Physics
Double Award
Double Award
Cambridge IGCSE
TOPIC 4B: MAGNETISM
A transformer is a very common magnetic device that 'transforms' voltages. For example, mains electricity at home is 230 V in the UK. To charge a phone battery takes just a few volts. Mains electricity voltage is too high, and you need a transformer to lower the voltage to a safer level. A transformer or similar device can also be found in the cable of all laptop chargers, again to lower the voltage.
A simple transformer consists of two coils wrapped around a soft iron core as shown in figure 1:
Figure 1: A simple transformer
When a voltage is applied to the primary coil, a current flows through it and a magnetic field is created, just like a solenoid. This strongly magnetises the soft iron core, and it becomes a temporary induced magnet. The rectangular core channels the magnetic field, so that nearly all of the field is transferred to the secondary coil. There is little magnetic field away from the core.
This magnetic effect passing through the secondary coil can be used to induce a voltage. However, as you should have found out in the last section, for induction to occur the wire must move through the magnetic field, or the field move through the wire. For this reason, an a.c. voltage is used. The changing current in the primary coil creates a constantly changing and moving magnetic field, which cuts through the secondary coil inducing an output voltage.
You may have noticed that this clever device has - so far - transferred electrical energy into magnetic energy, and then back to electrical. This all seems a bit pointless, until you realise that the voltage that is induced in the secondary coil depends on the number of turns in both coils. If there are more coils in the secondary coil than the primary, then the induced voltage is larger than the input voltage. The voltage has been increased. This type of transformer is called a step-up transformer. A step-down transformer reduces the voltage.
If there are double the number of coils on the secondary than primary, the output voltage is double the input voltage. This principle is written as a mathematical formula:
| input (primary) voltage [V] | = | number of primary turns |
| output (secondary) voltage [V] | number of secondary turns |
| Vp | = | Np |
| Vs | Ns |
Example:
A step down transformer is used to convert 120 V a.c. to 6 V. If the primary coil has 3000 turns, how many turns are needed on the secondary coil?
Answer:
| Vin | = | np |
| Vout | ns |
Substituting in values, this gives:
| 120 | = | 300 |
| 6 | ns |
Questions:
1. A phone charger uses a transformer with 5000 turns on the primary coil and 250 turns on the secondary. It is plugged into the mains supply with an input voltage of 230 V.
a) There are fewer turns on the secondary coil than the primary, so this is a step-down transformer.
b) The output voltage is the voltage on the secondary coil. Using the formula:| Vp | = | Np |
| Vs | Ns |
| 230 | = | 5000 |
| Vs | 250 |
| 230 x 250 | = | Vs |
| 5000 | 1 |
So Vs = 11.5 V
2. A very high voltage is required for 'spark' plugs that ignite the petrol in any car engine. The transformer used to do this has an input voltage of 12 V and an output voltage of 4800 V. The primary coil has only 10 turns.
Calculate the number of secondary turns required to produce 4800 V.
Using the formula:
| Vp | = | Np |
| Vs | Ns |
| 12 | = | 10 |
| 4800 | Ns |
| Ns | = | 4800 x 10 |
| 1 | 12 |
So Ns = 4000
A step-up transformer increases the voltage. This seems like an amazing energy producing device, providing 'extra electricity' for free. However, there is a catch that prevents this.
The formula for electrical power is P = I x V as covered in section 4.2e. We also know that energy cannot be created, from the law of conservation of energy (section 1.7a). Assuming the transformer is 100% efficient and no power is lost, then the output power must match the input power.
This means that if the voltage increases, the current must decrease so that the power remains the same.
This can be shown with this formula:
input power = output power
Vp x Ip = Vs x Is
[V] x [A] = [V] x [A]
Note that the formula above assumes the transformer is 100% efficient.
Example:
A step down transformer is used to convert 120 V a.c. to 6 V. If the input current is 0.1 amps, calculate the maximum output current, assuming the transformer is 100% efficient.
