Re: Cpacitor discharge

On Feb 14, 12:40 am, Andrew Edge <a_n_d_y_e_d_...@xxxxxxxxxxxxxx>
wrote:
On 13 Feb 2007 16:18:07 -0800, "Tom Bruhns" <k...@xxxxxxx> wrote:



On Feb 13, 12:05 pm, Andrew Edge <a_n_d_y_e_d_...@xxxxxxxxxxxxxx>
wrote:
On Tue, 13 Feb 2007 13:19:17 -0500, John Popelish <jpopel...@xxxxxxxx>
wrote:

Andrew Edge wrote:

The only way way you can get a current out of a capacitor is by using
a resistor.
(snip)

You might want to rethink that generality a bit. The
universe is more complicated than that.

It sure is. I am always willing to learn ... so give me an example.

Andy

In addition to what John wrote, consider LC filters and resonant tank
circuits. There may be some resistance in there, but the reactance of
the inductors generally dominates over the effects of the
resistance.

The Op is talking about discharging a capacitor. How do you normally
do that? I use a resistance. If you use LC filters or resonant tank
circuits well that is your choice. The case in consideration doesn't
mention any of your choices if I remember right. Besides I would
challenge that statement that the reactances dominate.
If resistance was not a factor in an LC circuit the circuit would
oscillate indefinitely AND that doesn't happen. Dissipation of energy
in the resistances takes place.
In the case the Op is considering, we talk about the reactance of
the capacitor, and it DOES NOT dominate over the resistance.

Consider capacitors used for power factor correction across a low-
frequency AC power line; the current is proportional to the rate of
change of voltage, driven by the power line. Half the time, current
is charging the capacitor and half the time, it's "coming out" and
discharging the capacitor. It would be very difficult to relate the
capacitor current to what happens in some resistor in that circuit,
and resistance is not required to adequately understand what's going
on in it.

It is quite a simple thing for an engineer. A Kirchoff loop equation
with the Voltage and Current phasors would get you there in a second.
Simulators use such equations. So why shouldn't we.
But that is another case. There is no outside Power supply in our
case ... discharge is the keyword. It is a sourceless circuit. The
equations and the relations are a completely different story.



You wrote in your previous posting about the exponential behaviour of
current and voltage in RC circuits, but that's all derivable from the
much more general

Q = C*V ----> i(t) = C*dv/dt + v*dC/dt
and of course v=i*R.

Yes for sure, otherwise how we would have ever got those relations.

Cheers,
Tom

Andy


Lessee...the OP stated that the current is constant, not i=v/R.

Lessee...you asked for examples that disproved your statement,
"The only way way you can get a current out of a capacitor is by using
a resistor." That seems pretty absolute to me, and not a reference to
the OP's question about the capacitor discharging at a constant
current. I think we've given you several good examples. Take 'em or
leave 'em.

As I re-read the OP's original posting, I think he's _very_ clear that
the current discharging the capacitor is constant, and he posts the
correct equations for calculating the discharge as a result. He does
ask about exponential discharge, but that would be only if the load
looks like R across the cap, and he's already stated it doesn't, by
writing that it's a constant current.

Cheers,
Tom


.

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