Ideal Transformer Equation

Equivalent forms

\frac{I_s}{I_p} = \frac{N_p}{N_s}

V_p I_p = V_s I_s \;\; \text{(ideal)}

Power in equals power out, so voltage and current trade places in exact inverse proportion — the electrical analogue of a lever.

Symbol	Name	SI unit	Dimension	Range
$V_s$	Secondary voltageoutput AC voltage across the secondary coil	V	M·L²·T⁻³·I⁻¹	0 – 1000
$V_p$	Primary voltage AC voltage applied to the primary coil	V	M·L²·T⁻³·I⁻¹	1 – 1000
$N_p$	Primary turns Number of turns in the primary winding	—	1	10 – 2000
$N_s$	Secondary turns Number of turns in the secondary winding	—	1	1 – 2000

Unit systems

Symbol	SI	CGS	Imperial
$V_s$	V	statV	V
$V_p$	V	statV	V
$N_p$	—	—	—
$N_s$	—	—	—

Where it holds

Ideal-transformer limit: perfect flux coupling, zero winding resistance, no core loss, and operation well below core saturation. Real transformers reach 95–99.5% efficiency; the ratio law degrades under heavy load or saturation.

Dimensional analysis

N_s/N_p is dimensionless, so $[V_s] = [V_p] = V \checkmark$

Discovery

Michael Faraday / William Stanley · 1885

Faraday's 1831 induction ring was the first transformer, but it took until 1885 for the ZBD engineers (Zipernowsky, Bláthy, Déri) and William Stanley to build practical closed-core transformers — the invention that won AC the War of the Currents.

Try this

How does the grid ship megawatts hundreds of kilometers and then hand your phone charger a gentle 5 volts?

A transformer has 400 primary turns and 20 secondary turns. With 240 V AC on the primary, what comes out — and what happens to the current?

Research status: stable

Real-world applications

Grid step-up $(25\,\mathrm{kV} \to 400\,\mathrm{kV})$ and neighborhood step-down transformers
Phone chargers and laptop power bricks
Microphone and audio impedance matching
Isolation transformers for medical equipment safety

Common misconceptions

Transformers do not create power — voltage steps up only as current steps down
They do not work on DC; a steady current induces nothing
High-voltage transmission exists to $cut I^{2}R$ line losses: same power at $100\times$ voltage means 10, $000\times$ less resistive loss

Experimental verification

Wind two known coils on a shared laminated core, drive one with a signal generator, and the measured voltage ratio tracks the turns ratio to within a few percent (leakage flux accounts for the gap). Grid transformers verify it at megawatt scale daily.

Derivation

The same flux

\Phi (t)

threads every turn of both windings.

Faraday's law per winding:

V_p = N_p d\Phi /dt

and

V_s = N_s d\Phi /dt

.

Divide:

V_s/V_p = N_s/N_p

.

Ideal energy conservation

(V_p I_p = V_s I_s)

then forces

I_s/I_p = N_p/N_s

.

Limiting cases

N_s = N_p

⟶

V_s = V_p

An isolation transformer: same voltage out, but galvanically separated circuits.

N_s ≫ N_p⟶ V_s ≫ V_p, I_s ≪ I_pStep-up: neon signs and X-ray supplies make kilovolts, paying with proportionally tiny current.

DC input

(f = 0)

⟶

V_s = 0

No changing flux means no induction — transformers are inherently AC devices. DC just heats the primary.

What if…

What if you connect a transformer to a DC battery?

A brief output blip at connection (flux changing from zero), then nothing — and the primary, being just low-resistance wire to DC, overheats.

What if the secondary is short-circuited?

The secondary current — and hence the reflected primary current — grows enormous, limited only by leakage inductance and winding resistance. Fuses exist for this.

What if the core saturates?

$d\Phi /dt$ stops tracking the primary voltage near the waveform peaks, the output distorts, and the magnetizing current spikes — the ideal ratio law fails.

1

Doorbell step-down

Given ·

V p:: 240
N p:: 400
N s:: 20

Find · V_s

Steps

Turns ratio $N_s/N_p = 20/400 = 1/20$
$V_s = V_p \times N_s/N_p = 240/20 = 12\,\mathrm{V}$
If the secondary delivers 2 A, the primary draws only 0.1 A — power balances at 24 W

Result ·

V_s = 240 \times 20/400 = 12\,\mathrm{V}

2

Transmission-line loss savings

Given ·

P:: 100000
V low:: 1000
V high:: 100000
R line:: 1

Find · Line loss at each voltage

Steps

At 1 kV: $I = P/V = 100$ A, loss $= I^{2}R = 10^{4} \times 1 = 10\,\mathrm{kW} (10$ %)
At 100 kV: $I = 1$ A, loss $= 1^{2} \times 1 = 1\,\mathrm{W} (0.001$ %)
Stepping voltage up $100\times$ cuts resistive loss by 10, $000\times$ — why grids run at hundreds of kV

Result · 10 kW lost at 1 kV vs 1 W lost at 100 kV

Faraday's Law of Induction→Self-Inductance of a Solenoid→Motional EMF→