How Can Capacitor Behavior Be Simplified for Beginners?

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In summary, the statements are asking for a translation of a capacitor's behavior at low frequencies in a circuit. The capacitor acts as a frequency-dependent switch, meaning its resistance varies with frequency. As the frequency approaches zero, the reactance of the capacitor also approaches zero. This results in a simple equivalent circuit that the source would "see", also known as the limiting form of the circuit. The input impedance of the circuit can also be determined by knowing how the capacitor's reactance behaves at low frequencies. A helpful way to understand this concept is to plot Vout/Vin as a function of frequency (jw) on log-log paper and observe where the "switch" may be considered open. However, the phrasing of the question may
  • #1
Duave
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PF,

I need two statements to be translated into a form that I understand as a novice. Can you help me please?

The statements are:


#1 As ω→0 sketch the limiting form of the circuit using the model that a capacitor acts as a frequency-dependent switch.

#2 What is the limiting value of the input impedance as ω→0?


Attempt at a solution:

ω= 1/RC; so as RC→∞, ω→0; and as RC→0, ω→∞

This is the only conclusion that I can come to.
Is there anyone that can write statements #1 and #2 in a different way so that I can know what they are asking?


Thank you
 
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  • #2
Duave said:
PF,

I need two statements to be translated into a form that I understand as a novice. Can you help me please?

The statements are:


#1 As ω→0 sketch the limiting form of the circuit using the model that a capacitor acts as a frequency-dependent switch.

#2 What is the limiting value of the input impedance as ω→0?


Attempt at a solution:

ω= 1/RC; so as RC→∞, ω→0; and as RC→0, ω→∞

This is the only conclusion that I can come to.
Is there anyone that can write statements #1 and #2 in a different way so that I can know what they are asking?


Thank you

We need to see the circuit diagram...
 
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  • #3
Here Is The Circuit

berkeman said:
We need to see the circuit diagram...

This is The circuit. It is an RC differentiator. I don't understand the jargon of statement #1 and #2. Here are some sources that I have looked at, in order to understand the statements, but I can't really use the sources yet as tools because I have nothing to confirm they are valuable as a tools.

Differentiator source information that I am using to try and piece some understanding together:

http://www.electronics-tutorials.ws/rc/rc_1.html

http://books.google.com/books?id=7v...fferentiator circuit frequency domain&f=false

http://en.wikipedia.org/wiki/RC_circuit

Thank you
 
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  • #4
Duave said:
#1 As ω→0 sketch the limiting form of the circuit using the model that a capacitor acts as a frequency-dependent switch.

#2 What is the limiting value of the input impedance as ω→0?

The term "frequency-dependent switch" is a bit misleading. A switch is usually something that's either on or off, open or closed. A capacitor on the other hand has a form of "resistance" to current flow which varies smoothly with frequency. The term for this is reactance. You should be able to find a formula that gives you the reactance of a capacitor with respect to frequency.

You're probably aware that capacitors block DC current. So you can likely guess how the reactance of a capacitor behaves as AC frequency tends towards zero.

So circuits can have both resistance and reactance. The two can be paired as a complex number called impedance. The real component is resistance while the imaginary component is the reactance. Your circuit has both resistance and reactance...

Regarding the second phrase involving input impedance, if you know how the capacitor's reactance behaves with frequency then you should be able to make a statement about the input impedance of the circuit...
 
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  • #5
gneill said:
The term "frequency-dependent switch" is a bit misleading. A switch is usually something that's either on or off, open or closed. A capacitor on the other hand has a form of "resistance" to current flow which varies smoothly with frequency. The term for this is reactance. You should be able to find a formula that gives you the reactance of a capacitor with respect to frequency.

You're probably aware that capacitors block DC current. So you can likely guess how the reactance of a capacitor behaves as AC frequency tends towards zero.

So circuits can have both resistance and reactance. The two can be paired as a complex number called impedance. The real component is resistance while the imaginary component is the reactance. Your circuit has both resistance and reactance...

Regarding the second phrase involving input impedance, if you know how the capacitor's reactance behaves with frequency then you should be able to make a statement about the input impedance of the circuit...


Gneill,

Thank you for your time. One last thing. Do you think that you can shed a little light on the phrase "...limiting form of the circuit..."? Thank you

Regards,
Duave
 
  • #6
Duave said:
Gneill,

Thank you for your time. One last thing. Do you think that you can shed a little light on the phrase "...limiting form of the circuit..."? Thank you

Regards,
Duave

It means, as the frequency tends to zero (so heading for DC), what does the circuit behave like from the point of view of the source (Vin)? In the limit, ω is essentially zero. What would be a simple equivalent circuit that the source would "see"?
 
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  • #7
A picture's worth a thousand words.

It may be helpful for you to go back to that voltage divider circuit you were involved with a few days ago.

Derive the term for Vout/Vin as function of frequency (jw) and plot it, jw on horizontal axis and Vout/Vin on vertical. Use log -log paper (or some program. I always liked Qbasic but it no longer runs on Windows because Microsoft is a bastion of mediocrity).
Now, if you plotted phase shift too you'd have what we used to call "bode plot" .

At what jw would you guess the "switch" might be considered to open?
[ hint - in my day that'd be when output is attenuated by approximately a decade ]
In my humble opinion this is a terribly phrased question. Teacher has asked you to read his mind, which is NOT how you instill confidence in beginners.

Plot it and see if it becomes any clearer. We learn by doing...

old jim
 
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  • #8
jim hardy said:
A picture's worth a thousand words.

It may be helpful for you to go back to that voltage divider circuit you were involved with a few days ago.

Derive the term for Vout/Vin as function of frequency (jw) and plot it, jw on horizontal axis and Vout/Vin on vertical. Use log -log paper (or some program. I always liked Qbasic but it no longer runs on Windows because Microsoft is a bastion of mediocrity).
Now, if you plotted phase shift too you'd have what we used to call "bode plot" .

At what jw would you guess the "switch" might be considered to open?
[ hint - in my day that'd be when output is attenuated by approximately a decade ]
In my humble opinion this is a terribly phrased question. Teacher has asked you to read his mind, which is NOT how you instill confidence in beginners.

Plot it and see if it becomes any clearer. We learn by doing...

old jim

...I agree...I think it was terribly phrased as well. I was working on an assignment, and I turned it in. After I turned it in, I found out that the professor wanted me to draw a circuit diagram...:confused::frown:
 
  • #9
Duave said:
...I agree...I think it was terribly phrased as well. I was working on an assignment, and I turned it in. After I turned it in, I found out that the professor wanted me to draw a circuit diagram...:confused::frown:

The thumbnail in your post 3 is indeed a differentiator. A frequency response plot, Vout / Vin vs frequency, would show breakpoint at w=1/rc = 1/(100 X 100e-6) = 100 radians/sec = 15.9 hz.

Below that frequency output decreases toward zero, above it output approaches input.
So one might think of that as a lazy switch with a gradual change of state, i suppose. But more precisely it's a high pass filter or differentiator.

One usually thinks of differentiators when dealing with non-sinewave signals such as pulses.
One usually thinks of filters when dealing with signals that carry information or power such as audio or radio , or power lines.

Did professor not even give you a schematic? Surely there was more than just those two statements.

Anyhow - tell him when you are confused by a problem statement.

I urge you to go through that exercise of calculating and plotting frequency response. You shouldn't just take my word for it.
Many of us have to learn the "What" before the "Why". At least i do.

old jim
 
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