# Capacitor and their behavior in AC

1. Jun 24, 2010

### dhruv.tara

I know how capacitors behave when connected to a dc voltage supply. For a simple RC circuit we have the time constant thing come up into the equations and then we can find the exponential fall of voltage and currents.

but what I don't get is the what would be the behavior of the same in a ac circuit. Can we determine it using juts the normal equations? If so then please confirm the result I get...

V(c) = V*sin(WT)/(C*Z*W)

where V(C) is the voltage across capacitor, VcosWT is the voltage of ac source. C is capacitance. Z is the impedance of RC circuit and W is the frequency. (given initial charge on the capacitor is 0)

Also please help me with the following extension of the problem: What difference would come up if we ground the circuit in the following manner. R and C are same, V is an AC source of waveform VcosWT...

Thanks...

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2. Jun 24, 2010

### sophiecentaur

Firstly, Earthing or not of any point in the circuit has no bearing on the matter because we are only considering potential differences and there is no other path to earth.

To appreciate the voltage across the capacitor it is convenient to think of the circuit as a potential divider.
V(C) will be a proportion of the applied voltage V
V(C) = V0 (Impedance of Capacitor)/(total Impedance of the circuit)
so
V(C) = V0(1/jωC)/(R+1/jωC)
which seems to be what you have written.
The resulting voltage will be phase shifted as well as reduced in amplitude.
Using a source voltage V=V0sin(ωt) or V=V0cos(ωt) will make no difference as the two waveforms are identical - merely phase shifted.

3. Jun 24, 2010

### lifeattthesha

Linear tech has a FREE spice engine called LT spice or something like that. Sometimes it is helpful to simulate these things if you do not understand what is going on. In general if you look at the voltage across the capacitor, say with an oscilloscope, then you swept the frequency of your source from close to DC up to some higher value you would find that the voltage waveform will decrease as the frequency goes up. That circuit is commonly called a low pass filter because of that property. The value of the R and C will determine what frequencies are not passed or where in the frequency domain the cut off is. This can readily be seen by performing an AC analysis on the circuit and sweeping the source from below the cut off frequency to above the cutoff frequency. The outputs are normally looked at in a log scale with frequency on the bottom.
L

4. Jun 24, 2010

### dhruv.tara

okk.. thanks everyone...

@ lifeattthesha... I have matlab in my pc... can circuits be stimulated in that software? Or can you mention some others that are readily available and easy to learn?

@ sophiecentaur: I wanted to know just the same that earthing will not have any effect on the circuit.
But please clarify me on this one...? In case I earth two points in a wire then the wire would become short and in such a case it will effect the circuit...? Right?

5. Jun 24, 2010

### lifeattthesha

you could do it in matlab but you would need to write the governing equations then solve them. To hard for someone stupid like me. I would take the easy way out and just use a tool designed to do just that. A spice tool is probably best. And a free schematic spice tool is even better.

go here: http://www.linear.com/designtools/software/ltspice.jsp

and you can download the LTspiceIV.exe for free , you do not have to register to get it. The entry is schematic based and it will generate the net list for you. Pretty easy to learn. plus you simulate many other standard circuits.

6. Jun 24, 2010

### sophiecentaur

I may have misunderstood your question.Yes, of course, joining two points to Earth is applying a short circuit but attaching one point to Earth will have no effect.

7. Jun 25, 2010