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As the title says, why does that not happen. As far as my understanding goes, an induction cooker is a coil of conductive wire through which AC is pumped. This generates an oscillating magnetic field around the coil, which in turn generates a DC current in the pot sitting on the cooker. Since the pot is not made of a superconductor, the current flowing through it heats it up, and that's where we get the electricity from.

Now this is where i get confused. First, yes, there's a current, but is there a potential difference between any points, or are the electrons just really excited ? If there's a potential difference between any 2 points in the pot, then why does touching it not give us an electrical shock?

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    \$\begingroup\$ Ever touched a 9V battery? \$\endgroup\$ – PlasmaHH Dec 2 '16 at 9:26
  • \$\begingroup\$ are you saying the amperage is too small? or that the voltage induced is so small that is doesn't have the power to penetrate the skin? .. Sorry if these questions don't make sense... i'm here to learn \$\endgroup\$ – omu_negru Dec 2 '16 at 9:27
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    \$\begingroup\$ the "shock" you get from touching electricity is usually because of high voltage, snice your body resistance is more or less the same. \$\endgroup\$ – PlasmaHH Dec 2 '16 at 9:32
  • \$\begingroup\$ A bit over-simplified you can see at as many many turns on the primary coil in your stove and just one turn in your secondary (your pot) so the induced voltage is very low to start with. Think of @PlasmaHH 9 V battery but with 500 A behind it. Touch it and nothing happens. Put your cooking pot against the poles and it heats up significantly. \$\endgroup\$ – winny Dec 2 '16 at 11:27
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    \$\begingroup\$ "which in turn generates a DC current in the pot sitting on the cooker" - - - Just to set the record straight, AC is generated (induced) in the pot, Not DC. \$\endgroup\$ – Marla Dec 2 '16 at 14:52
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The induced current in the pot (pan), is on the bottom of the pot where you are not likely to touch. In fact you would have to lift the pot in order to touch where the current is being induced.

By lifting the pan up and away from the induction coil, you will reduce the induced current in the pan (and reduced voltage, which is low already).

Very very little current would be induced in the side of the pot (where you can easily touch).

Comments made above correctly state that the induced voltage is low and that induced current is high (all relative of course).

"First, yes, there's a current, but is there a potential difference between any points"

You could measure a potential difference (voltage) by measuring on the bottom of the pot between two points that are on opposite sides of the pot bottom. The voltage will be very low (not going to address the value of voltage here).

Thus, if you were to able to touch the bottom of the pan in 2 places (difficult to do with pan sitting on induction heater), on opposite sides , you might get a mild shock.

And last,

"which in turn generates a DC current in the pot sitting on the cooker"

Just to set the record straight, AC is generated (induced) in the pot, Not DC

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  • \$\begingroup\$ I guess that more than answers my question, faulty formulated as it is. Thanks for the patience \$\endgroup\$ – omu_negru Dec 2 '16 at 16:42

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