# The Resistance in an electric kettle [closed]

I have an electric kettle with two parallel heating elements, both with resistance $$R_0=2\,\Omega$$. Both can be independently shut on and off. There is also another resistor with resistance $$R$$. Upon using the kettle, I notice that there is no difference whether I use one or two heating elements, the same amount of water is heated in the same amount of time. How big is $$R$$? $R_0$" />

I have absolute no clue how to tackle this problem.

• Think about it. The parallel combination of the two Ro resistors is 1 ohm. If there was no resistance R and your kettle were connected to a 120 vac outlet with both switches closed, if would draw 120 Amperes or 14,400 watts!!. What does that tell you R must be compared to Ro? But what I am really curious about is what does the manual say about the purpose of the two switches? May 4 at 20:04
• Does the heat from R also go into the kettle? Do any of these resistors have inductive reactance? I'm guessing that R is a current limiting inductor. May 4 at 20:13
• @BobD The OP is German, so the VAC will be 220 V, even worse! ;-)
– Gert
May 4 at 20:28
• Calculating the actual value of $R$ isn't possible without additional information.
– Gert
May 4 at 20:29
• @Gert Yes I am German and tired ;) Can I assume that the electric current is constant? May 4 at 20:34

The parallel combination of the two Ro resistors is 1 ohm. If there was no resistance R and your kettle were connected to a 120 vac outlet with both switches closed, if would draw 120 Amperes or 14,400 watts!!. What does that tell you R must be compared to Ro? But what I am really curious about is what does the manual say about the purpose of the two switches?

Is it even possible to find a concrete value for R, without imagining what outlet I am using? Because I now have the value of R=(RoU2+(-RoU1)/2)/(-U2+U1) (can I use LaTex in Comments?) Which would mean R highly depends on the Voltage I am using. Or is it wrong to assume that I, the amperage, is constant? About the manual, I have no idea.

You really can't get a concrete value for R simply on the basis of your subjective stated observation "I notice that there is no difference whether I use one or two heating elements". You can only say that R should result in a small enough difference in power dissipated for you not to notice the difference between one or two resistors Ro connected.

In any case, as I understand it most electric kettles consume about 1500 watts power. That's about 12.5 amperes for a 120 vac source. Using Ohm's law we can calculate the resistance R assuming a full load current of 12.5 amps.

$$I=\frac{120}{R+1}=12.5A$$

Which gives you R = 8.6 Ohms.

So if you have one Ro resistor switched on the total resistance would be 10.6 Ohms for a total power of 1358 watts and if you had two Ro resistors turned on the power would be 1500 watts, for a difference of 142 watts, or about 10% difference. Can't say, however, whether or not you would "notice" the difference without doing actual measurements of the time it takes to heat the water.

Hope this helps. Assume both switches on the $$R_0$$ closed.

With the presence of $$R$$ the device can then be seen as a voltage divider, because:

$$U=U_1+U_2$$

$$U_1=\frac{U}{R+R_0}\times R$$

$$U_2=\frac{U}{R+R_0}\times R_0$$ The power dissipated by the $$R_0$$ is then:

$$P_2=\frac{U_2^2}{R_0}$$ $$P_2=\frac{U^2}{(R+R_0)^2}\times R_0$$ So the higher $$R$$, the lower $$P_2$$. Without a power rating ($$P_2$$) no value for $$R$$ can be determined.