A transformer changes an alternating voltage using two coils around an iron core. For an ideal transformer, Vp ÷ Vs = Np ÷ Ns, and the power in equals the power out: Vp × Ip = Vs × Is.
This lesson follows comparing generator and motor energy transfers and belongs to induction and transformers.
How does a transformer work?
An alternating current in the primary coil produces a changing magnetic field in the iron core. The changing field passes through the secondary coil and induces an alternating voltage there. This is the idea from explaining induced voltage.
The size of the secondary voltage depends on the number of turns on each coil. More turns on the secondary gives a step-up transformer. Fewer turns gives a step-down transformer.
What are the stated assumptions?
The equations in this lesson hold only for an ideal transformer. No energy is lost as heat in the coils or the core, and all the magnetic field from the primary passes through the secondary. An exam question will say “assume the transformer is 100% efficient” or “ideal”.
Real transformers are slightly less than 100% efficient, so a real secondary power is a little smaller than the primary power. The practice set returns to that point.
Worked example 1: find the secondary voltage
A fictional charger has a primary coil of 1200 turns connected to a 240 V supply. The secondary coil has 60 turns. (Invented example data.)
Step 1, choose the equation: Vp ÷ Vs = Np ÷ Ns.
Step 2, rearrange: Vs = Vp × Ns ÷ Np.
Step 3, substitute: Vs = 240 × 60 ÷ 1200 = 14 400 ÷ 1200 = 12 V.
Step 4, check: the secondary has fewer turns, so a smaller voltage is expected. 12 V is less than 240 V, so the answer is sensible. The ratio is 20 to 1.
Worked example 2: find the turns, then the current
A fictional step-up transformer takes 12 V on a primary of 100 turns and gives 180 V. (Invented example data.)
Secondary turns: Ns = Np × Vs ÷ Vp = 100 × 180 ÷ 12 = 18 000 ÷ 12 = 1500 turns.
Suppose the secondary supplies 0.50 A. For an ideal transformer, power out = 180 × 0.50 = 90 W. Power in is also 90 W, so the primary current is 90 ÷ 12 = 7.5 A.
The primary current is larger, because the primary voltage is smaller and the power is the same.
The mistake to watch for
Mistaken working: Vs = Vp × Np ÷ Ns = 240 × 1200 ÷ 60 = 4800 V (for the charger above).
The student put the turns the wrong way round.
The correction: write the equation as Vs ÷ Vp = Ns ÷ Np if it helps, so the secondary is on top on both sides. Then use a sense check. A coil with fewer turns cannot give a bigger voltage, so 4800 V from a step-down charger is impossible.
Check yourself
Try these, then open each answer.
1. A transformer has 460 primary turns and 46 secondary turns, with a 230 V primary supply. Find the secondary voltage.
Show answer
Vs = 230 × 46 ÷ 460 = 10 580 ÷ 460 = 23 V. It is step-down, as expected.
2. A 240 V supply feeds a primary of 2000 turns. The secondary should give 6 V. How many secondary turns are needed?
Show answer
Ns = Np × Vs ÷ Vp = 2000 × 6 ÷ 240 = 12 000 ÷ 240 = 50 turns.
3. An ideal transformer steps 240 V down to 12 V. The secondary supplies 0.50 A. Find the primary current.
Show answer
Power out = 12 × 0.50 = 6.0 W. Power in = 6.0 W. Ip = 6.0 ÷ 240 = 0.025 A.
Where this leads next
The transformer only works on a changing current, so continue with why alternating current is relevant. Then apply the ratio to a transmission loss problem.
Choosing the right rearrangement under time pressure is a skill that improves with guided repetition. Our teachers can do this with you in online one-to-one Physics tuition.