# RC / CR Circuits

Discussion in 'Homework Help' started by conquer_confusion, Jan 19, 2010.

1. ### conquer_confusion Thread Starter New Member

Jan 19, 2010
2
0
Im very new to this electronics game and i've found this site's resources very helpful (better than most books i've found actually), but i'm still struggling to get my head round some very basic concepts i believe.

I have conducted an experiment for an assignment observing how RC and CR circuits can act as low pass and high pass filters. The results confirm that they do of course.

Im stuck on the how, and it relates to a general difficulty im having understanding AC current and voltage, and capacitors mainly. I would really appreciate it if someone could answer some questions for me because reading has not really cleared it up and just introduced more ideas before im ready for them.

Here's what i think i know (which is very little ) and I could be woefully mistaken on much of it, but here goes: A capacitor stores charge and also therefore energy / voltage by an accumulation of electrons on one of its plates or both alternately given an AC power supply, the essential reason for their characteristic of blocking DC and passing AC (?)

In an RC circuit with a DC supply (for starters) the values of the resistor and capacitor determine how long the capacitor will take to charge. This becomes exponentially slower because as there is a build up of charge on a capacitor plate (the one closest to the negative terminal?) and the voltage drop across the capacitor increases, simulataneously the current through the resistor decreases therefore so does the voltage (V=IR) which is supplying the charge to the capacitor (?).

When the capacitor is fully charged, depending on the value of its capacitance, the flow of electrons in the circuit is none so there is no current anywhere then presumably (?). But the electrons in the capacitor are still attracted to the positive terminal of the power supply maintaining the charge / stored energy of it (?)

Questions:

If the power supply is disconnected then the capacitor will discharge also exponentially.

What's going on here then? With the power supply disconnected and the electric field of the circuit removed, presumably the charges from the other side of the plate (the resistor side) return to that part of the circuit where they were previously more opposed to as the charged plate diffuses back towards where the negative terminal was. But why an exponential discharge? Would you still see this exponential discharge in the absence of any resistor?

What happens in an CR circuit with a DC supply?

And secondly (this is where im really perplexed), why does the position of the resistor matter at all if the current is alternating?

I hope someone can help, i desperate to get to grips with this stuff, not just to pass exams etc. but to understand and hopefully one day apply it! Many thanks for reading at least.

Pat

2. ### Fraser_Integration Member

Nov 28, 2009
142
6
The position of the resistor matters because of where you measure the filtered voltage from (in between the two components). This forms a voltage divider, and depending on where the resistor is, either low frequencies, or high frequencies will be passed.

I have to go to my Digital class now , but I will try to answer your query on capacitors today.

3. ### Fraser_Integration Member

Nov 28, 2009
142
6
The capacitor only discharges when the cell is shorted out, not disconnected. A charged capacitor will hold it's charge for a long time if there is nowhere for the electrons to flow, but when you give them a path (by shorting the terminals) they flow out quickly. The reason it is exponential is because initially there is a large attraction due to the separation of positive and negative charge, and as some electrons flow to the positive source, that attraction gets less and less.

In an RC circuit with a DC supply, initially some current would flow in the circuit whilst the capacitor charges up. But when the capacitor has the same potential difference as the battery/cell, they cancel each other out and no current flows.

The attached picture shows why it matters where the position of the resistor is to AC. This took me ages, lol. Hope it helps.

• ###### filters.jpg
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4. ### conquer_confusion Thread Starter New Member

Jan 19, 2010
2
0
Many thanks for taking the time to do that, it has definitely been helpful and after reading a few more things it is starting to make a little more sense now.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,923
1,381
Current in the capacitor is proportional to the rate of change in voltage.
I=C*du/dt=C*ΔU/Δt
So, the faster the voltage change, the larger the current flow through the capacitor. So we can say that "resistance" of a capacitor (we call it reactance Xc) decreases when frequency of AC signal increases.
For F=0Hz-->Xc=∞; F=∞ -->Xc=0Ω.
Thanks to this property we can look at filters as a voltage dividers.
For example a low-pass filter consists a resistor and capacitor connected in series. The output voltage is taken from capacitor.
This circuit should be treated as a voltage divider where the role of the variable resistance is take by capacitor. For low frequencies the capacitor has a high reactance so almost the entire input voltage is transferred to the output.
When input frequency increase the reactance of a capacitor decreases and thus causes output voltage to increase.
At frequency at which reactance of a capacitor is equal to the value of resistor the output voltage is equal Vout=0.707*Vin
And this happens at frequencies equals

$F=\frac{1 }{2*\Pi*R*C}$

The divider's voltage ratio Vout/Vin of a voltage divider that contain two resistors doesn't depend on frequency, because the resistance of resistors does not change with frequency.
In our case the divider's voltage ratio change with frequency.
Because R is unchanged but capacitor reactance Xc is change with frequency.

http://en.wikipedia.org/wiki/Low-pass_filter#Electronic_low-pass_filters