Answer:
Vp x Ip = Vs x Is
The output current we need to find is the current from the secondary coil, Is.
Substituting in values gives:
120 x 0.1 = 6 x Is
So Is = 12 / 6 = 2 A
Alternative method: Some people like to use ratios to calculate values for transformers. In this example, the transformer reduces the voltage by a factor of 20 (120/6). This means the current must increase by a factor of 20 so that the power remains the same.
0.1A x 20 = 2 amps.
A typical exam question on transformers asks you to use the first formula to calculate voltages or number of turns required to produce a required voltage. Once you have done this, you can use the second formula to see how this affects the current. For example, if a step up transformer increases the voltage by a factor of 10, then the output current will be reduced to one tenth so that the input power is the same as the output power.
Questions:
3. A school power supply is used to run a heating element for an experiment. The heating element runs on 12 V, and has a power output of 60 W.
The transformer used in the power supply has 240 V input and 800 turns on the primary coil.
Calculate:
a) Using the formula:
| Vp | = | Np |
| Vs | Ns |
| 240 | = | 800 |
| 12 | Ns |
| Ns | = | 12 x 800 |
| 240 |
So Ns= 40
b) We know P = I x V,
so 60 = I x 12
Therefore I = 5 A
c) Using the second transformer equation Vp x Ip = Vs x Is gives:
Vp x Ip = Vs x Is
240 x
Ip = 12 x 5 (from part b)
Therefore:
| Ip | = | 12 x 5 |
| 240 |
To distribute electricity around any country, a national system of power lines is needed, called a national grid. These power cables transmit power from the power stations to our homes, schools and offices. However, some power is lost in the cables. As with any resistor, the cables heat up when a current passes through them. (see section 4.2e). This means the lower the current, the less power is lost as wasted heat in the cables.
For this reason, it is better to transmit the power with a high voltage and low current. A step-up transformer is placed between the power station and the power cables to increase the voltage, up to as high as 400 000 V!
Figure 2: Power cables in the national grid
This high voltage is far too dangerous for homes and offices, and a local step-down transformer is needed to reduce the voltage back to safe levels.
Remember that the power transmitted is P = I x V. You can have a high current and low voltage, or a low current and high voltage, and still transmit the same power. A low current is the most efficient.
Any transmission cable as shown in figure 2 will have a resistance. The power lost in the cable depends on the current flowing through this resistance, and can be calculated using the following formula:
power lost as heat = (current)2 x resistance
P = I2 R
[W] = [A]2 x [Ω]
Example:
A long 4 km cable has a resistance of 200 Ω. A current of 10 A passes through it. Calculate the heat loss per km in the cable.
Answer:
The power lost as heat will be found using P = I2 R
So P = 102 x 200 = 20 000 W (or 20 kW)
However this is the total power lost over 4 km, so the power lost per km is 20 kW ÷ 4,
So power lost per km = 5 kW
Questions:
4. The diagram below shows the main stages in a local power distribution system:
a) State the name of the type of transformer used at X and Y.
b) Explain why these transformers are required.
a) X is a step-up transformer, Y is a step-down transformer.
b) A step-up transformer increases the voltage and decreases the current. A lower current leads to less power loss in the cables. (The high voltage and low current makes the transmission more efficient).
The step-down transformer is required to reduce this high voltage back down to safe levels for domestic use.
5. The live wire in a long mains extension cable has a resistance of 1.2Ω and delivers a current of 13 A.
a) Calculate the power lost in the cable when carrying a 13 A current.
b) Suggest why it is dangerous to keep the cable coiled when carrying 13 A current.
a) The power lost is calculated using P = I2 R,
So P = 132 x 1.2 = 202.8 W (or 203 W to 3 sig. figs)
b) This heat could be trapped in the inner part of the coil and eventually melt the plastic insulation. This could cause a fire/short circuit/overheat the coil (or any other safety related issue).
Now test your understanding using this quiz on Induction and Transformers:
