RC circuit question

Hi all,
I'm reading The Art of Electronics and they have a circuit that shows an arbitrary resistor and a capacitor connected in parallel (no source
connections). They also stated that according to the capacitor rules, we get:
C*(dV/dt) = I = - (V/R)
I can't seem to comprehend the - (minus sign) in front of (V/R). Can anyon explain this? Show it mathematically?
Thanks!
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The discharge current is in the opposite direction to the charging current. If the charging current is considered to be in a positive direction then the discharge current will be in a negative direction. Need more info on the type of circuit you are looking at.

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If I may, avoid the maths at first and think physics.
Assume that the capacitor is charged at time 0 to a pd V, and has the resistor connected across it. The capacitor will discharge as electrons flow through the resistor and, as this happens the pd across the capacitor will decrease, ie dV/dt is -ve. , pd is decreasing as time increases.
The rate of discharge is a function of V, R, and C, related as:
|dV/dt| = V/(RC) (note the | | ) , RC being the normal time constant.
From the above, we know dV/dt is -ve so, dropping the | | :
dV/dt = -V/(RC)
A simple transposition gives:
C dV/dt = -V/R
--
73
Brian
G8OSN
www.g8osn.org.uk
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You are quoting a relationship from A of E Section 1.13. If you go back and re-read the previous Section (1.12), your question will be answered. If no, post more on what confuses you.
It helps to know enough calculus to go from:
v = 1/C [Integral] idt
To:
i = C(dv/dt)
... as a hint.
I will agree that the authors of A of E were a bit sloppy in this introductory chapter, by not making a distinction between i vs. I and v vs. V, although they may have covered their unique use of symbol conventions in an earlier section of the book.
I really like the book from a practical electronics design standpoint, but it certainly is not the best possible source from which to learn fundamental electrical basics.
Obviously I own my own copy, and refer to it often, still for a thorough comprehension of basic circuit theory, nothing is as helpful as a good text covering only that subject. I use Walsh & Miller's "Electric Circuits", but likely more recent and better ac/dc circuits texts are now available. Still, be forewarned that a basic working skill in calculus is not optional, but needed to fully comprehend this fundamental material.
Harry C.
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Comes from Kirchoff's laws - the sum of any voltages round a closed loop is zero. The voltage going up the capacitor is positive, that coming down the resistor as you go on round the loop, is negative.
Lenz's is the only know breaker of Kirchoff!

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Oops! Faraday's Law. Close, though!

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You really don't have a clue, do you?
Do you know the difference between potential difference, EMF, and current.
I'm not surprised the idea of frequency is to much for you.
Chimera
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If you have something of a technical nature to offer to the discussion, then do so.
Why not, for example, contribute to the discussion in an adult manner by replying to the OP and offering your own explanation in reply to her expressed difficulty?
If all you can come up with is rather silly and infantile comments as you do below then you do yourself and your reputation no favours.

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Mr Raey has already given a very good explanation which I notice you haven't dared to challenge.
You should read it, it is very good. Unlike you, Mr Raey understands the topic and can explain things.
Now, tell me about Lenz's law.
Chimera
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If you have something of a technical nature to offer to the discussion, then do so.
Why not, for example, contribute to the discussion in an adult manner by replying to the OP and offering your own explanation in reply to her expressed difficulty?
If all you can come up with is rather silly and infantile comments as you do below then you do yourself and your reputation no favours.

haven't
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That is a poor explantion of Kirchoff's Law.
Better to think in terms of EMF and Potential Differences with KVL and treat the energy source, in this case the capacitor, as having the EMF and the resistor as having the PD across it then:
Vemf +Vpd =0
Having said that, applying KVL does't really answer the question asked, for which you need to think in terms of dV/dt, for which I refer you to my other posting.

Care to expand on that?
Brian
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WTF are you talking about? Read the guys posts, you might learn something.

Yours Airy, is a very poor explanation of Kirchoff's Law. In fact, one of the worst I've ever seen.

Actually, Mr Reay appears to be very knowledgeable and to be able to explain things very well.
You should try reading his posts.
Chimera
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If you have something of a technical nature to offer to the discussion, then do so.
Why not, for example, contribute to the discussion in an adult manner by replying to the OP and offering your own explanation in reply to her expressed difficulty?
If all you can come up with is rather silly and infantile comments as you do below then you do yourself and your reputation no favours.

explain
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Airy R. Bean wrote:

its your reputation that is in question airy
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On Sun, 18 Jan 2004 08:54:51 -0000, "Brian Reay"

The sum of the voltages around a circuit = Vt. The subtraction of the voltages around a circuit = Common potential.

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I disagree. It is zero. (Save for Lenz's Law)

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Do you even know what Lenz's law is?
I fancy not.
Chimera
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If you have something of a technical nature to offer to the discussion, then do so.
Why not, for example, contribute to the discussion in an adult manner by replying to the OP and offering your own explanation in reply to her expressed difficulty?
If all you can come up with is rather silly and infantile comments as you do below then you do yourself and your reputation no favours.

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Huh? Aren't you confusing current and voltage just a bit?
Also, you don't seem to have quite yet mastered Kirchoff's Law or understand capacitors. Keep working on it, because it is very useful and there is still hope for you!
Harry C.
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