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14-Alexandrite

## Circuit equation with PRIME7

Hello everyone

I attached tow work sheets with PRIME7.

I was very surprised about I1 ...

Is this a bug? Or is this joke?

ACCEPTED SOLUTION

Accepted Solutions
21-Topaz II
(To:LucMeekes)

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

9 REPLIES 9
20-Turquoise
(To:ssato)

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

Tokoro.

14-Alexandrite
(To:ttokoro)

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.

20-Turquoise
(To:ssato)

21-Topaz II
(To:ssato)

14-Alexandrite
(To:-MFra-)

Hello MFra san

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.

23-Emerald III
(To:ssato)

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

Success!
Luc

21-Topaz II
(To:LucMeekes)

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

14-Alexandrite
(To:LucMeekes)

Hello LucMeekes

But your solution V1/I1 seems to differ somewhere...

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

14-Alexandrite
(To:ssato)

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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