4. Transformers and Transmission

TransformersTransformers are a major component of the electrical distribution system. They

enable energy losses in transmission lines to be reduced.

are used to change household power supply (240 Volts) to usable voltages in appliances

have the circuit symbol

What are Transformers?A transformer consists of three major components:

an iron core shaped in U or square shape a primary coil wound around one side of the iron a secondary coil wound around the other side of the iron

The alternating current in the primary coil generates an alternating magnetic field in the iron core. This alternating field passes through the secondary coil and induces an EMF in the secondary coil.

The ratio of the voltage across the secondary to the voltage across the primary coil is equal to the ratio of the turns

Voltage steppingThe purpose of a transformer is to either step up or step down a voltage. When stepping up a voltage, the output voltage is greater than the input voltage. In a perfect transfer of magnetic flux is assumed then energy (power) is conserved.

Input power = Output power

It can be seen that in a step-up transformer, the number of secondary turns exceeds that in the primary coil, and the current is stepped down by the same factor by which the potential difference is stepped up.

Almost all electrical appliances other than lights and motors connected to a domestic supply use a transformer (usually step-down)

Measuring AC VoltageWhen comparing DC and AC power supplies, it is necessary to carry out conversions to make a true comparison. Below is a regular sinusoidal AC voltage, as could be produced by a generator.

EMF

The Root Mean Square (RMS) voltage is the square root of the mean of the square of the potential difference. This complicated value is used because it gives the value of the DC potential difference that has the same heating effect as the alternating potential.

In the case above, the DC potential difference that would transfer the same energy is given by

where Vo is 339V. In Australia the RMS is equal to 240V.

Any AC current can be given an RMS equivalent value in exactly the same manner. When doing calculations involving power always use the RMS values, unless otherwise stated.

Household electricity supplyTwo wires come from the power pole to the house. The ACTIVE wire is 240 VRMS. The NEUTRAL wire is ) V and is connected to the ground. The cable from the pole goes first to a mains connection which contains a fuse in the active wire. Then it is connected to the switchboard or fusebox. A main switch is placed in the active wire so that the power can be easily cut in case of emergency. Several lighting and power circuits originate at the switchboard and each has a fuse or circuit breaker (again in the active wire) so that individual circuits can be isolated and cut.

The neutral is connected to the neutral bar in the switchboard which is earthed. The neutral bar has connections to each power and lighting circuit. A third wire, an earth, also goes to each circuit.The earth wires are connected into the ground, often by a waterpipe or metal stake. They serve as a safety feature in that the earth wire is connected to the metallic case of an appliance. If this case accidentally becomes electrified, the current flows harmlessly to earth and possibly blows a fuse in the process. The current will flow to the earth, because it will always take the path of least resistance.

Resistance of power lines is proportional to the length of the lines. The longer the wires, the more important the transmission losses become. The larger diameter of the wire, the less the resistance, but the greater the weight, cost and the higher the supports need to be. It is paramount to reduce the power losses in cables.

PLoss = VLoss ITransmission

But, VLoss = Itransmission RPLoss = I R

339 V

240 V

Time (secs)

-339 V

0.01 0.02

The peak-to-peak voltage VPP is defined as the difference between the maximum and minimumIn this case, +339 to – 339 = 679V

The peak voltage is 339V

Transmitting at high voltages and relatively low current reduces power losses. This requires stepping up the voltage at the power station and stepping it down for industrial and domestic use.

In order to transmit large blocks of power (for Melbourne up to 5 109W) over transmission lines, it is necessary to use a high voltage (500kV), only then can the current be kept down to a reasonable level (1 104A). If the power was transmitted at 250V it would require a current of 2 107 amps to supply the power to Melbourne.

The power is transmitted in AC current because of its guaranteed voltage level over long distances. AC is also advantageous in its ability to be transformed either up or down. In contrast DC current is unable to be transformed necessitating transmission of large amounts of power, which would increase the energy loss resulting from large voltage drops.

At various stages, power is taken from the system and rectified into 600 V (trams) or 1500 V DC (trains) for public transport. Power is also transformed into other AC voltages for specific uses in industry.

Power losses in transmission lines need to be considered in some questions. The power is not actually lost, it is degraded into heat, which becomes unusable. The power supplied to the consumer is equal to the power supplied at the start minus the power lost in transmission.

