My Stove

I'm new to electronics/electricity, and I've been reading up on ohm's law, and trying to get a feeling for how current, voltage, and power relate in electric circuits, and I ran into something that confuses me. I wonder if I can get some help...

I was cooking some dinner yesterday, and I was looking at the glowing red stovetop cooking elements last night. I thought "there must be a lot of current flowing through those elements to generate that much heat and to get them to glow red hot". So, once the element cooled down, I hooked my ohmmeter up to the element, and it registered 24 ohms. I thought, that must be reasonable. A low resistance means that there is more potential to generate heat rapidly and heat the element. Then I wondered:

"There are wires under the stovetop which run to those elements...surely they must be glowing red from all of the current, because they're part of the same circuit...and current is the same at all points in a series circuit..."

I ripped the stovetop off, and looked inside...what I saw surprised me! The wires weren't particularily thick (looked like 8 gauge wire or so), and went to flimsy plastic coated connectors. *HUH*?! I would have thought that the high current flowing through the elements would also be flowing through the connectors, and would definately melt them in a matter of less than a minute...Anyone have any idea how this can be?

The other thing that puzzles me: I removed the heating element, and placed my voltmeter across the leads that connect to the element. I turned the variable control on, and the voltage at low heat registered at 240 high heat...240 volts. The only guess I could make was that maybe they are switching the current on and off very rapidly at the same voltage through the control module. So quickly that the meter wouldn't detect it...but maybe some kind of a square wave where they have a broader peak on high heat, and narrow peak at low...just an idea. I suppose switching the heat on and off very quickly could work, and might have an advantage over simply varying the magnitude of the voltage from 0-240 volts.

Anyway, even when the switch is on high, with a supposed broad square wave peak, there would be enough current flowing through that circuit to melt the rest of the circuitry inside of the stove, not to mention the controls.

Anyone have any idea how this can be?

- Jamie

The Moon is Waxing Crescent (47% of Full)

Reply to
Midnight Oil
Loading thread data ...

I'll do what I can.

Power is produced by current passing through a voltage drop. Resistance is what causes the voltage drop. In fact another way to say ohms is volts per ampere. So power is volts times amperes. But ohms law relates volts and amperes to resistance by Ohms = Volts per ampere. We can combine these two formulas to solve for power in terms of resistance and either current or voltage. Power = voltage squared divided by resistance = current squared times resistance.

The last one is probably the easiest to apply to the range circuit. The pieces of wire that share the element current probably have only a tiny fraction of an ohm. Lets say 10 milliohms for the sake of discussion. If 240 volts is applied to the element, it will pass about 10 amperes current. The power dumped into the element would be about 10^2*24=2400 watts. The same current passing through 10 milliohms (.01 ohm) of wiring will dump 10^2*.01=1 watt of heat into the wire.

It is more likely a slow cycle of off and on. the controls often have thermal relays and a small heater. Different positions of the control produce different amounts of heat, which vary the time it takes before the thermal relay kicks off the power to both the element and the small heater. If the small heater is a low resistance type in series with the element, the cycle will never time out if the element is removed.

The controls are very simple. As long as the on off cycle is quicker than the cool off time of the element, you will not notice the cycle. But it takes place in seconds.

Reply to
John Popelish

Power dumped is proportional to volts across a resistance times the current through it. The wires have a resistance very much lower than the element, so they drop very little voltage (like, tiny fraction of a volt), while the elements drop almost all of the 240 line volts. Lets sat the wiring has a resistance of 10 milliohms (.01 ohm). The current through the series combination of wiring and element is

240/(24 + .01) or about 10 amperes. The power dumped into the elements is 240 volts times 10 amperes or about 2400 watts. The wiring drops about 10*.01 = .1 volt when 10 amperes passes through it. That means it receives a power of .1 volt times 10 amperes or 1 watt of heat.

Most electric ranges have very simple time cycle controls based on a small heater (either a high resistance one in parallel with the element, or a low resistance one in series with it) and a thermal relay that switches off when it gets warm. The cycle time is usually a few seconds on and a few seconds off in the middle of the dial. The element has enough thermal mass to average these pulses out so you don't notice the cycle. If yours is based on a low resistance series heater, it will not get hot if you pull the element out, so the cycle never times out.

Reply to
John Popelish

Thanks to you and John for helping me to understand this...once you guys explained it that way...the lights came on. Thanks!

- Jamie

The Moon is Waxing Crescent (48% of Full)

Reply to
Midnight Oil

John P went into great depth on this, which I think you don't really need.

The difference is "resistance."

Ah! Here's where you can do the test yourself, and really get a feel for "resistance" - measure the resistance of the copper wire, from the end where it plugs into the heating element to the end where it plugs into that flimsy plastic connector.

