# My Stove

Discussion in 'Electronic Basics' started by Midnight Oil, Dec 8, 2005.

1. ### Midnight OilGuest

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
volts...at 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)

2. ### John PopelishGuest

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.

3. ### John PopelishGuest

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.

4. ### Rich GriseGuest

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

5. ### Midnight OilGuest

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)

6. ### Pooh BearGuest

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.

Graham

7. ### Peter BennettGuest

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.

8. ### Jasen BettsGuest

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

9. ### James DouglasGuest

Dang, you were really interested in this!

10. ### Rich GriseGuest

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

11. ### Big Mouth Billy BassGuest

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.
<snip bad assumptions>

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.

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