# Capacitor (applications)

1. Jul 29, 2009

### jeff1evesque

In the following diagram: http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/lab/expt2/rc.gif; I believe it is a Low-pass filter. But within the picture, does current come in from both-top/bottom left portion of the wires, and exit the top/bottom right portion of the wires?

In another forum, a person mentioned to me that the capacitor acts like a short. But if both the top/bottom left wire is signal coming in, then the plates have no potential difference. So if this is true, how can it be storing charge- more specifically, how will current go through the plates?

What happens when we switch the position of the capacitor with the resistor in the diagram given in the link above?

Thanks,

JL

2. Jul 29, 2009

### queenofbabes

The signal is carried in the form of a potential difference between the top and bottom wire (that's Vin). At high frequencies, the capacitor will indeed act like a short, so Vout = 0, but at low frequencies the signal is passed along to the right hand side of the circuit (almost) unchanged. (Smaller magnitude, but the waveform is preserved)

3. Jul 29, 2009

### jeff1evesque

Got it, thanks.

4. Jul 29, 2009

### jeff1evesque

Actually, I have a question, does the resistor act like a short as well when the frequency increases? If so, does this effect what we were talking about earlier?

Thanks,

JL

5. Jul 29, 2009

### queenofbabes

If it is a pure resistance, then it's not affected by frequency...

6. Jul 29, 2009

### jeff1evesque

Cool. Do you mind explaining this with respect to high pass filters (where the capacitor and resistor are interchanged by the original low-pass filter)? Since the resistor is a pure resistor, for some reason I am thinking high-pass is similar to low pass filters?

7. Jul 29, 2009

### queenofbabes

I'm not quite sure what you mean. In both cases the key component is the capacitor, the resistor isn't doing very much. For high pass filter, the capacitor is attached directly to Vin, so only at high frequencies will it "act like a short" and allow the signal through.

8. Jul 30, 2009

### jeff1evesque

Do you know what is meant by the voltage dropping across the resistor for the low pass filter? I think I've gotten the concept down besides that.

9. Jul 30, 2009

### queenofbabes

You always get a voltage drop across a resistor right?

10. Jul 30, 2009

### jeff1evesque

Is there any chance you could explain why ...more voltage will drop across the resistor?

Thanks

11. Jul 30, 2009

### rl.bhat

As frequency increases, the capacitive reactance Xc, which is 1/ωC, decreases. Hence total resistance in the circuit decreases. Due to that the current in the circuit increases.
If the current increases through the resistance what happens to the voltage across it?

12. Jul 30, 2009

### jeff1evesque

Current increases through the capacitor since it is shorted at high frequency- I understand that. The capacitor has impedance, which is just as you defined, and that decreases with frequency. But how is there more current through the resistor?

13. Jul 30, 2009

### rl.bhat

The resistance and capacitor are connected in series across the voltage sourer. If the total resistance across the voltage source decreases, the current in the circuit increases. consequently the current through resistance increases.

14. Jul 30, 2009

### jeff1evesque

So voltage drop corresponds to current flowing through the resistor easier?

15. Jul 30, 2009

### cepheid

Staff Emeritus
Many resistors obey Ohm's Law, if that's what you're asking. V = IR. In this equation, V is the voltage across the resistor (i.e. the different in electric potential between its two terminals). The symbol I is the current through the resistor. If electric current is flowing through the resistor, a certain amount of energy is lost due to heat. That's the whole point: a resistor hinders or resists the flow of electric current. That is the reason why the voltage drops.

16. Jul 30, 2009

### jeff1evesque

So more energy is lost through the resistor because more current goes through it- thus it heats up more and more voltage drop?

Last edited: Jul 31, 2009
17. Jul 30, 2009

### cepheid

Staff Emeritus
Yeah that's the physical explanation, in somewhat vague terms. But you don't even really have to worry about what's happening in detail. Just look at Ohm's law. If R is constant, and I increases, what happens to V?

18. Jul 31, 2009

### jeff1evesque

I was reviewing some notes, and it's very simple. Voltage Drop simply refers to the concept of potential difference through a given material- in our case a pure resistor. In the specific case of Low-Pass filters, the capacitor is shorted at high frequency, and the resistor- since it is a pure resistor- behaves with it's ability to drop voltage.

And according to rl.bhat,
Which means as the impedance [in DC we say resistance, and AC we say impedance] decreases in the overall circuit, more current flows, $$I = VR,$$ hence more voltage "drops" across our resistor, $$V = IR$$.

Thanks guys for helping me.

Jeffrey Levesque