http://elecraft.85.s1.nabble.com/OT-WWV-Time-Transmissions-tp7664561p7664642.html
>
> You neglect to fill in the missing part where voltage is causing the
> current which is causing the heat.
>
> The use of a magnetic core can increase the strength ofmagnetic field
> <
https://en.wikipedia.org/wiki/Magnetic_field>in anelectromagnetic
> coil <
https://en.wikipedia.org/wiki/Electromagnetic_coil>by a factor
> of several hundred times what it would be without the core. However,
> magnetic cores have side effects which must be taken into account.
> Inalternating current
> <
https://en.wikipedia.org/wiki/Alternating_current>(AC) devices they
> cause energy losses, calledcore losses
> <
https://en.wikipedia.org/wiki/Core_losses>, due tohysteresis
> <
https://en.wikipedia.org/wiki/Hysteresis_loss>andeddy *currents*
> <
https://en.wikipedia.org/wiki/Eddy_current>in applications such as
> transformers and inductors. "Soft" magnetic materials with
> lowcoercivity <
https://en.wikipedia.org/wiki/Magnetic_coercivity>and
> hysteresis, such assilicon steel
> <
https://en.wikipedia.org/wiki/Silicon_steel>, orferrite
> <
https://en.wikipedia.org/wiki/Ferrite_(magnet)>, are usually used in
> cores.
>
> Sudden high swr issues at high power are caused by insulation breakdown.
>
> The heat is directly proportional to the current producing it. You
> half the current and therefore halve the heat.
>
> However this does not apply to voltage as current flow cannot be taken
> for granted. talking only about voltage, as voltage can in a current
> equation.
>
> You can halve the voltage and may have little heat due to the voltage
> breakdown no longer in place.
>
> Magnetism is created by current, Magnetism cutting a conductor
> produces current which produces heat.
>
> Voltage only defined, never has and never will be responsible for heat
> production P =I^2 R .
>
> *Dielectric loss*quantifies adielectric material
> <
https://en.wikipedia.org/wiki/Dielectric_material>'s inherent
> dissipation of electromagnetic energy (e.g. heat).^[1]
> <
https://en.wikipedia.org/wiki/Dielectric_loss#cite_note-1> It can be
> parameterized in terms of either the*loss angle*/δ/or the
> corresponding*loss tangent*tan /δ/. Both refer to thephasor
> <
https://en.wikipedia.org/wiki/Phasor>in thecomplex plane
> <
https://en.wikipedia.org/wiki/Complex_plane>whose real and imaginary
> parts are theresistive
> <
https://en.wikipedia.org/wiki/Electrical_resistance>(lossy) component
> of an electromagnetic field and itsreactive
> <
https://en.wikipedia.org/wiki/Reactance_(electronics)>(lossless)
> counterpart
>
> It is the *current* induced by this electromagnetic field, not
> cancelled by back emf, that causes heat.
>
>
> On 3/9/20 12:25 pm, David Gilbert wrote:
>>
>>
>> OK ... I'm going to make this simple for you. Picture a material
>> (like a capacitor) with a lossy dielectric, and then apply a high RF
>> voltage across it. The dielectric passes a current as the result of
>> the voltage, and the lossiness of the dielectric generates heat. With
>> a lot of voltage the heat generated can be considerable.
>>
>> Now then, whether you want to acknowledge it or not, a ferrite core
>> IS a lossy dielectric and can get hot when you put a high enough RF
>> voltage across it INDEPENDENT OF THE CURRENT FLOWING THROUGH THE
>> WINDING AROUND THE CORE. You can find innumerable references to the
>> dielectric losses of ferrite materials if you just bother to do some
>> internet searching. This is NOT an insulation breakdown issue ...
>> not at all.
>>
>> All of this can happen as soon as power is applied to the system
>> containing the ferrite. Certainly the rate of temperature rise will
>> be dependent upon how much voltage is applied, the frequency of it,
>> and the dielectric loss characteristics of the particular ferrite,
>> but VSWR is the voltage we're talking about here and that becomes
>> relevant immediately upon application of power.
>>
>> I suspect that you will dig in your heels and continue to dispute
>> this basic physics, but at least I hope others here will understand
>> things better than you do.
>>
>> Dave AB7E
>>
>>
>>
>> On 9/2/2020 5:52 PM, Adrian wrote:
>>> I say that your response is completely false and you are missing
>>> basic electricity facts. The high voltage becomes an issue when
>>> insulation breaks down, and then *current *starts
>>>
>>> to flow through the fault path converting to emf & heat directly and
>>> via induced current resulting ; P = E X I*. *Without the current
>>> the heat does not occur, it is basic physics, and
>>>
>>> the heat is directly proportional to the current. Voltage can exist
>>> without current, but current cannot exist without voltage. Heat
>>> produced is directly proportional to the current whether
>>>
>>> it be in the intentional circuit path, or fault path caused by high
>>> voltage insulation breakdown..
>>>
>>>>
>>>> In addition, your statement that only current in the balun circuit
>>>> can produce heat is completely false. High voltage RF can create
>>>> major core heating due to dielectric losses in the ferrite core
>>>> independent of the magnitude of current flow in the tuning circuit.
>>>> Several discussions on the TowerTalk reflector have pointed this
>>>> out over the years for baluns and common mode chokes in ham radio
>>>> applications.
>>>>
>>>> Dave AB7E
>>>>
>>>>
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