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RC circuit question

C

CD

Jan 1, 1970
0
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!
 
A

Airy R. Bean

Jan 1, 1970
0
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!
 
B

Brian Reay

Jan 1, 1970
0
CD said:
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?

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
www.amateurradiotraining.org.uk for FREE training material for all UK
amateur radio licences
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to learn more about amateur radio
 
B

Brian Reay

Jan 1, 1970
0
Airy R. Bean said:
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.

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.
Lenz's is the only know breaker of Kirchoff!

Care to expand on that?


Brian
 
A

Airy R. Bean

Jan 1, 1970
0
Mr.Reay would appear to be cyber-stalking once again...

It's not an "explantion" of anything.

It is, however, a good _explanation_ of Kirchoff's Law,
especially to someone who appeared to be not well-versed
in technical matters and to whom an overly technical
approach would not have helped.
..
If Mr.Reay wants to score points, then he'd better not
make a fool of himself (once again!)!

It's noticeable that you did not respond to the questioner in
the first instance, but only did so to satisfy your paranoid
obsession that you have with me. For example, what was the
purpose in your campaign about bestiality with sheep? QED.
 
E

EEng

Jan 1, 1970
0
The sum of the voltages around a circuit = Vt.
The subtraction of the voltages around a circuit = Common potential.
 
G

Gregory Cohen

Jan 1, 1970
0
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.
 
C

Chimera

Jan 1, 1970
0
Airy R. Bean said:
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.

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
 
A

Airy R. Bean

Jan 1, 1970
0
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.
 
C

Chimera

Jan 1, 1970
0
Airy R. Bean said:
Mr.Reay would appear to be cyber-stalking once again...

WTF are you talking about? Read the guys posts, you might learn something.
It's not an "explantion" of anything.

It is, however, a good _explanation_ of Kirchoff's Law,
especially to someone who appeared to be not well-versed
in technical matters and to whom an overly technical
approach would not have helped.
.

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

If Mr.Reay wants to score points, then he'd better not
make a fool of himself (once again!)!

Actually, Mr Reay appears to be very knowledgeable and to be able to explain
things very well.

You should try reading his posts.

Chimera
 
C

Chimera

Jan 1, 1970
0
Airy R. Bean said:
I disagree. It is zero.
(Save for Lenz's Law)
Do you even know what Lenz's law is?

I fancy not.

Chimera
 
A

Airy R. Bean

Jan 1, 1970
0
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.
 
A

Airy R. Bean

Jan 1, 1970
0
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.
 
A

Airy R. Bean

Jan 1, 1970
0
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.
 
C

Chimera

Jan 1, 1970
0
Airy R. Bean said:
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?

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
 
P

PCK

Jan 1, 1970
0
Airy said:
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.
its your reputation that is in question airy
 
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