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-MFra-Answer
21-Topaz II
March 22, 2021

The circuit is an Antoniou inductance simulator. In fact, the result gives an impedance proportional to CR² which dimensionally is an inductance (FΩ² = H).

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9 replies

ttokoro
ttokoro21-Topaz I
21-Topaz I
March 22, 2021

I want to know the electric circuit that you want to solve. Including the points with voltages and current flows of the elements.

Tokoro.

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t.t.
ssato15-MoonstoneAuthor
15-Moonstone
March 22, 2021

Hello ttokoro san

 

I attached circuit diagram and some equations.

 

I have been able to obtain I1, V2 and V4 from the file "sample1" I attached earlier.

I have two questions.

 

  1. V2 remains in the solution of I1 called from SLOVE.
  2. The only difference is where SOLVE calls I1.

 

The solution is different when symbolic I1 is called again.
The solution of V2 is substituted into one of them to obtain the solution of the equation.
On the other hand, V2 remains.

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ttokoro
ttokoro21-Topaz I
21-Topaz I
March 22, 2021

image.png

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t.t.
-MFra-21-Topaz II
21-Topaz II
March 22, 2021

ssato answer.jpgssato answer.jpg

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ssato15-MoonstoneAuthor
15-Moonstone
March 22, 2021

Hello MFra san

 

Thank you for your advice.

 

I assume V1 is the source voltage and a known constant.

So I think I1, V2 and V4 are unknown and can be solved by three equations.

In fact, Ver. 15 gives 3 unknowns without problems.

It looks like a PRIME 7 problem.

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LucMeekes23-Emerald IV
23-Emerald IV
March 22, 2021

How about:

LucMeekes_0-1616418163191.pngLucMeekes_1-1616418182051.png

The circuit presents itself as an impedance to the voltage source. That impedance is:

LucMeekes_2-1616418236604.png

Success!
Luc

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-MFra-21-Topaz IIAnswer
21-Topaz II
March 22, 2021

The circuit is an Antoniou inductance simulator. In fact, the result gives an impedance proportional to CR² which dimensionally is an inductance (FΩ² = H).

Translate
ssato15-MoonstoneAuthor
15-Moonstone
March 22, 2021

Hello everyone

 

This circuit operates as an inductance"L" when an ideal operational amplifier is used.

 

L=C1*R1*R3*R5/R4

 

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