Understanding Short Circuit Current Paths in Electrical Circuits

AI Thread Summary
The discussion centers on understanding short circuit current paths in electrical circuits, particularly in relation to Thevenin's theorem. It highlights that during a short circuit, resistance is effectively zero between terminals, impacting the current flow through circuit elements. Participants express confusion about how potential difference affects current through a resistor when terminals are connected by resistanceless wires. Clarifications are made regarding the concept of potential difference (p.d.) and its implications for current flow. Overall, the conversation emphasizes the need for deeper contemplation of these electrical principles.
madafo3435
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Homework Statement
I want to explain why when calculating the short-circuit current between the terminals (see the attached figure), no current takes the route through resistor R.
Relevant Equations
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1605395148922.png

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My reasoning is as follows: with the short circuit present, a 0 resistance can be considered between the terminals. Then, since the only objective of the short circuit current ##I_{SC}## is to give information about the Thevenin equivalent resistance and the open circuit voltage (the existence of these values is given by thevenin's theorem). The presence of the short circuit is considered for only an instant, so the only current that is measured is the one that arrives due to the resistor ##2R##, that is, at the instant of this presence of the short circuit, the current that flows through circuit element ##R## does not reach the terminals, so we can assume that no current takes the path through circuit element ##R##.

I do not know if this is correct, and if it is not, could someone enlighten me with this inconvenience?
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I don't follow your explanation, but if those two terminals at the bottom are connected then the two junctions connecting to R and the positive pole of the battery are all connected by resistanceless wires and so are t the same potential and there us no p.d. to drive current through R.
 
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epenguin said:
I don't follow your explanation, but if those two terminals at the bottom are connected then the two junctions connecting to R and the positive pole of the battery are all connected by resistanceless wires and so are t the same potential and there us no p.d. to drive current through R.
Your reasoning confuses me. It is not clear to me which elements have the same potential, and I do not see why this leads to no current flowing through R. I also assume that p.d. refers to partial discharge, but I am not related to that concept.
 
P.d. = potential difference.
Everything connected by a resistanceless wire is at the same potential.
 
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epenguin said:
P.d. = potential difference.
Everything connected by a resistanceless wire is at the same potential.
ohh I get it, I need to think a lot more about it since it is not obvious to me. thanks for your help
 
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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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