place all the resistance switches in their open positions for zero load current. Note that the windings are connected for voltage step-up operation (220 V primary to 380 V secondary).

3. Please your circuit wiring approved by the instructor before proceeding.

4. a. Turn on the power supply and slowly advance the voltage output control knob while noting the transformer secondary current meters I₁, if the windings are properly phased, no load or secondary currents should be flowing.

b. Adjust the power supply voltage to 220 V ac as indicated by the voltmeter connected across the wattmeter.

5. a. Gradually increase the load R_L until the load current I_L equals 250mA ac. Check to see that the input voltage is exactly 220 v ac.

b. Measure and record the load voltage, load current, transformer secondary currents and the input power.

    E_L = 340 V ac

    I_L = 0.24 A ac

    I₁ = 0.12 A ac

    I₂ = 0.12 A ac

    Pin = 95 W

c. Return the voltage to zero and turn off the power supply.

6. a. Calculate the load power,

 E₁ (340) × I_L (0.24) = 81.6 W

b. Calculate the circuit efficiency,

c. Calculate the transformer losses

Pin (95) – Pout (81.6) = 13.4 W

d. Calculate the power delivered by transformer-1.

I₁ (0.12) × E_L (340) = 40.8 W

e. Calculate the power delivered by transformer – 2.

I₂ (0.12) × E_L (340) = 40.8 W

7. Is the load reasonably distributed between the two transformers? Yes.

Review Questions:

1. Show how you would parallel connect the transformers to the source and the load in Figure – 2. Windings 1 to 2 and 3 to 4 are rated for 2.4 KV ac and windings 5 to 6 and 7 to 8 are rated for 400 V ac.

Figure – 3

In a parallel connection of two transformers, should be properly connected with regard to polarity. The percentage impedance should be equal in magnitude. The regulation must be same.

2. The efficiency of transformers which supplies a pure capacitive load is zero. Explain.

Ans: We know the efficiency of a transformer,

For capacitive load, Pout = VI cosθ = VI cos 90^o = 0

3. Name the losses which cause a transformer to heat up.

Ans: The I²R losses that means copper losses and core losses are cause a transformer to heat up.

4. How does the efficiency of our Single-Phase Transformer compare to the efficiency of DC Motor/Generator? Explain.

Ans: In DC generator or motor; there is a friction loss which plays an important role on efficiency calculation. But in Transformers there is no friction loss, so transformer is more efficiency than a dc motor or generator of same rating.