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If you plug in an electrical charger or an adapter of an electronic device such as a mobile phone or a laptop into the power socket, but you don't actually attach the device into the charger, the charger does not consume significantly as much electrical energy as it would if the device were connected and drawing power.

Is it the same way for an induction cooktop? Does it also only consume significant amount of energy when a cookware is actually on top of it and getting heated?

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    $\begingroup$ I'd like to mention, that if the charger/adapter has a transformer, it will still consume electricity. That aside, you can think of it as a switch. When you plug something in, the circuit is closed and electricity can flow. When nothing is plugged in, the circuit is open because nothing is connecting the positive and the negative leads on the charger. $\endgroup$
    – CoilKid
    Commented Jul 12, 2015 at 0:25
  • $\begingroup$ The fact that none of the posted answers (including the accepted one) include a circuit diagram is somewhat disturbing. $\endgroup$
    – DanielSank
    Commented Jul 12, 2015 at 17:15
  • $\begingroup$ @DanielSank How is it disturbing? If you don't include a form of the switch Earnie described, it's really just an inductor. Sure if you add various controls to increase efficiency it gets more complicated, but it's a pretty simple device when you look at it. $\endgroup$
    – CoilKid
    Commented Jul 12, 2015 at 19:04
  • $\begingroup$ This may be of interest... (From Wikipedia) Note the power supply and the control circuit. $\endgroup$
    – CoilKid
    Commented Jul 12, 2015 at 19:06
  • $\begingroup$ @CoilKid In my eight years of working on circuit oriented physics projects I find that without a diagram information given in discussion of circuits is often misconstrued. $\endgroup$
    – DanielSank
    Commented Jul 12, 2015 at 19:09

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I disagree with the other two answers, or at least believe they are not very clear.

The important point is that, as you suggested, a induction cooker only consumes significant amounts of energy when a pot is actually on top of it.

There are, of course other losses, but without any metal object in the vicinity, the cooktop is like a transformer without a load, which (in the ideal case) does not dissipate any energy.

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Induction cooktops contain electromagnets below each pot or pan station. When a station is switched on, electric current flows through wire wrapped around an iron core. In order for magnetic flux to be induced in the iron core, the electric current must constantly change, so the current must alternate. The iron core concentrates the magnetic flux generated by the electric current, and a magnetic field is created on the cooktop.

If you place a ferromagnetic pot or pan above the electromagnet, the changing magnetic field induces electric current in the pot or pan. The induced current swirls around the pot or pan and dissipates its energy against the electrical resistance of the pot in the form of heat.

If there is no pot or pan on top of the electromagnet, no electric current will be induced, but the electromagnetic field will still exist, and the circuit will still draw current according to Ohm's law

$$I = V / R$$

where $I$ is current, $V$ is potential difference across the circuit, and $R$ is resistance.

Induction cooktops must have time changing current in order to generate a magnetic field, so the simple Ohm's law must be altered to include capacitance and inductance. The result is that resistance is replaced by impedance.

Nevertheless, the idea is the same, and even without inducing electric current above it, the electromagnet will draw current which is directly proportional to the voltage input, and indirectly proportional to the impedance. The amount of current drawn by the electromagnet when there is no pot being heated will be minimal, as Daniel pointed out in his answer.

Induction cooktops have automatic circuits which shut off the electromotive force to a cooking station if there is no pot or pan on it. But in the absence of such a shut-off, current would continue to be drawn, though an insignificant amount with no pot being heated.

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  • $\begingroup$ For the record, you posted while I was typing. Also, I like your answer better. :) $\endgroup$
    – CoilKid
    Commented Jul 11, 2015 at 23:48
  • $\begingroup$ Would you comment on this another answer that seemingly contradicts yours? $\endgroup$
    – laggingreflex
    Commented Jul 12, 2015 at 15:27
  • $\begingroup$ @laggingreflex: I agree with Daniel's answer, and I should have made clear that although energy will be consumed, the amount of energy consumed will be significantly less than if a pot is being heated. I will edit my answer. $\endgroup$
    – Ernie
    Commented Jul 12, 2015 at 16:33
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I believe an induction cooktop works by creating an alternating magnetic field, which heats any ferromagnetic materials close to the inductor.

If you have an inductor switched on, but you're not heating anything, the inductor is still producing a magnetic field. It takes energy to sustain such a field, even if there's nothing being heated.

Also, even if the cooktop inductor does not appear to get hot, any wire with current passing through it gives off heat energy. The amount of energy given off depends on the amount of current, and the material the wire is made from.

So no, it would consume energy as long as it is switched on.

Edit:

Without a pot/pan or other ferromagnetic material near the inductor, it will consume significantly less energy than it would while actively heating a pan.

It also appears from Ernie's answer that that type of cooktop can have an automatic switch-off that I did not know about. If it does indeed have a switch-off, then it would be safe to assume that as long as it is functional, the cooktop will not be using large amounts of energy.

*Note:

Even with the switch-off, if it has one, it would be inadvisable to rely on it to turn your stove off for you. You should always turn your stove completely off before leaving it unattended, even if it used very little energy without a pot on it. That said, I doubt that was the point of your question.

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  • $\begingroup$ Would you comment on this another answer that seemingly contradicts yours? $\endgroup$
    – laggingreflex
    Commented Jul 12, 2015 at 15:28
  • $\begingroup$ @laggingreflex Yes it will consume significantly less energy without a load, and I probably should have mentioned that. However, that doesn't change the fact that as long as it is switched on, it will use energy. I don't believe his answer invalidates mine, just suggests that clarification is in order. $\endgroup$
    – CoilKid
    Commented Jul 12, 2015 at 16:46

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