Let's consider a transformer. The ideal transformer equation is:

$$\frac{V_1}{V_2} = \frac{I_2}{I_1}$$

Suppose the secondary is open (that means \$I_2 = 0\$): it will result \$I_1 = 0\$. How is this physically possible?

I'd say that, also in case of absence of losses, there must be a current in the primary circuit. In fact, the primary circuit is a normal closed circuit, (voltage source closed on transformer primary winding), so I'd say that there will be the current that will flow if the secondary circuit does not exist.

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    \$\begingroup\$ there must be a current in the primary circuit that's what we call "magnetizing current". Also think about the reactance of the primary winding. \$\endgroup\$ Jun 25, 2020 at 5:58
  • \$\begingroup\$ @RohatKılıç There's no such thing as "magnetizing current" in an ideal transformer. It has an infinite inductance and will resist any current building up in one winding unless there is a simultaneous current in another winding. \$\endgroup\$ Jun 25, 2020 at 13:34
  • \$\begingroup\$ @Andyaka Nope. Redacted. :) \$\endgroup\$
    – JYelton
    Jun 25, 2020 at 17:28
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    \$\begingroup\$ There's no such thing as an "ideal transformer". \$\endgroup\$
    – Hot Licks
    Jun 25, 2020 at 21:58

3 Answers 3


The ideal transformer equation is: \$\frac{V1}{V2} =\frac{I2}{I1}\$

That's fine but then you go on to consider a non-ideal transformer: -

Suppose the secondary is open (that means I2=0): it will result I1=0. How is this physically possible?

Now you are considering a non-ideal transformer by mentioning "physically possible", hence: -

$$\frac{V1}{V2} ≠\frac{I2}{I1}$$

there must be a current in the primary circuit.

For any practical transformer there will be current flowing in the primary when the secondary is open circuit. It's called magnetization current. The primary at this point is just an inductor (and also remains an inductor under secondary loading conditions). The magnetization current is what induces the secondary voltage (under load or off-load). Magnetization current is due to magnetization inductance, \$L_M\$: -

enter image description here

Picture from here.

  • \$\begingroup\$ Also if we re-arrange equation to get rid of the division by zero $$V_1 I_1 = V_2 I_2$$, then we see that as $$I_2$$ tends toward zero $$V_2$$ will tend toward infinity. Until the other factors (see this answer) have an effect. There for secondary potential difference will go up, but cap out because of losses, etc. \$\endgroup\$ Jun 25, 2020 at 22:13
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    \$\begingroup\$ I’m sure if you put that into an answer it might get a serious number of down votes. \$\endgroup\$
    – Andy aka
    Jun 25, 2020 at 23:10
  • \$\begingroup\$ can you tell me what is wrong? \$\endgroup\$ Jun 26, 2020 at 15:15
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    \$\begingroup\$ @ctrl-alt-delor Because it's nonsense to say that \$V2\$ will tend to infinity if \$I2\$ tends towards zero. Transformers work by induction and largely speaking, the secondary voltage (V2) will be constant and determined only by the primary voltage. \$\endgroup\$
    – Andy aka
    Jun 26, 2020 at 15:19

A transformer with an open secondary is just an inductor like any other with similar construction. Since no current flows in the secondary, it has no influence or interaction with the magnetic field produced by the primary. A typical mains power transformer has a primary inductance on the order of several henries.

  • \$\begingroup\$ Is this high inductance value that causes the transformer equation to say that I1 = 0? \$\endgroup\$
    – Kinka-Byo
    Jun 25, 2020 at 4:07

An ideal transformer would have an infinite magnetizing inductance. A simple (linear) model of a real transformer which takes into account winding resistance and magnetizing inductance as well as leakage inductances and core losses looks like this:

enter image description here

Where there is an ideal transformer Np:Ns. Rp/Xp represent the primary resistance and leakage reactance, and X's/R's represent the transformed secondary resistance and leakage reactance. Xm is the magnetizing inductance (the main factor we're talking about here) and Rc represents core losses.

When the secondary is open, the ideal transformer and X's/R's are no longer of any consequence, but there are still Rp, Xp, Rc, Xm.

Usually the magnetizing inductance is chosen so that the magnetizing current is about 1/10 of the full-load current, so if the transformer will draw 1A at 220VAC it will draw about 100mA with the secondary open.

Of course that current is mostly reactive so the lost power is relatively small (due to the \$I^2R\$ current in the primary resistance and the core losses).

A more sophisticated model would take the nonlinear behavior of the core into account, but this is adequate for many purposes when the transformer is operated in steady-state and within ratings.


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