Does a wire in an electrical cord of an electrical kettle have a higher or lower resistance than the heating element inside the kettle?

Answer 1

The heating element in the electric kettle will be primarily pure resistive. Power in that case is calculated

P = E*I = 250V*I = 3000 W

So the current is

I = 3000 W/250 V = 12 Amps

If there is anything else (lights, radio, PC, etc.) on the circuit that may be taking some current you are very likely to go over 13 A frequently. I don't know how accurate you can expect a fuse's blow point to be. Perhaps +/- 10%? Doesn't seem very trouble free.

Even if the circuit is dedicated to the kettle, consider this idea. We don't know exactly what conditions the manufacturer was considering when specifying the wattage. Probably operating conditions, when the element was hot. Then Ohm's Law tells us that the resistance R is

R = E/I = 250 V/12 Amps = 20.8 Ohms

But the element's resistance would be lower at first when it is cold.

My calculations indicate that if the element is made of nichrome wire, and if the temperature change is 200 degrees C, the hot resistance of 21 Ohms is probably about 1.7 Ohms higher than when at room temperature. If that is true, when you first turn it on, Ohm's Law says that the initial current should be 12.95 Amps.

Regardless of this analysis, remember that the fuse requirement is based on the size of the wires in the wall. You don't want those wires to get hot. Don't put in a bigger fuse than what was there originally.

Answer 2

The wire in the cord has much lower resistance than the heating element. The heating element is a resistive or resistance heating element. Resistance in the quality of a substance or material that causes it to limit current flow, and it get heated up in the process. The heating element has all but the smallest fraction of the resistance in the circuit, so the heat, that thermal energy that get the water hot, is generated by the resistance of the heating element.