Transformers and Efficient Power Transmission

Changing voltage without creating energy

A transformer changes an alternating potential difference, often called voltage. Its primary coil receives energy from the input supply; its secondary coil transfers energy to the output circuit. The coils are electrically separate: energy passes between them through a changing magnetic field in an iron core, rather than current flowing directly from one coil into the other.

A step-up transformer produces a higher output voltage than its input voltage. A step-down transformer produces a lower output voltage. Neither creates energy. Increasing the voltage does not mean that the transformer can deliver more power than it receives.

Conservation of energy links voltage and current

Power is the rate of energy transfer. An ideal transformer is 100% efficient, so all the energy supplied to it is transferred electrically to the output circuit. Its input and output powers are therefore equal:

$$P_{\text{input}} = P_{\text{output}}$$

Using the electrical power relationship $P = VI$ gives:

$$V_p I_p = V_s I_s$$

Here, $V_p$ and $V_s$ are the potential differences across the primary and secondary coils, measured in volts (V). and are the currents in those coils, measured in amperes (A). Each voltage–current product is a power in watts (W).

For a step-up transformer, the secondary voltage is greater. To keep the product $VI$ unchanged, the secondary current must be smaller. For example, if the voltage doubles, the current halves. Conversely, a step-down transformer lowers the voltage, so its output current is greater: halving the voltage doubles the current in the ideal model.

Real transformers transfer some energy to their surroundings, so their electrical output power is slightly less than their input power. The equal-power equation uses the ideal-transformer assumption.

Using the transformer equation

Keep input quantities together and output quantities together. To calculate an unknown primary current, rearrange the equation:

$$I_p = \frac{V_s I_s}{V_p}$$

For example, a travel adapter steps up a 115 V alternating supply to the 230 V required by a hairdryer. The hairdryer draws an output current of 5 A. Assuming the transformer is 100% efficient:

$$I_p = \frac{230 \times 5}{115} = 10\text{ A}$$

The output voltage is twice the input voltage, while the output current is half the input current. Both sides transfer power at $1150\text{ W}$, so the calculation is consistent with conservation of energy.

Why electricity is transmitted at high voltage

Transmission cables have resistance. When current flows through them, energy is transferred to the thermal energy stores of the cables and their surroundings. This heating wastes some of the energy that should reach consumers.

For a given power being transmitted:

$$I = \frac{P}{V}$$

Raising the transmission voltage therefore reduces the current needed to transfer the same power. The power wasted through cable heating is:

$$P_{\text{loss}} = I^2R$$

Here, $R$ is the resistance of the transmission cables in ohms (Ω). Because the current is squared, reducing it has a large effect. If the current is halved while cable resistance stays the same, the heating power falls to one-quarter. Less energy is wasted each second, making transmission more efficient.

The National Grid uses step-up transformers before long-distance transmission and step-down transformers near consumers. The voltage is raised for efficient transmission, then reduced to safer, usable levels.

Transformer power

For an ideal, 100% efficient transformer:

$$V_pI_p = V_sI_s$$