Yes - the element has more resistance than the copper wire.

No, they have a thermostat. Even when you turn it to "low", it simply turns on the element - that's your 240. But in "low", when the stove is intact, it senses when the element is giving off "low" heat, and turns it off. It goes on and off like that, at a lower duty cycle for low heat, but on a timescale that's commensurate with the thermal inertia of the element.

So you're almost right. :-) An ordinary light dimmer adjusts the duty cycle on a per-cycle basis, but there's no need to do that with stove- sized heating elements. (or it'd be too expensive, when a thermostat will do).

Well, I answered this in advance. It's just on-off, very much like the thermostat for the house furnace.

Yeah - the conductors are made of copper, which has a very low resistance, and will allow the current to easily flow to the load, which is the heating element. That's where the work is done (making heat), and they use copper (and sometimes aluminum) to transfer the energy from the power plant to your heating element.

Hope This Helps! Rich

Reply to
Rich Grise

The wires within the elements are made of a relatively high resistance material (usually Nichrome), while the supply wires are of much lower resistance copper. The resistance of the copper wire from the control to the element will likely be a small fraction of an ohm - probably too small to read with your meter - so it will dissipate very little power.

The control may be a thermostat - it may measure the element temperature, and if the temperature is below its setpoint, will apply full power, then turn off when the element reaches the testpoint.

Peter Bennett VE7CEI 
email: peterbb4 (at)        
 Click to see the full signature
Reply to
Peter Bennett

To melt a connector requires heat.

The heat generated is I^2*R.

Connectors are made to have low contact resistance. Hence the heat generated by current passing is low. So no melted connector.

The heating *element* however has intentionally high resistance compared to the wire ( and connectors ) so most of the heat in the circuit is generated there - where it's needed.


Reply to
Pooh Bear

Dang, you were really interested in this!

Reply to
James Douglas

that's why they use teflon or asbestos inulation on the wires under stovetops :)

but seriously the thing is that the wires have enven lower resistance say 0.24 ohms.

so when the power is turned on most of the voltage drop is in the element so it gets hotter than the wires (which only get warm)

they actually switch it quite slowly, try connecting a 100W light bulb there instead of your meter.

P = I x V in any series circuit I is the same in every part of the circuit. V however differs in proportion to the resistance of the part so parts with low resistance like the wires get less of the voltage (and therfore don't get as hot)

parts with higher resistance (like the element) get more voltage and get hotter.

In a parallel circuit it's the other way round everything gets the same voltage and things with low resistance get more current and thefore get hotter

Bye. Jasen

Reply to
Jasen Betts


He's just got a script that reads out the phase of the moon like a clock, that he executes for his .sig file. In my "Wacko" persona, I have an executable .sig file which picks a random quote from the "fortunes" directory.

i.e., once he's got the program installed, it just automagically comes up with the current phase of the moon and he doesn't have to do anything at all. :-)

Albeit, it is more fun to just look at the actual moon, but then you have to remember if it's waxing or waning. :-)

Cheers! Rich

Reply to
Rich Grise

On Wed, 7 Dec 2005 20:53:12 -0600, via , Midnight Oil spake thusly:

What is the background thinking behind that conclusion? How did you reason that a low resistance gives more potential for rapid heat generation? Your answer will allow an insight into how well you understand how power is generated.

This thought shows that perhaps you don't quite grasp how current produces heat (or watts, to be more precise).

You are the Hulk, and I claim my pair of Marvel Comics boxer shorts.

Heat (or watts, or power) is generated by the combination of current and voltage (P = I * E, yep, that spells pie). The relationship of voltage and current to resistance is critical (E = I * R, Ohm's law), and allows the formula to be alternatively expressed as P = I^2 * R (current squared times resistance).

Now on to your stove, which I hope is not too badly damaged. Add all the resistances of the circuit together, calculate the total current, then apply the power equation to each bit of the circuit you are concerned about. In this case, let's assume 23.98 ohms for the element, and a hundredth ohm each leg for the wires and connectors.

240 / 24 = 10A total (Ohm's law), 240 * 10 = 2400 watts total (the PIE equation). That's plenty hot, and will steam your espresso and fry your eggs and bacon. The wires and connectors, however, are generating only 0.02 * 100 = 2W total (current squared times resistance), 1 watt each leg. Significant, but not hot in comparison to the total, and not hot enough to melt but the flimsiest of wires.

As others have noted, the heat is controlled by a thermostat, which changes sensitivity according to the heat setting. As others have apparently not yet noted, you are getting constant voltage because you have no heating element (you removed it), and thus no heat source to trigger the thermostat. The stove sees a cold element, and will not switch off until the heat expected by the thermostat setting is reached.

Reply to
Big Mouth Billy Bass

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.