A transformer trades voltage for current while conserving power

Transformers — AQA A-Level Physics

For an ideal transformer (100% efficient), the powerThe rate of energy transfer. Measured in watts (W). input equals the powerThe rate of energy transfer. Measured in watts (W). output. You cannot get more energyThe capacity to do work. Measured in joules (J). out than you put in.

$$V_p I_p = V_s I_s$$

$V_p$: primary voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference. (V)
$I_p$: primary currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). (A)
$V_s$: secondary voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference. (V)
$I_s$: secondary currentThe rate of flow of charge. Measured in amperes (A). (A)

This is conservation of energyEnergy cannot be created or destroyed, only transferred from one form to another. The total energy of a closed system remains constant. applied to the transformer. PowerThe rate of energy transfer. Measured in watts (W). $in = power out$.
If a step-up transformer increases the voltageThe energy transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. by a factor of 10, the currentThe rate of flow of charge. Measured in amperes (A). must decrease by a factor of 10. The product V x I stays the same.
If a step-down transformer halves the voltage, the current doubles.
You are not creating energy. You are trading voltage for current (or current for voltage) at constant power.
This is why a phone charger (step-down) outputs low voltage but can deliver a higher current than the mains provides.

Worked Example

An ideal step-up transformer has a primary voltage of 230 V and a secondary voltage of 11 500 V. The primary current is 26 A. Calculate the secondary current.

Show Solution

List known values

Primary voltage: $V_p = 230 \text{ V}$
Secondary voltage: $V_s = 11\,500 \text{ V}$
Primary current: $I_p = 26 \text{ A}$

Write the power conservation equation

For an ideal transformer:

$$V_p I_p = V_s I_s$$

Rearrange for Is

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

Substitute and evaluate

$$I_s = \frac{230 \times 26}{11\,500} = \frac{5980}{11\,500} = 0.52 \text{ A}$$

The voltage was stepped up by a factor of 50, so the current has been stepped down by a factor of 50. That checks out: $26 \div 50 = 0.52$ A.

Answer

$I_s = 0.52$ A

Common Mistake MEDIUM

Students often: Thinking a step-up transformer gives you 'more power' because the voltage is higher.
Instead: Power is conserved. Higher voltage means proportionally lower current. $P = VI stays$ the same. No free energy.