Transformer Equations
Transformers are devices that use electromagnetic induction to change the voltage of an alternating current (AC) supply. They are crucial for adapting mains voltage for electronic devices and for transmitting electrical power efficiently over long distances.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Transformers require a changing magnetic field to function, which is why they only work with alternating current (AC), not direct current (DC).
- In a step-up transformer, the secondary coil has more turns than the primary (ns > np), resulting in a higher output voltage and a lower output current.
- In a step-down transformer, the secondary coil has fewer turns than the primary (ns < np), resulting in a lower output voltage and a higher output current.
- For an ideal transformer (100% efficient), the electrical power input to the primary coil equals the power output from the secondary coil.
- High-voltage transmission lines are used to minimise power loss. By stepping up the voltage, the current is reduced, and since power loss in wires is proportional to I2, this reduction has a large effect on efficiency.Ploss = I2 × R
› Why does this happen?
Why Transformers Need Alternating Current (AC)
A current flowing through the primary coil creates a magnetic field in the iron core. To induce a voltage in the secondary coil, this magnetic field must be constantly changing. An alternating current (AC) is needed because its direction and magnitude are continuously changing, which creates the required changing magnetic field. A direct current (DC) from a source like a battery creates a steady, unchanging magnetic field. This steady field cannot induce a voltage in the secondary coil, which is why transformers do not work with DC.
How the Number of Turns Changes the Voltage
The changing magnetic field from the primary coil is channelled through the iron core to the secondary coil. The amount of voltage induced in a coil is directly proportional to the number of turns of wire it has. This is because each turn of the coil has a voltage induced in it by the changing magnetic field, and these voltages add together. So, if the secondary coil has twice as many turns as the primary, the total induced voltage will be twice as large. This is why the ratio of voltages is equal to the ratio of turns.
Why Current Decreases When Voltage Increases
This is a direct result of the conservation of energy. For an ideal (100% efficient) transformer, the power put into the primary coil must equal the power that comes out of the secondary coil. Since Power = Voltage x Current, if the voltage is stepped up (increased), the current must be stepped down (decreased) by the same factor to keep the power constant. For example, if the voltage is doubled, the current must be halved so that the input power (Vp x Ip) equals the output power (Vs x Is).
Formulae
Vp / Vs = np / ns To relate the primary (p) and secondary (s) voltages (V) and the number of turns (n) in the coils. This allows calculation of any one quantity if the other three are known.
Vp × Ip = Vs × Is For an ideal (100% efficient) transformer, to state that the input power (Vp × Ip) equals the output power (Vs × Is). This is used to find an unknown current or voltage.
Definitions
- Step-up Transformer
- A transformer that increases AC voltage. The secondary coil has more turns of wire than the primary coil.
- Step-down Transformer
- A transformer that decreases AC voltage. The secondary coil has fewer turns of wire than the primary coil.
- Primary Coil
- The input coil of a transformer which is connected to the source of the alternating voltage.
- Secondary Coil
- The output coil of a transformer in which a changing voltage is induced by the magnetic field from the primary coil.
Worked example
An ideal transformer is used to step down a voltage of 12 kV from a transmission line to 240 V for household use. If the transformer supplies a power of 4.8 kW to the house, what is the current in the primary coil connected to the transmission line?
- 1
Identify the given values:
Vp = 12 kV = 12000 V, Vs = 240 V, Powerout = 4.8 kW = 4800 W - 2
The transformer is ideal, so input power equals output power:
Pin = Pout = 4800 W - 3
The formula for input power is Pin = Vp × Ip.
- 4
Rearrange the formula to solve for the primary current, Ip:
Ip = Pin / Vp - 5
Substitute the values:
Ip = 4800 W / 12000 V - 6
Simplify the fraction by cancelling zeros:
Ip = 48 / 120 - 7
Further simplify by dividing both numerator and denominator by their greatest common divisor, 24:
Ip = (48/24) / (120/24) = 2 / 5 - 8
Convert the fraction to a decimal:
Ip = 0.4 A
Answer: 0.4 A
Common mistakes
- ×Inverting the turns or voltage ratio. Remember that voltage is directly proportional to the number of turns (Vp/Vs = np/ns), but current is inversely proportional (Is/Ip = np/ns). A simple check: if voltage steps up, current must step down.
- ×Forgetting to square the current for power loss calculations. The power dissipated as heat in a cable is P = I2 × R, not I × R. This means halving the current reduces power loss by a factor of four.
- ×Applying transformer equations to DC circuits. Transformers do nothing with a steady DC input as there is no change in magnetic flux to induce a current in the secondary coil.
No-calculator tips
- ✓Focus on ratios and factors instead of full calculations. In the example Vp=12000V, Vs=240V, the voltage step-down factor is 12000/240 = 1200/24 = 50. Therefore, the current must be stepped up by a factor of 50.
- ✓Simplify fractions and ratios before you multiply. For a turns ratio of 400:8000, immediately simplify it to 1:20. This makes subsequent calculations with currents or voltages much easier.
- ✓Use standard form (powers of 10) to handle large numbers. For example, calculating with 25 kV and 250 kV is easier when you see them as 2.5 x 104 V and 2.5 x 105 V, clearly showing a factor of 10 difference.