Re: Giancoli Problem

[Gary Turner <turner@MORNINGSIDE.EDU>]

On Thu, 7 Mar 2002 17:10:04 -0600, Peter Schoch <pschoch@NAC.NET> wrote:

> OK, I'm stumped and need a bit of help.  The problem is in Physics for
> Scientists and Engineers by Gaincoli (the latest ed.), Chapter 26 number
> 83.
>
> They give a circuit with 4 resistors and 1 capacitor, something like
> this:
>
> -----------|------------|
>                R2              R1
>                |                  |
>                -----C------
>               |                  |
>               R4              R3
> ----------|------------|
>
> They then connect this to a 12V battery and ask what the voltage across
> the C will be when it is fully charged.  Then they ask what the time
> will be for the C to discharge to 0.05 of its max. value when
> disconneted.  (At least that is a quick recap.)
>
> So, here's my question(s):  aren't  R1 and R2  in parallel and R3 and R4
> are in parallel?  So, shouldn't you combine them first to get an
> equivalent resistance and then treat them in series with the C?  Then,
> for large time the I (current)  through the capacitor is 0, and you
> should be able to relate the V through the capacitor to the voltage
> drops across the two equivalent resistors you just found, correct?
It's a tricky problem.  Wherever you have reistors in parallel with jumpers
across the parallel legs, you really have to use Kirchoff's Rules.  In this
case, as you pointed out, the current through the capacitor will be 0
(fully charged), so the current passing through R1 and R3 must be
identical, also R2 and R4 (but possibly different to R1 and R3).  Both
pairs (R1-R3, R2-R4) must drop 12V across them.  You could then find the
potential (relative to a fixed reference, such as +12V - 0V at the battery)
at each end of the capacitor.  The potential difference across the
capacitor will then be the difference between those values.

During discharge, R2 and R1 will form a series pair in parallel with the
series pair R3 and R4.  1/R = 1/(R1+R2) + 1/(R3+R4)

[The current will flow from the capacitor, split at the T (between R1 and
R3 ?-depending on which side is high potential).  Some will go up and
around, some will go down & around.  The current will recombine at the
other T (between R2 and R4 ?) and back to the capacitor.]

Then ln(0.05)=-t/RC, t=-RC*ln(0.05).

Does this work?


> But if I work the problem that way, I get nowhere near the "book
> answer".  So, I'd appreciate anyone who can give me an insight as to
> where my logic is wrong.
>
> Thanks,
> Peter Schoch