What happens if the secondary of an ordinary transformer is left open?

Question

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.

Answer 1

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\$: -

Picture from here.

Answer 2

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.

Answer 3

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:

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.