1
$\begingroup$

Skin effect causes the currents to flow more on the surface layers of conductor and less on the middle. This increases efective resistance felt.

But does this effect also reduce the current that will flow (compared to scenario at same voltage amplitude, but now DC with the circuit that never had any reaktance components)?

I am asking this because in induction heating if you increase the frequency, the skin depth decreases which increases the resistance.

Losses are calculated by P=I^2*R, which means that losses would increase if we increase the frequency because resistance increased. But wouldn't increase in resistance because of skin effect reduce the currents, so the overal losses would now be smaller?

I am also aware of the effect that increasing frequency increases emf. But if we ignore the increase in emf, would the current and so losses increase or decrease, or would the current stay the same?

Improve this question
$\endgroup$
2
  • $\begingroup$ Isn't the answer sel-evidently "yes" since resitance/impedance is by definition the ratio of voltage to current. Can you clarify why there is a dilema? $\endgroup$
    – JMLCarter
    Commented Oct 24, 2017 at 8:42
  • $\begingroup$ I thought so. It is a dillema because I am not completely sure if that resistance is only used for calculation of the losses, or it is an actual resistance. This is because The current distribution is forced by changing magnetic field into this distribution. Why I am also confused is because in induction heating if we decrease the skin depth, the power Will increase even though the resistance Will increase (and because of this current should decrease). And because losses are dependant on the current squared, this doesn't make sense since the losses should decrease instead of increase. $\endgroup$
    – MaDrung
    Commented Oct 24, 2017 at 10:09

1 Answer 1

Reset to default
1
$\begingroup$

Your use of $P=I^2R$ is giving the wrong answer here, because I and R are interdependent. You cannot increase R without a (more significant) decrease in $I^2$.

Instead, use $P=\frac{V^2}{R}$

As (rms) V is constant, this gives the true relation to power of a change in resistance in this case.

As you expected, an increase in resistance will lead to a reduction in power loss.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Ok. Then consider an induction heating system which has a coil trough which the alternating current flows. If we put on a pot with low resistance and low permeability (which doesn't reduce skin depth as much) like aluminium, the resistance will be very low. This implys that for a given voltage the induced current should be big, which means the losses should be big. But the effect is just the opposite. The losses in aluminium are very small compared to those in materials with small skin depth (hysteresis losses contribute only up to 10 % of complete losses so they can not be the reason). $\endgroup$
    – MaDrung
    Commented Oct 24, 2017 at 11:01
  • $\begingroup$ What I find in literature is that the resistance of the pot must be big in order for the biggest losses to occure in the pot. So if they decrease the skin depth, the resistance increases which increases the losses. But then this would mean that if I put in there a plastic pot it would have the biggest losses because of highest resistance. This is why I think it's related to the skin depth and that the current remains the same, even though the effective resistance increases. But I think I've reached the edge of your expertise, so I will not press further. $\endgroup$
    – MaDrung
    Commented Oct 24, 2017 at 11:32
  • $\begingroup$ Can you clarify are you talking about cooking, i.e. a cooking pot, or electronics, i.e. a potentiometer? anyway it seems the pot is not the same thing as the induction coil. $\endgroup$
    – JMLCarter
    Commented Oct 24, 2017 at 11:33
  • $\begingroup$ In this case I am talking about cooking hobs and cooking pots. $\endgroup$
    – MaDrung
    Commented Oct 24, 2017 at 11:35
  • $\begingroup$ The current in the cooking pot/pan is induced by magnetic coupling, not by a voltage. The pan base has to be thicker than the material skin depth in order for eddy currents to form before it will heat. THe heating is the consequence of the differential in resitance through the pan base depth as quantified by the skin depth. $\endgroup$
    – JMLCarter
    Commented Oct 24, 2017 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.