Re: AC sine wave: What does increasing the frequency do?

From: John Larkin (jjlarkin_at_highSNIPlandTHIStechPLEASEnology.com)
Date: 11/28/04 

Date: Sat, 27 Nov 2004 16:53:47 -0800

On Sat, 27 Nov 2004 17:52:36 -0600, John Fields
<jfields@austininstruments.com> wrote:

>On Sat, 27 Nov 2004 15:18:08 -0800, John Larkin
><jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:
>
>>On Sat, 27 Nov 2004 15:01:06 -0800, John Larkin
>><jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:
>>
>>>The zero crossings obviously can't move, since there can be no current
>>>anywhere in this setup when the line voltage is zero. But the time of
>>>peak current is not simultaneous with the voltage peak, because the
>>>filament resistance varies with time and doesn't peak at the voltage
>>>peak. This is not a paradox, because harmonics are present to make
>>>everything work out. If you were to measure the Fourier fundamental
>>>component of current, *that* would lag the voltage.
>> ^^^
>>
>>Oops, lead. Resistance is higher in the last half of each cycle as
>>compared to the first half. Current leads: It looks capacitive.
>
>---
>Since there are two heating events per cycle, ISTM like it should be
>that resistance is higher in the last half of each _half_ cycle than
>in the first half,

Right, the resistance varies in half-cycles, at 120 Hz, just like the
temperature and light output do.

> but it still looks resistive because current is
>staying precisely in phase with voltage, since where resitance is
>gonna be or where it was doesn't matter. What does matter is what's
>the resistance right now and what's the voltage across it right now.

Phase shift has to be measured over time. No instantaneous measurement
of a circuit can identify a phase shift, even a circuit with real
capacitors. "Gonna be and where it was" is fundamental to a
time-referenced measurement. What matters is how the current waveform
looks compared to the voltage waveform, and a point measurement isn't
a waveform.

>There's no Xl or Xc in the circuit, and without a reactance the
>impedance will be entirely resistive with no difference in phase
>between E and I.

We're not getting anywhere on this, are we. Do you propose that a
triac dimmer, driving a resistive load, runing at 50% conduction
angle, has no current-versus-line-voltage phase shift? Even though all
the load current flows in the last half of each cycle? That seems like
a phase shift to me.

John

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