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In a water heater stove or arc welder, the most basic electric circuit is formed when one has a source (AC or DC) and 2 wires going to and from the heating element, in the case of the heater and stove, or to the gap where the arc is formed, in case of the welder.

When the arc forms it creates a very low resistance path in air, for example, allowing for relatively large current. The same thing, just without the arc, is happening in the heating element. Heating element has a relatively low resistance allowing for large currents through the element and the element itself giving out large amounts of heat.

The question is why doesn't the whole circuit go hot as the arc or the heating element, why does this temperature increase happen only at the specific places in the circuit, since the same current is flowing through the whole circuit? is it because the wires need to be of much less resistance then the described elements or is there another explanation? Thanks.

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The power dissipated by a DC current is the product of current and voltage, not just current: $P = IV$. Ohm's law allows us to replace $V$ with $IR$, where $R$ is the resistance of the component. As a result, we have $P = I^2R$.

Now in a given series circuit in steady state, $I$ is fixed and the same everywhere. Thus the relative power dissipated by different parts is proportional to their resistance. Thus the higher resistance components are the ones that heat up, and indeed the arc and the electric heating element are higher resistance than the wires feeding them.

What might be confusing is that their resistances can't be too large. At some point, if a part of the circuit is too resistive, the overall resistance of the circuit is too high, and the fixed-voltage power supply won't induce enough current to flow. That is, as we increase $R$ for a component, there is a secondary effect of decreasing $I$ in most setups. If you blow out the arc or remove the heating element, the circuit now has an extremely high resistance governed by the air gap, but $I$ will be so low that $I^2R$ will actually be negligible.

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  • $\begingroup$ Thanks I was thinking in the same direction, but its kind of confusing at first glance it seems weird :) $\endgroup$
    – adoion
    Sep 3, 2015 at 19:47

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