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My Stove

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

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  1. Midnight Oil

    Midnight Oil Guest

    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

    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

    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

    Anyone have any idea how this can be?

    - Jamie

    The Moon is Waxing Crescent (47% of Full)
  2. 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. 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 Grise

    Rich Grise Guest

    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!
  5. Midnight Oil

    Midnight Oil Guest

    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 Bear

    Pooh Bear Guest

    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.

  7. 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
    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 Betts

    Jasen Betts Guest

    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

  9. Dang, you were really interested in this!
  10. Rich Grise

    Rich Grise Guest

    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"

    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. :)

  11. 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
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