Is a 1A Current to the Right Equivalent to -1A to the Left?

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A current of 1A to the right is conventionally considered equivalent to -1A to the left, as the signs indicate direction rather than actual negative current. This convention helps clarify the flow of current in diagrams and calculations. When measuring current, the positive value indicates the direction chosen for reference, while negative values denote the opposite direction. Understanding this convention is crucial for accurately analyzing current at junctions in electrical circuits. The discussion emphasizes the importance of direction in current measurement and calculation.
QueenFisher
a very simple question that i have no idea about:
can you say that a current of 1A to the right is also a current of -1A to the left?
 
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Well, I don't really know, but that's what happen when you invert the inputs of an amperemeter.

So, it's either a device convention to tell you the current go the other way or intensity is really counted negatively.

In fact, I don't really know neither, but I'm sure a guru is going to answer us :p

Kyon
 
After a quick google search, I think I've found the answer: using + and - for current intensity is just a convention to indicate the direction of the current.

Positive intensity if the current go in the same direction as you decided on your drawing, negative otherwise. Just a convention.

When you calculate intensity, I guess you should get positive results only.

Kyon.
 
well, i was wondering about it in the context of working out current into/out of junctions, and it seemed important

cheers!
 
Thread 'Variable mass system : water sprayed into a moving container'

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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