How this instrument works
An ideal transformer links two coils of wire around a shared iron core, and it works because both coils see the same changing magnetic flux. Faraday's law says the voltage induced in a coil equals its number of turns times the rate of change of that flux, so the volts-per-turn figure is identical on both sides: Vp/Np = Vs/Ns. Rearranged, that equality is the turns-ratio formula, Vs = Vp × (Ns/Np) — more secondary turns than primary steps voltage up, fewer steps it down.
The formula's shape follows directly from that shared flux: turns count is just a multiplier on one common induced voltage-per-turn, so doubling Ns while Np stays fixed doubles Vs. Because an ideal transformer is lossless, it also conserves power, which forces current to move the opposite way: Is = Ip × (Np/Ns). A winding with ten times the turns carries a tenth of the current at ten times the voltage — the product Vp×Ip equals Vs×Is throughout.
Real transformers only approximate this. Winding resistance dissipates a little power as heat, the core loses energy to hysteresis and eddy currents, and some flux leaks out instead of linking both coils — effects the ideal model sets to zero. The practical consequence is voltage regulation: a real secondary sags a few percent below the ideal figure once current is drawn, so a supply built to this formula should be checked under actual load, not judged from a no-load meter reading alone.
- Enter Primary voltage — the AC voltage actually applied to the primary winding, in volts.
- Enter Primary turns — the number of wire turns wound on the primary coil, Np.
- Enter Secondary turns — the number of turns wound on the secondary coil, Ns.
- Read Secondary voltage — computed as Vp × (Ns/Np); a result above the primary means a step-up winding, below means step-down.
- For the current side of the trade, remember Is = Ip × (Np/Ns): a higher secondary voltage always means a lower available secondary current.
Worked example — sizing a 10:1 step-up for a 1,200 V test set
A test technician needs a bench step-up supply for insulation-resistance testing, targeting about 1,200 V AC from an ordinary 120 V wall circuit. The transformer on hand has 100 turns on its primary and 1,000 on its secondary, a 10:1 ratio. Entering Primary voltage = 120 V, Primary turns = 100, and Secondary turns = 1000 gives Secondary voltage = Vp × (Ns/Np) = 120 × (1000/100) = 120 × 10 = 1,200 V, exactly the withstand voltage the test procedure specifies.
Because an ideal transformer conserves power, that voltage gain comes at the cost of current: a 5 A draw on the 120 V primary corresponds to only 0.5 A available at 1,200 V on the secondary, since 120 × 5 = 600 W equals 1,200 × 0.5 = 600 W. The technician sizes the secondary wiring for that lighter 0.5 A load rather than the primary's heavier 5 A — the detail that trips up anyone new to reading transformer nameplates.
Questions
Why does the secondary voltage depend only on the turns ratio?
Because both windings encircle the same iron core and see the identical changing magnetic flux. Faraday's law makes the induced voltage per turn equal on each side, so the only thing that can differ is how many turns each winding has — the ratio Ns/Np is the whole story for an ideal transformer's voltage relationship.
Does this formula account for real-world losses?
No — it describes the ideal case only. Winding resistance, core hysteresis and eddy-current losses, leakage flux, and magnetizing current are all set to zero. A real transformer's secondary voltage sags a few percent below this ideal figure once load current flows, an effect called voltage regulation, so treat the result as a no-load target rather than a guarantee under load.
What happens to current when voltage steps up?
It drops in the same proportion the voltage rises, because an ideal transformer conserves power: Is = Ip × (Np/Ns). A 10:1 step-up that turns 120 V into 1,200 V also turns a 5 A primary current into just 0.5 A on the secondary — ten times the voltage, one-tenth the current, the same 600 W throughout.
Can this same formula size a step-down transformer?
Yes, the relationship is symmetric. Put more turns on the primary than the secondary and Vs comes out lower than Vp — 1,000 primary turns against 100 secondary turns, fed with 120 V, returns 12 V. The formula doesn't care which winding is called primary; it just multiplies by the turns ratio you give it.
Is the turns ratio the same as the impedance ratio?
No, and this is the mistake that trips up audio and RF work: impedance scales with the square of the turns ratio, not the ratio itself. A winding with Ns/Np = 10 multiplies voltage by 10 but multiplies impedance seen from the secondary by 100, which is why matching transformers are specified by impedance ratio, not turns count alone.
Why is it called an ideal transformer specifically?
Because it assumes perfect coupling and zero losses: no winding resistance, no core losses, no leakage flux, and infinite core permeability so no magnetizing current is needed to establish the flux. Real transformers are engineered to approximate these conditions closely, which is why the simple turns-ratio formula predicts actual behavior so well near rated load.