Example 1
U1 = 230 V, n1 = 1150, n2 = 60 -> U2 = 12 V.
Type U1 [V], n1 and n2. The calculator computes U2 = U1·n2/n1: 230 V, 1150 turns and 60 turns is 12 V, and 230 V, 500 and 1000 is 460 V. Zero primary turns does not divide.
Turns n2: n2 = n1 U2/U1. Induction: |ε| = ΔΦ/Δt.
Primary voltage U1 (V), Primary turns n1 and Secondary turns n2. The result shows up here.
An ideal transformer keeps the voltage ratio equal to the turns ratio: U2 = U1 · n2/n1. From 230 V, 1150 primary turns and 60 secondary turns you get 12 V. From 230 V, 500 and 1000 you get 460 V, because the secondary has more turns. From 120 V, 600 and 60 you get 12 V again.
Three fields: primary voltage U1 in volts, then the turn counts n1 and n2, unitless. The result U2 is in volts. The calculator also shows the ratio n2/n1, so you see at once whether this is a step-down or a step-up. A comma and a period in 230.5 are the same U1.
n1 must be greater than zero, because you divide by the primary turns. Zero n1 will not run. n2 may be larger than n1: then U2 > U1, as with 500 and 1000 on 230 V.
This is a lossless model. U1/U2 = n1/n2. Efficiency η = P_out / P_in is a separate page, once you know the powers. There is no current, copper, or core here: only the ratio.
If you know both voltages and n1 and you want n2, open the secondary-turns page. Faraday |ε| = ΔΦ/Δt lives on induction: there a changing flux, here turns and voltages.
After U2 you can move to Ohm or to P = U·I if you want current on a winding. This calculator stops at secondary volts.
U2 = U1 · (n2 / n1)
U1 > 0, n1 > 0, n2 > 0. Ideal model, no losses.
Ideal ratio U2/U1 = n2/n1. From 230 V, 1150 primary turns and 60 secondary turns you get 12 V. No-loss model.
U1 = 230 V, n1 = 1150, n2 = 60 -> U2 = 12 V.
U1 = 230 V, n1 = 500, n2 = 1000 -> U2 = 460 V.
U1 = 120 V, n1 = 600, n2 = 60 -> U2 = 12 V.
U1 = 230 V, n1 = 920, n2 = 48 -> U2 = 12 V.
U1 = 230 V, n1 = 2300, n2 = 50 -> U2 = 5 V.
U1 = 230 V, n1 = n2 = 800 -> U2 = 230 V.
U1 = 230 V, n1 = 1150, n2 = 120 -> U2 = 24 V.
U1 = 400 V, n1 = 800, n2 = 460 -> U2 = 230 V.
Secondary voltage is 12 V. 230 × 60 / 1150. That is a step-down from mains to a supply.
You get 460 V. More secondary turns than primary raises the voltage, lossless model.
Voltage U1 [V], turns n1 and n2 with no unit. Result U2 [V]. This is an ideal model, with no winding ohms.
No. The calculator only uses the turns ratio. Efficiency, magnetizing current and copper drop sit outside the formula.
On the transformer voltage-ratio page when it is on the hub. Here the unknown is U2, not the turn count.
Yes. 230.5 and 230,5 are the same U1 [V]. A comma and a period mean the same value.
On the electromagnetic-induction page. There EMF from dΦ/dt. Here only the turns and voltage ratio.
Yes. Then U2 is larger than U1: a step-up. 1000 on 500 at 230 V gives 460 V.
Because you divide by n1. Zero primary turns does not make a transformer, and the calculator refuses it.
Yes. U2 = U1. Galvanic isolation in the ideal model does not change the volt number.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.