Reciprocity theorem, not working

[nForce]

I have build this circuit in LT spice:

I have measured the current Ia, it is 0.453 mA.

Then I have change the voltage source and the branch where I had earlier measured the current. And it is 1 mA.

How? It is a bilateral circuit. Thank you for your help.

[IanB]

I am not understanding your question.

In the circuit as shown you have total loop resistance of 11/3 kilohms. The inductor L1 is not relevant with DC and can be ignored.

The supply voltage is 5 V DC. Therefore the total loop current is 3/11 x 5 = 1.364 mA.

The current through R4 is 1/3 of the total, so 1/3 x 3/11 x 5 = 5/11 = 0.456 mA. This is what you have in the simulation.

Now, what are you changing in the circuit for the next part? I am not clear on that.

[nForce]

Yes what you said is correct.

I have switched the R4 with V1 (voltage source).

[IanB]

Yes what you said is correct.

I have switched the R4 with V1 (voltage source).

OK, if we do that then the total loop resistance is now 5 k in parallel with 1 k, which gives 5/6 k. That makes the total loop current 6 mA. This current will split so that 5/6 of it flows through R3, and 1/6 of it will flow through R1, R2 and R4. So the current through the R1 branch will be 1 mA.

[nForce]

Yes I know it's 1 mA. But why the theory isn't consistent with the practical circuit.

https://electronicspani.com/reciprocity-theorem/

[IanB]

If you want the reciprocity theorem to work, you must just move the voltage source and the current measurement. You must not move the resistor R4 because that changes the topology of the circuit.

[IanB]

Suppose you put V1 where Ia was measured, then the old V1 becomes a short circuit.

Now R1 and R2 are in parallel with R3 giving 3/4 kilohms, and this is in series with R4 giving a total loop resistance of 11/4 kilohms.

The total loop current is therefore 5 x 4/11 = 20/11 mA.

If we measure the current through R1 (where V1 used to be), then it will be split 1/4 through this branch and 3/4 through R3.

Therefore the current through R1 is 1/4 x 20/11 = 5/11 mA, the same answer as previously.

[nForce]

I get it now, thank you.