Typically, the power is sent from the generating station at voltages as high as 500kV.

As it arrives at a terminal station it is stepped down to 66kV and then down to 11kV at a substation.

Pole transformers further step it down to 415 and 240 volts for domestic use.

Examples or class discussionAn electrician has imported an electric light globe that is designed to operate with an RMS voltage of 110 V.When operating at this voltage it has a resistance of 55 ohm. The globe has a power rating of 220 W. Two methods are considered that will allow the light globe to be used with a mains supply with an RMS voltage of 240 V.In method 1 a resistor of 65 ohm is placed as shown below in Figure 1, so that when used with an RMS supply voltage of 240 V, the RMS voltage across the light globe is 110 V.

Figure 1

In method 2 a transformer is used to convert the RMS voltage of 240 V to an RMS voltage of 110 V, as shown below in Figure 2. Assume the transformer is ideal.

Figure 2There are 1440 turns on the primary coil of the transformer.

1997 Question 1How many turns are on the secondary coil of the transformer?

1997 Question 2What is the RMS current in the primary coil of the transformer?

1997 Question 3What is the power supplied from the mains for the two different methods?

A 4.0-kW electric generator is used to supply power to an electric motor which is located some distance away. The RMS voltage at the output of the generator is 400 V, and the RMS current supplied to the motor is 10 A. The total resistance of the two cables, AC and BD, is 3.0 . The situation is shown below.

1998 Question 2How much electric power is dissipated in the cables?

An engineer suggests that by connecting a transformer (T1) at the generator output, and another (T2) at the motor input, a larger fraction of the generated power could be transferred to the motor. The plan is shown below in Figure 3.

1998 Question 3Explain why this procedure increases the power transferred to the motor. Your answer should clearly show the physics involved. Indicate what type of transformer (step-up or step-down) should be used in each location.

After installation of the transformers, the RMS current in the transmission wires is 0.5 A. Assume the transformers are ideal.1998 Question 4What is the RMS voltage between the terminals A and B in Figure 3 now?

1998 Question 5How many turns of wire are in the primary of transformer T1 if the secondary consists of 5000 turns? Show your working.

Answers for discussion questions1997 Question 1The ratio of the number of turns in the primary coil : secondary coil is the same as the input voltage : output voltage. Remember always think about the 'voltages'.

=

x =

x = 660 turns.

Check that your answer makes sense by asking the following questions. Is it a 'step-down' or 'step-up' transformer? In this case it is a step-down transformer because the output voltage is less than the input voltage. This then means that the number of turns in the secondary coil must be less than the primary coil.

1997 Question 2We are told that the transformer is 'ideal'. This means that the power in = power out.So (V × I)in = (V × I)out. The power out is the rating of the globe. It is rated as 220W, so the power in must equal 220W.So Pin = 220 = V × I

= 240 × I I = 220/240

= 0.92 Amp

1997 Question 3.In Method 1, some power is used in the resistor, this is given by V2/R.

V = 240 - 110 = 130 Volt (sum of voltages drops around the circuit = 240 V)

So power used in resistor = 1302/ 65 = 260W. total power used in circuit = 260 + 220

= 480W

In method 2, the power loss is 220W. Because the globe is the only appliance that is using power.

1998 Question 2If the Power loss is given by i2R

P = 102 × 3 = 300 W = 3.0 × 102 W (correct to 2 sig. figs)

1998 Question 3Power losses due to heating in transmission lines are given by P = i2R. To minimise this we need to minimise I. Since the power input is P = V × I, this means we need to increase V as much as possible. We use a step-up transformer at 'T1'. The step-up transformer steps up the voltage but steps down the current, resulting in a lower electric current in the cables and reduced power loss within the cables.We then need to use a step-down transformer 'T2' at the motor, to reduce the voltage to the original value, 400 V.

1998 Question 4If the initial current was 10A, and the new current is 0.5A, then the current has decreased by a factor of 20. This means that the potential difference between A and B must increase by a factor of 20. This is because in ideal transformers the power in = power out. So VIin = VIout.

So the new voltage across AB will be 400 × 20 = 8,000 V= 8.0 × 103V

1998 Question 5The voltage across AB increases by a factor of 20. So it must be a step-up transformer. This means that the number of turns in the primary will be 5000/20 = 250 turns.