Capacitor and their behavior in AC

In summary: So, it is not suitable for this problem.In summary, the conversation discusses the behavior of capacitors in circuits, specifically in AC circuits. The voltage across the capacitor can be determined using the equations V(c) = V*sin(WT)/(C*Z*W), where V(c) is the voltage across the capacitor, VcosWT is the voltage of the AC source, C is capacitance, Z is the impedance of the RC circuit, and W is the frequency. It is also mentioned that earthing a point in the circuit will not have any effect on the overall behavior. Various simulation tools, such as MATLAB, LTspice, and Multisim, are also suggested for further analysis and understanding.
  • #1
dhruv.tara
46
0
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
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
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
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
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
dhruv.tara said:
okk.. thanks everyone...


@ 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?

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
dhruv.tara said:
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?

...

Yes, just use simulink and "draw" the circuit. The toolbox "SimElectronics" should contain all the components you need, except scopes. Tutorials are wide available.

Altough NI multisim would be a better choice.
 
  • #8
But SimPowersystems of Simulink/Matlab won't give you correct results. It too have R, C and DC sources. If you simulate a simple RC circuit with DC source in SimPowersytems, Capacitor is instantaneously charged, no matter how much value of R is kept. I think it is a bug in the software. I don't know about SimElectronics but it contain only variable capacitors, not fixed.
 

1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When an AC voltage is applied across the capacitor, it causes the plates to accumulate opposite charges, resulting in an electric field between them. This electric field stores the energy, which can then be released when the capacitor is discharged.

2. How does a capacitor behave in an AC circuit?

In an AC circuit, the voltage across a capacitor alternates between positive and negative values as the AC current changes direction. When the voltage is positive, the capacitor charges up, and when the voltage is negative, the capacitor discharges. This results in a time delay between the voltage and current in the circuit, known as the capacitive reactance, which is dependent on the frequency of the AC signal and the capacitance of the capacitor.

3. What is the difference between a capacitor in an AC circuit and a DC circuit?

In a DC circuit, the voltage remains constant, so the capacitor charges up to a specific voltage and then remains charged. In an AC circuit, the voltage is constantly changing, causing the capacitor to continuously charge and discharge. Additionally, in a DC circuit, the capacitor behaves like a short circuit once it is fully charged, while in an AC circuit, it acts as an open circuit due to the alternating voltage.

4. How does the frequency of the AC signal affect the behavior of a capacitor?

The capacitive reactance of a capacitor is inversely proportional to the frequency of the AC signal. This means that as the frequency increases, the capacitive reactance decreases, allowing more current to flow through the capacitor. As a result, the capacitor charges and discharges more quickly, and its ability to store energy decreases.

5. What are some common applications of capacitors in AC circuits?

Capacitors are commonly used in AC circuits for power factor correction, filtering, and tuning. They are also used in AC coupling circuits to block DC signals while allowing AC signals to pass through. In addition, capacitors are essential components in electronic devices such as radios, televisions, and computers.

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