# Does a transformer draw only as much power on its input as it used on the output? If so, how?

In an ideal transformer, the output power is said to be equal to the input power, with voltage/current being multiplied or divided by the turns ratio between input and output.

Now, I understand that this is the ideal case, and there can be losses.

I would perfectly understand that the output power is at most the same as the input power.

Does the current drawn by the transformer on its input change when the load on its output changes?

My (probably very naïve) understanding of a transformer is that we have a primary which is basically generating a magnetic flux, and then the secondary is absorbing this magnetic flux and generating current from that.

If we take out the secondary, or the secondary circuit is open, or the secondary only needs very little power, wouldn't the primary still generate the flux and still draw the same amount of current/power, and the flux is just "lost"?

I'm picturing this in my mind like a radio: a radio transmitter still draws the same amount of power, whether there are receivers or not, how many, etc. How is a transformer different? How does the generated flux vary with the output load (if it does)?

My (probably very naïve) understanding of a transformer is that we have a primary which is basically generating a magnetic flux, and then the secondary is absorbing this magnetic flux and generating current from that.

The flux is not in the least bit absorbed (or altered) for an ideal or semi-ideal transformer.

The primary is still an inductor and, for a given terminal voltage applied, it generates a flux in the core that remains static as an AC entity. The secondary generates an output voltage based on the rate of change of flux. Current flows in the secondary when a load is connected.

We call the current that produces flux magnetization current. This is unaffected by the load on the secondary.

Does the current drawn by the transformer on its input change when the load on its output changes?

Yes it does. The magnetization current remains as it was when there is no load but, an extra current is drawn into the primary in order to service the secondary load current. The flux produced by that extra current is completely cancelled by the flux produced by the secondary load current. What remains is the magnetization flux.

If we take out the secondary, or the secondary circuit is open, or the secondary only needs very little power, wouldn't the primary still generate the flux and still draw the same amount of current/power, and the flux is just "lost"?

The primary will still generate the same alternating flux and take the same magnetization current but, because it's a reactive component there is no power associated with it.

For a non-ideal transformer (one that has imperfect coupling between windings) the magnetization flux will slightly reduce under secondary load current but, maybe that's a step beyond your question.

Yes, an unloaded transformer draws a certain amount of "magnetizing" current that creates a certain level of magnetic flux in the core.

In very simple terms, when you start to put a load on the secondary, the current in the secondary winding is in a direction that cancels out some of the flux in the core. The primary winding "sees" this drop in flux and draws additional current to make up the difference.

A model of a real transformer typically shows this as a large inductor that represents the magnetizing current, in parallel with a "transformed" impedance that represents the additional current caused by the secondary load. (There are usually also additional components in the model that represent things like the DC resistance of the primary winding, leakage inductance, parasitic capacitance, etc.)

A transformer with no load is essentially an inductor. The magnetic field it creates in the core opposes the current that creates it, limiting the current for a certain (AC) voltage. The current can be large, but as it is 90 degrees out of phase with the voltage, no power is drawn (in the ideal case).

This flux could also create current in a second coil wrapped around the same core. If a resistive load is applied to the secondary, it will draw current based on the magnetic flux. This leaves less energy in the flux to oppose the current in the primary. It also changes the phase relationship between the primary voltage and current, resulting in the same power (ideally) that the secondary load consumes.

Most power transformers have efficiency in the high 90%s, so there will be some loss (both resistive and hysteretical) both unloaded and loaded, but it's typically negligible until you're pushing enough power to challenge the cooling mechanisms.

• The magnetic field in the core does not oppose the current that creates it and, nether is the current out of phase with the voltage by 180 degrees. I think you need to brush up on transformer theory and consider re-writing your answer. The magnetization flux is also not depleted of energy by the load current. This answer is becoming ridiculous. Commented Mar 26 at 14:28
• There is also another anomaly: If a resistive load is applied to the secondary, it will draw current based on the magnetic flux <-- this is an impossibility; the current drawn is based on the secondary voltage and ohms law but, the secondary voltage is based on the rate of change of flux. Commented Mar 26 at 18:54

Before answering your questions I'd like to explain the simplified real model of a transformer. Because if you understand it, it'll get easier for you to get the answer by yourself:

Image nicked from my answer here.

NOTE: The parasitic elements such as leakage inductance and winding DC resistances are neglected for simplification.

The TX in the red dashed rectangle is the ideal transformer which has 100% efficiency i.e. input power equals to the output power. The ideal transformer has infinite magnetising inductance(s) therefore it doesn't draw any extra current, hence the 100% efficiency.

$$\\mathrm{L_p}\$$ is the primary magnetising inductance that you measure with an LCR meter from the primary when the secondaries are open.

The definitions of marked currents above:

• $$\\mathrm{i_M}\$$: Primary magnetising current
• $$\\mathrm{i_L}\$$: Primary energy-related current
• $$\\mathrm{i_P}\$$: Total primary current which is the vector sum of $$\\mathrm{i_L}\$$ and $$\\mathrm{i_M}\$$ above.

Now if there's an alternating voltage across the primary, there's going to be a magnetising current determined by the Faraday equation:

$$V_p = N \ A_e \ \frac{\Delta B}{\Delta t} = L_p \ \frac{\Delta i_M}{\Delta t}$$

$$\A_e \cdot \Delta B\$$ product is the generated magnetic flux, $$\\Phi\$$, as a result of the applied voltage and $$\\Delta i_M\$$ is the drawn magnetising current by the magnetising inductance.

This magnetising current $$\i_M\$$ is independent from the presence of the load i.e. even if there's no secondary load the primary magnetising current will be drawn (assuming the frequency stays the same). This magnetisation is needed in order for the energy transfer.

So, whilst there's a magnetisation, if you connect a load to the secondary there's going to be an energy-related current $$\i_L\$$ set by the turns ratio and the secondary load current.

Does the current drawn by the transformer on its input change when the load on its output changes?

Yes. Since the total primary current is the vector sum of the energy-related current and the magnetising current mentioned above, the total primary RMS current will be $$\i_{P(rms)}=\sqrt{i_{M(rms)}+i_{L(rms)}}\$$. If there's no load then the primary input current will be equal to the magnetising current.

If we take out the secondary, or the secondary circuit is open, or the secondary only needs very little power, wouldn't the primary still generate the flux and still draw the same amount of current/power, and the flux is just "lost"?

As explained above, even without any load there's going to be a magnetising current, and therefore a magnetisation (flux swing or flux density swing). Flux will not be lost, but the core needs to be reset every cycle so the flux makes plus/minus swings. This will create a power loss on to the parasitic components and the core itself.

How does the generated flux vary with the output load (if it does)?

It doesn't, unless the frequency changes with load (e.g. resonant converters). Think of the flux as something like a "carrier" for the energy transfer. The flux remains the same regardless of the power demand (unless, as I said, the frequency changes).

• I think you should be using $\Phi$ and not $B$ Commented Mar 27 at 16:13
• @Andyaka thanks. Fixed. Commented Mar 27 at 16:44