Ideal transformer no-load, load, ratios

An ideal transformer has no winding resistance, no leakage flux, no core loss and infinite core permeability (so the magnetising current is negligible). Its two windings have N1 (primary) and N2 (secondary) turns.

(i) No-load operation

  I0 ≈ 0    +-------+  
 o--+ ( ) Φ ( )+--o  open circuit
 V1 |N1  core  N2|  V2 = E2
 o--+ ( )   ( )+--o

• The primary V1 is applied; the secondary is open.
• A tiny magnetising current Iμ flows, lagging V1 by 90°. It sets up the alternating flux Φ in the core.
• The flux links both windings and induces emfs in them by Faraday's law: E1 = −N1 dΦ/dt, E2 = −N2 dΦ/dt.
• With no resistance, E1 = V1 (it opposes V1 as a counter-emf) and V2 = E2.
• Since both windings link the same flux:

E2/E1 = V2/V1 = N2/N1 = K (voltage ratio, K = transformation ratio)

(ii) Loaded operation

• When a load is connected, the secondary current I2 flows, and produces an mmf N2 I2 that opposes the core flux.
• The primary draws an additional current I1 to cancel it. As the flux (and hence the magnetising mmf) stays unchanged, the net mmf is zero: N1 I1 − N2 I2 = 0.

I2/I1 = N1/N2 = 1/K (current ratio)

Power relation

V1 I1 = V2 I2: input VA equals output VA, because the transformer is ideal.

Step-up: K > 1 (voltage rises, current falls). Step-down: K < 1.