Define
N_{p} = Number of turns in the primary coil
N_{s} = Number of turns in the secondary coil
Φ _{p} = Magnetic flux in the primary coil
Φ _{p} = Magnetic flux in the secondary coil
Suppose the following conditions are satisfied:
Remarks
Secondary coil is on open circuit
 The primary coil behaves as a pure inductor.
 The only current flowing is the primary current (I_{p}).
 The flux linking the two coils is caused by I_{p}. This primary current is called magnetizing current (I_{p,M}).
 The primary voltage (V_{p}) leads the magnetizing current (I_{p,M}) by π/2.
 Average input power = 0
Resistive load R is connected to the secondary coil
 The secondary current (I_{s}) becomes nonzero.
 I_{s} is always in phase with V_{s}, because the load is a pure resistor.
 The flux linking the two coils is still caused by I_{p,M}.
 Besides the manetizing current, the primary current has one more component (call it I_{p,L}), which is antiphase with the secondary current I_{s}, i.e.,
I_{p} = I_{p,M} + I_{p,L}, where I_{p,M} and I_{p,L} are π/2 out of phase.
By Lenzâ€™s law, I_{s} must flow in a direction such that the change of magnetic flux in the core is reduced.
However, the change of flux in the core always produces a voltage in the primary coil to have the same value as the a.c. source voltage (since the input loop is resistanceless).
To maintain this, the primary current eventually becomes larger to restore the original magnetic flux, compensating the opposition due to the secondary current.
Eventually, the core flux is still produced by I_{p,M},while the fluxes produced by I_{p,L} and I_{s} cancel each other out.
 The primary voltage (V_{p}) leads the primary current (I_{p}) by an angle θ = tan^{1}( R/X_{s}), where X_{s} is the reactance of the secondary coil.
 Average input power = V_{p,rms}I_{p,rms}cos(θ)
If the load R is decreased,
 I_{p,L} will be further increased, but I_{p,M} remains the same
 the phase angle between the primary voltage and the primary current will be further reduced, approaching zero.
When the load R << X_{s} (reactance of the secondary coil),
 I_{p,L} >> I_{p,M}, hence I_{p} ≈ I_{p,L}.
 the primary voltage is (nearly) inphase with the primary current.
Any tume, V_{p}I_{p,L} = V_{s}I_{s}
Now,V_{p}I_{p} = V_{s}I_{s}
Therefore, I_{p} : I_{s} = V_{s} : V_{p} = N_{s} : N_{p}
It is noteworthy that
 the voltage ratio V_{s} : V_{p} = N_{s} : N_{p} holds under the two assumptions (i) coils are resistanceless, and (ii) no flux leakage. No matter the secondary coil is open or not, the resistance of R is large or small, the voltage ratio holds.
 the current ratio I_{p} : I_{s} = N_{s} : N_{p} holds when, besides the transformer satisfying the above two assumptions, the resistance of the load R must be sufficiently small (R << X_{s}).
