Is induction cooker a pure resistance appliance? Can we say that part of the induction cooker is used for heat dissipation, because this part is very small, so it can be ignored? So it can be regarded as pure resistor?

Is induction cooker a pure resistance appliance? Can we say that part of the induction cooker is used for heat dissipation, because this part is very small, so it can be ignored? So it can be regarded as pure resistor?


The induction cooker generates an alternating magnetic field through the components of the electronic circuit board. When the furnace surface is placed at the bottom of the iron containing cooker, the cooker cuts the alternating magnetic line of force and generates an alternating current (i.e. eddy current) in the metal part at the bottom of the cooker. The eddy current makes the iron molecules in the cooker move at high speed and irregularly, and the molecules collide and rub with each other to generate heat energy



Do you mean that as long as the resistance of my induction cooker is certain, even if the power is large, the power consumption is also certain? That is, the active power of the watt hour meter remains unchanged?


When the power is high, the active power is increasing, of course, the power consumption is increasing. The household watt hour meter is an active meter, and it will not react to the reactive power. The reactive power is of little significance to individuals. The high reactive power indicates that the quality of power supply is not good, and the power loss is increasing, which is related to the power supply department, and this parameter is only concerned by the power supply department



An electric kettle marked with "220V 1000W" can generate heat when it works normally for 5min___ J. Its resistance in normal operation is____ Ω
It's the same as the working TV___ Yes.


An electric kettle marked with "220V 1000W" can generate heat when it works normally for 5min_ 300000__ J. Its resistance in normal operation is__ 484__ Ω. It's the same as the working TV__ And_ It's connected



What is the resistance of a bulb marked "220V 400W" under normal working conditions and what is the current through the bulb


R = u ^ 2 / P = 220 V ^ 2 / 400 W = 121 Ω
I=P/U=400W/220V