Re: Question about capicitors
- From: Alan B <nannerbac@xxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 03 Oct 2006 16:25:32 -0700
On Tue, 3 Oct 2006 18:36:44 +1000, in message
<452220da$0$5105$afc38c87@xxxxxxxxxxxxxxxxxxxx>, "Johnny Boy"
<NoSpam@xxxxxxxxxx> scribed:
In the math
model, one doesn't see direct current until the capacitor is charged and
the transients settle.
Correct me if I'm wrong, but I would have thought that ALL you'd see IS
direct current, while the cap is initially charging, and that from a source
with a low resistance, at the instant that the voltage is applied, the
capacitor "appears" to be a short.
Yes to the last statement. Taking a simple series circuit containing a
capacitor: at t = t0, applying a "DC" source, the capacitor drops zero
voltage (axiom: the voltage cannot change instantaneously across a
capacitor), and hence will at that instant draw a current equal to the
voltage source divided by whatever resistances are present. The voltage
across the cap will then rise - as the current falls - logarithmically
until the voltage stabilizes at the "DC" calculated level (tn - in this
example, that level would be zero). The current is a curve - a transient -
between t0 and tn, and as such isn't calculated by steady-state DC
equations. That's what I mean when I say that until tn, there is no direct
current in the circuit, rather there is only transient current.
.
- References:
- Question about capicitors
- From: Dana
- Re: Question about capicitors
- From: Alan B
- Re: Question about capicitors
- From: Johnny Boy
- Question about capicitors
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