how to calculate impedance

[kareno]

I want to find the impedance in this circuit.
Is it possible to calculate the capacitor as 1 / jwc after removing the voltage from this circuit?
I am not sure because it is confusing.
Thank you for letting us know how to do the calculations with and without the dashed part of the circuit.

 

[KlausST]

Hi,

What do you mean by "removing the voltage"?

Do you have to calculate it in general, or for a (not) given frequency and waveform?

(A forum is not meant that others do your job/homework, but we will help to rectify your mistakes. Thus
Show your ideas, show what you have done so fare. Show your calculations.

Klaus

 

[kareno]

thank you for the reply
It was my first time, so I was not good at posting. Sorry.

Eliminating the voltage meant that I could ignore the voltage when calculating the impedance.


When I calculated it my way, it came out like a picture.
If I use w instead of frequency, I am wondering this is correct.

thank you!

 

[KlausST]

Hi,

If I use w instead of frequency, I am wondering this is correct.

"w" still includes frequency, since w = 2 x Pi x f

Eliminating the voltage meant that I could ignore the voltage when calculating the impedance.

I still don't understand. How is voltage involved in impedance calculation?

Klaus

 

[kareno]

Thank you

I am wondering how to handle the voltage.
The equation posted in my comment above is to find the impedance using only resistors and capacitors. is this right?
If not, please let me know how to solve it.

 

[pancho_hideboo]

I am wondering how to handle the voltage.

It is very easy thing.
Your problem is no more than Thevenin Equivalent circuit theorem.

 

[kareno]

It is very easy thing.
Your problem is no more than Thevenin Equivalent circuit theorem.

oh then can i consider voltage as short circuit? right?
then is my calculation right?

veryvery thanks

 

[c_mitra]

I am not sure because it is confusing.

We understand. But once you have made a schematic, you have progressed a lot.

Now consider the impedance just like an AC resistance.

You apply a voltage (AC at some given frequency) and you measure the corresponding current (you need to measure again the AC current).

Yes, you need to define two points in the above circuit where you will apply the AC voltage. So you need to measure the AC impedance between two given points in the diagram.

Just like the DC resistance of a circuit depends on the points you select to put the test probes (the two pointed things in your multimeter) on.

Fortunately your circuit has only resistors and capacitors. Voltage sources contribute zero impedance to the overall circuit.

Yes, the impedance of the capacitor is 1/(jwc) and the rest is simply series and parallel combinations.

Please work it out (long hand way) and put it out here!

Where is the confusion, by the way?

 

[kareno]

I understood that the power in this circuit is DC voltage, so it is excluded from the impedance calculation. So, can i calculate the dc part as a short circuit? Like the above calculation?

 

[c_mitra]

Like the above calculation?

I do not follow your calculation in post #3. Can you please work out in detail?

 

[kareno]

I do not follow your calculation in post #3. Can you please work out in detail?

The total impedance was calculated using the capacitor's impedance of 1 / jwc and the resistance of R.
The picture is how I calculated it.

 

[BradtheRad]

The usual tactic is to imagine all supply rails are 0v. Then calculate time constants, series branches, parallel branches, etc. Because time constants don't change regardless of supply voltage, whether capacitors are charging or discharging.

The top and bottom ends of your schematic should have a wire joining them, if that's the entire circuit.

 

[kareno]

The usual tactic is to imagine all supply rails are 0v. Then calculate time constants, series branches, parallel branches, etc. Because time constants don't change regardless of supply voltage, whether capacitors are charging or discharging.

The top and bottom ends of your schematic should have a wire joining them, if that's the entire circuit.

Does that mean that the impedance cannot be obtained because this circuit does not represent the whole part? If I give the value of each resistor or capacitor, can I find the impedance?

 

[c_mitra]

The picture is how I calculated it

It looks right.

Yes, you can put voltage sources as zero impedance and current sources as infinite impedance (with imaginary part being zero).

It does not matter if the voltage or current sources are DC or AC.

Perhaps you are curious why we ignore the current flowing in parts of the circuit due to the voltage or current sources.

The reason is simple: we define the impedance as the partial derivative dV/dI at a given potential, frequency, temp and whats not...

I hope you get the point.

 

[BradtheRad]

Does that mean that the impedance cannot be obtained because this circuit does not represent the whole part? If I give the value of each resistor or capacitor, can I find the impedance?

Since your diagram is the main focus, then what I said is not absolutely necessary about the top and bottom ends of your schematic should have a wire joining them.

Since you specify DC supply (not AC), you can calculate the amount of time for the circuit to reach stable equilibrium after an abrupt change of DC volt level. Conventionally this takes 5 time constants.

 

[albbg]

The total impedance was calculated using the capacitor's impedance of 1 / jwc and the resistance of R.
The picture is how I calculated it.View attachment 158778

What is the purpose of your impedance calulation ?
You calculation is correct and gives the impedance of the network when excited at a given angular frequency "w" at steady state.
If, instead, you have to determine the transient response of you network you have to do a different calculation.

 

[c_mitra]

Does that mean that the impedance cannot be obtained because this circuit does not represent the whole part? If I give the value of each resistor or capacitor, can I find the impedance?

One important point. Impedance Z is usually a function of w (i.e., 2*pi*f) but not of time. You may be tempted to consider that Z varies during a capacitor charge or discharge process but that is not true.

The reason for this is not obvious to many students: that w (this is greek omega; the angular frequency) is defined in the reciprocal time domain. In other words, w and t are conjugate variables or w*t is dimensionless. Thus Z(w,t) is mathematically meaningless.

You will encounter w*t very often while doing transforms. They are useful when you want to transition from frequency domain to time domain (or vice versa).

Hope this is now a little bit clearer.

 

[Ratch]

I want to find the impedance in this circuit.
Is it possible to calculate the capacitor as 1 / jwc after removing the voltage from this circuit?
I am not sure because it is confusing.
Thank you for letting us know how to do the calculations with and without the dashed part of the circuit.View attachment 158756

What's confusing about it? Just include the impedance of the voltage or current source of the circuit for your total calculations. Impedance is only defined for sinusoidal excitations.

pZ in the attachment if a parallel impedance function. The computer did all the grit work to rationalize the impedance equation of the circuit. Notice the impedance of the circuit when the frequencies are zero and infinity. The peak voltage or current values themselves do not influence the the values of the impedance.

Ratch