+---------------
|
[email protected] wrote:
| >>>It's current that causes heat rise (I^2R).
| >>... and voltage (accuracy of voltage control method), and frequency.
| >But isn't it the curent that does the heating?
|
| No, current only flows based upon the difference in
| potential voltage. It is a result not cause.
+---------------
No, it takes *both* current & voltage to get heat!!
Look, it's like any other mechanical potential -- gravity,
springs, whatever. You have potential energy -- a difference
in height, compression of a spring, voltage between two points
in a circuit -- and then you have *work* (actual energy consumed),
which is the exertion of a force across some "distance" of potential
energy. When a rock falls off a shelf, its potential energy of
position is converted first into kinetic energy (increasing speed
as it falls), and then into thermal energy when it hits the floor.
When the rock *slides* at a near-constant speed down a rouch ramp,
the same total thermal energy is released -- it's just that the
conversion from potential energy to thermal energy is more continuous.
Likewise, when an electron is sitting on one side of a charged
capacitor, it's not doing any work -- its energy is all potential.
It is not until the electron flows "downhill" from the higher
(negative) potential to the lower that its potential energy
is converted to heat (in a series of very many small increases
of kinetic energy followed immediately by conversion to thermal
energy as it bounces around inside whatever conductor it's in).
"Current" is just a count of the number of electrons flowing
per unit time (Q/t), so the more current (I) is flowing across
a larger voltage (V) the greater the dissipated thermal energy (E),
that is, E = I * V * t. Or the more familiar P = I * V, if you
want power instead of total energy. But it takes *both* voltage
and current to get heat!
Frequency only comes into it if you're charging and discharging
capacitors. The number of electrons (in Coulombs) or "mass" entrained
in a capacitor is C * V, and the voltage is V, of course, so the
total available potential energy ("mass" * "height") is C * (V^2)
[or 0.5*C*V^2, I forget]. Charging the cap stores energy in the cap
[*and* dissipates energy in the wires leading to the cap]; discharging
a cap dissipates energy in the wires leading away from the cap [due
to the current (charge/second) flowing across a potential voltage].
One full change/discharge cycle dissipates some specific amount of
energy [never mind the exact constants], which is roughly linearly
proportional to the value of the capacitance times the square of
the voltage. Since each cycle dissipates a fixed amount of energy,
the more cycles per second you have, the more *power* (energy/second)
you dissipate. Q.E.D.
-Rob
.
