Removing the reactance

[alaios]

Hello,
I have a load that has an impedance with a real and reactive part. For example
X=70+40j

I want to connect in parallel a circuit to remove the reactance. Make all the impedance real.
I have found already that it is easier to work with admittance and use a capacitor since then the reactances will be considered in series.

I also know the formula on calculating which capacitor I should use (for a specific frequency) to get the reactive capacitance I need.

What bugs me though is that since my load X and my capacitance in parallel I can add them like they are in series. The problem is that I need in my equation to have a minus sign so the +40j with my new element in series will give me a nice zero.

Can you please help me understand where I can find this minus part that will make my impedance only real?

Thanks a lot.
Regards
Alex

 

[KlausST]

Hi,

I don't think this is possible with passive circuits.
Maybe with limited effect for a dedicated frequency.

But with an active (amplifier) circuit you may be able to compensate the reactance to get only "real" behaviour...up to a limited frequency.

Klaus

 

[c_mitra]

An interesting question: I have to ponder.

Simply, you need to add a series reactance of -40j. That will do the trick.

I do not know why it is easier to work with admittance but that is not relevant here. It will be simply the same equation in a different form: Y=0.868-0.496j

But they are no more resistance and reactance; they have also been magically transformed into conductance and susceptance (whatever that may mean). But reactance and susceptance are not reciprocally related. They are different beasts altogether.

Likewise the resistance and conductance are reciprocally related only for a circuit with no reactance. But this is only mathematical jugglery. The basic physics remains the same.

Because of the positive sign, I presume the reactance is inductive and you will need to use a capacitor to remove this positive 40j term. This will work only at a fixed frequency and the point will be oscillatory (pole or a zero).

It may be wise to learn to live with it.

 

[FvM]

Maybe with limited effect for a dedicated frequency.

Describing circuits with reactance values clarifies that it's single frequency/small band problem, because there's no circuit element with frequency independent complex impedance. Respectively there's no circuit element to compensate the reactant part.

Just a typical AC network exercise question.

The answer to the question is simple and can be found in any text book covering AC networks.

The usual way is to convert the reactance to admittance, equivalent to replacing the series circuit by a parallel circuit. Add a parallel reactance with inverse value to compensate the reactant part, e.g. a capacitor to compensate the inductive impedance/admittance component.

 

[alaios]

An interesting question: I have to ponder.

Simply, you need to add a series reactance of -40j. That will do the trick.
....

Because of the positive sign, I presume the reactance is inductive and you will need to use a capacitor to remove this positive 40j term. This will work only at a fixed frequency and the point will be oscillatory (pole or a zero).

It may be wise to learn to live with it.

Very good point that I have missed. Since I have a plus sign of +40j it is inductive and then a capacitance in series will do (of course as you said this is for a single frequency). Where I can learn more about the topics of inductance and reactance and their plus minus on the imaginary parts of impedance?


Thanks again.
Alex

- - - Updated - - -

The answer to the question is simple and can be found in any text book covering AC networks.

The usual way is to convert the reactance to admittance, equivalent to replacing the series circuit by a parallel circuit. Add a parallel reactance with inverse value to compensate the reactant part, e.g. a capacitor to compensate the inductive impedance/admittance component.

thanks a lot for that reply. This is exactly the part that I have missed.
Regards
Alex

 

[FvM]

I do not know why it is easier to work with admittance but that is not relevant here. It will be simply the same equation in a different form: Y=0.868-0.496j

That's admittance normalized to magnitude 1, the actual admittance is
0.00615 - 0.01077j

Respectively the compensating susceptance is +0.01077, or expressed as reactance -92,85.

Translated to 50 Hz, the series circuit of 70 Ohm and 0.127 H can be compensated by a parallel capacitor of 34.3 µF.

- - - Updated - - -

Sorry, I have confused real and imaginary part

That's admittance normalized to magnitude 1, the actual admittance is
0.01077 - 0.00615j

Respectively the compensating susceptance is +0.00615, or expressed as reactance -162.6.

Translated to 50 Hz, the series circuit of 70 Ohm and 0.127 H can be compensated by a parallel capacitor of 19.6 µF.

Please check the calculation yourself.

 

[c_mitra]

the actual admittance is 0.00615 - 0.01077j

Sorry, I goofed up.

The usual way is to convert the reactance to admittance...

admittance is related impedance and not reactance; only reactance cannot be converted to admittance.

(Or, am I still messed up??)

 

[FvM]

admittance is related impedance and not reactance; only reactance cannot be converted to admittance.

True of course, I meaned to say converting (complex) impedance to (complex) admittance, as we both did.