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If someone wants to find the total resistance of a circuit he has to deactivate the independent current sources.

As an ideal independent voltage source has zero resistance and the ideal independent current source has infinite resistance, either working or not, why do we need to deactivate them? Can't we just leave them be and just consider their resistances as zero and infinite respectively?

Simply put do we need to disable them so no current can flow in the circuit (for some reason) or we just do it as a visual aid in order to point out that those element has those resistance "values" in an attempt to simplify the circuit?

My question is purely theoretical and it doesn't have to do with real circuits and resistance measurements with a multimeter.

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  • \$\begingroup\$ What do you mean by deactivate? \$\endgroup\$ – user253751 Dec 2 '16 at 0:44
  • \$\begingroup\$ @immibis Deactivating an ideal voltage source is replacing it with a short circuit and an ideal current source with an open curcuit. \$\endgroup\$ – Adam Dec 2 '16 at 0:46
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    \$\begingroup\$ Whats the difference of a circuit not being connected and having "infinite resistance" ??? \$\endgroup\$ – Jorge Aldo Dec 2 '16 at 2:20
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    \$\begingroup\$ no difference of a circuit not being connected and having "infinite resistance" \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 2 '16 at 2:22
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    \$\begingroup\$ @TonyStewart.EEsince'75 I don't care about real measurements of any kind I have point that out in my question. \$\endgroup\$ – Adam Dec 2 '16 at 2:32
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I am assuming you are talking about a "linear" circuit.

You don't have to deactivate the internal sources.

You can "measure" (by simulation, if you're talking about a circuit that only exists hypothetically) the output current (remembering that current is taken as positive when it flows in to the port) and voltage with any two different loads connected (say two different resistors, or two different voltage sources, or two different current sources). That will give you two points on the I-V curve of the output port. Through these you can draw a line, and the slope of that line is the equivalent resistance of that port.

If you're talking about a nonlinear circuit then you can not disable the internal sources. And you must choose two loads that do not disturb the operating point significantly. Then this method will give you the differential resistance near the operating point.

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  • \$\begingroup\$ Can't I just combine the resistances and use zero and infinite resistance for ideal voltage and current sources respectively? I mean to do what you would normally do but with the sources turned on. \$\endgroup\$ – Adam Dec 2 '16 at 3:12
  • \$\begingroup\$ Using zero and infinite resistance for the ideal sources and ignoring their voltage/current values is exactly the same as disabling them. \$\endgroup\$ – The Photon Dec 2 '16 at 3:32
  • \$\begingroup\$ I know, as I mentioned in the comments, that it is the same thing...What I want to understand is if disabling has any other benefit or is just a way to visualize the fact that those elements have zero and infinite resistance? I mean do we need to disable them so no current can flow in the circuit (for some reason) or we just do it as a visual aid in order to simplify the circuit? \$\endgroup\$ – Adam Dec 2 '16 at 3:37
  • \$\begingroup\$ Nobody makes you redraw it. You just need to get the slope of the IV curve. However you get it, as long as you get the same number, nobody but you will ever know. \$\endgroup\$ – The Photon Dec 2 '16 at 3:44
  • \$\begingroup\$ I mentioned redrawing as an example not that you have to do it! So there is no reason for not wanting current through the circuit? Disabling the sources is just a way of simplification and you can do exactly the same procedure with sources on or off? Do I get it right? \$\endgroup\$ – Adam Dec 2 '16 at 3:50
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From a circuit theory point of view any bipole may be modelled as Thevenin or Norton equivalent. In math

\$V=R_\text{Th}I+V_\text{Th}\$ or

\$I=G_\text{No}V+I_\text{No}\$

note that each above quantity may either be scalar (single port circuit) or vectors of currents and voltages at ports and R or G matrix (multiport circuits)

\$I_\text{No}\$ and \$V_\text{Th}\$ model internal generators and if disabled (open current generators and short voltage ones) just yields naught reducing the above to

\$V=R_\text{Th}I\$ or \$I=G_\text{No}V\$, easy enough you can now force current or voltage and then measure the counterpart to get equivalent resistance(conductance) seen at port(s).

\$ R_\text{Th}=\frac{V}{I}\$ or \$ G_\text{No}=\frac{I}{V}\$

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The usual method of determining an unknown circuit linear resistance is done by R=Voc/Isc for open circuit voltage and short circuit current.

Open circuit is circuit violation of an ideal independent constant current (CC) source or sink, as this would require infinite voltage.

Hence, this is the reason why the CC cannot be ON to measure resistance in a theoretical ideal circuit.

enter image description here

it is not possible to force an independent current source to take up a current which is different from its defined value

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  • \$\begingroup\$ I dont understand members who downvote correct answers without anything intelligent to comment. Who are you? Either improve the answer with a comment or ask a question or offer constructive criticism. otherwise -1 only indicates your incompetence or character. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 3 '16 at 14:54

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