Java Physics



Electric Transformer

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  • Here, the transformer is assumed to have (1) no flux leakage, and (2) no resistance in the two windings.
  • The inductances of the two windings are NOT assumed to be infinite. However, we can assign some large inductances (but not infinite ) to them by selecting "Large winging inductances".
  • Please note that it is pointless to compare the amplitude of a voltage and that of a current when they are dispalyed simultaneously on the screen. They have different units.



Internal Links :

Electromagnetic Induction (Applet)

RLC series a.c. (Applet)

Exact Mathematical Solution of Simple Transformer (Pdf)

Screenshots of waveforms from a real transformer (Web)

Run Applet Online

Download Executable Jar File



Np = Number of turns in the primary coil
Ns = 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:

  1. The two coils are resistanceless.
    Voltage across the primary coil, pd primary coil .....(1)
    Voltage across the secondary coil, pd secondary coil .....(2)

  2. There is no flux leakage, i.e., Φ p = Φ s .....(3)
From (1), (2) and (3), we get voltage ratio


  1. When the secondary coil is on open circuit,

    1. the primary coil behaves as a pure inductor.
    2. the only current flowing is the primary current (Ip).
    3. the flux linking the two coils is totally caused by Ip.
    4. the primary voltage (Vp) leads the primary current (Ip) by π/2.
    5. average input power = Vp,rmsIp,rmscos(π/2)= 0, where rms denotes the root-mean-square values
    6. average output power = 0.
  2. When a resistive load R is connected to the secondary coil,

    1. the secondary current (Is) becomes nonzero.
    2. Is is always in phase with Vs, because the load is a pure resistor.
    3. the flux linking the two coils is now caused by Ip and Is.
    4. the primary current (Ip) is larger than when the secondary coil is open.

      By Lenz’s law, Is 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.

    5. the primary voltage (Vp) leads the primary current (Ip) by an angle θ < π/2.
    6. average input power = Vp,rmsIp,rmscos(θ)
  3. If the load R is decreased,

    1. the primary current will be further increased.
    2. the phase angle between the primary voltage and the primary current will be further reduced, approaching zero.
  4. When the load R is much smaller than the reactance of the secondary coil,

    1. the two conditions have already implied there is no energy loss, and now
    2. the primary voltage is (nearly) in-phase with the primary current.
      ∴ input Power = output Power
      ∴ VpIp = VsIs
      ∴ Ip : Is = Vs : Vp = Ns : Np
  5. It is noteworthy that

    1. the voltage ratio Vs : Vp = Ns : Np 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.
    2. the current ratio Ip : Is = Ns : Np holds when, besides the transformer satisfying the above two assumptions, the resistance of the load R must be small (R << Xs).




  • The author (Chiu-king Ng) has the copyright on all the simulations in this website.
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  • Last Update:2017-6-5