# Parellel RC with DC voltage source?

Discussion in 'Electronic Basics' started by E. Thomson, Dec 1, 2004.

1. ### E. ThomsonGuest

This should be a simple analysis. I am analyzing a parallel RC
circuit, with a DC voltage source. I have not yet studied AC analysis.
The circuit looks like this:

_________
| | |
V R C
| | |
|____|____|

V is really V*u(t), so the voltage is turned on at t=0.

I want to calculate Vc and Vr. On one hand, Vc and Vr should be the
same, as they are in parallel. On the other hand, Vc cannot jump
to voltage V instantaneously. I have not been able to set up the
differential equation to get a solution with any time-varying
behavior at the capacitor.

Thanks for any help, especially with setting up the equation to get
this right.

2. ### CFoley1064Guest

Subject: Parellel RC with DC voltage source?
If the R and C are in parallel, the voltage rise across the R and C will be the
same. Also, the rise in voltage across the R and C will be entirely dependent
on the internal resistance of the DC voltage source and the residual resistance
of the wire.

I believe you're thinking about R and C in series. This is covered here:

http://people.sinclair.edu/nickreeder/eet155/mod02.htm

Actually, if you can find the internal resistance of your voltage source, you
can use this information to calculate the voltage across the R and C for any
given time with this information, too.

Good luck
Chris

3. ### John PopelishGuest

The circuit has only two nodes, so it can have only a single voltage
difference between those two points. You have arbitrarily defined the
voltage between those two nodes, and it will appear across all paths
between those nodes. However, an instantaneous step change in the
voltage across the capacitor will require an impulse of current
(infinite magnitude, zero duration).

4. ### John LarkinGuest

In theory, the charging current is infinite, which differential
equations don't like. EE's use the concept of an impulse of current, a
pulse of current of zero width but finite charge (the "delta"
function, Laplace transform = 1) This is sort of like dividing by
zero, and is mathematically iffy.

In real life, there's always some resistance and inductance in the
charging path; if you include either or both of these, the circuit
becomes mathematically tractable.

John

5. ### E. ThomsonGuest

(CFoley1064) wrote
entirely >dependent on the internal resistance of the DC voltage
source and the residual >resistance of the wire.

OK, so in other words using an ideal source it is correct that there
should be no time varying dynamics in the voltage, other than at t=0.

No, that problem is relatively easy to solve, and inspired me to
explore other possibilities for theoretical interest. In reality if I
build it with a breadboard there is small R in series with V, so the
circuit as I've drawn it is mpossible to build. So, basically, I'd get
a delta function for current through the capacitor, which would charge
up instantaneously, and never discharge.

Thanks.
Eric  