Conventionally the direction of current is taken from positive terminal to negative terminal but the actual flow of electrons is from negative terminal to positive terminal and the actual direction of current is from negative to positive terminal then, Why the direction of current is taken from positive to negative terminal???
it is like asking why is positive charge called positive or negative charge called negative its purely a matter of convention according to SI system always remember that conventional current flow direction is opposite to the flow of electrons our life would have been very easy if faraday while assigning names to the two varieties of charges made negative charge positive or used any other pair of opposites
Some science guy called Benjamin Franklin done tests on electricity and he found that electricity goes from the higher potential to the lower. So, he reasoned that electric go from positive to negative.
but the Modern reserch shows that the electrons flow from lower to higher potential then why the conventional current is from higher to lower potential?
but the charges were assigned the names of positive and negative by faraday instead of benjamin franklin
the electric current flowed from high to low. kinda like diodes. faraday? oh man.
y??
electrons flow from places with high density of it to someplace of low density of it.
hmm 10x
you're welcome
@KhiZ ! Hi, this was a point of confusion for me for a while, too. It REALLY works out though. So, current is the flow of charge. One measure of this is the number of charges that cross an area per time. Amperes measure current, and amperes are often measured as charge per time, like coulomb per second. Right now, you're thinking electrons move. That is true! If negative charge moves, the current is negative in that direction. So...|dw:1377984621467:dw| Then, the charge is negative. \(current = \dfrac{\Delta Q}{\Delta t}=\dfrac{Q-Q_0}{t-t_0}\). Time stays positive, and the \(\Delta Q\) just gets more negative. So the current to the left is negative. The current in the right direction is opposite, and so is positive! If the current in my drawing is \(-1A\), then the current going to the right is \(1A\).
|dw:1377985120276:dw|
This brought about the "right hand rule" vs the "left hand rule" The military training in their instruction and manuals define current as the flow of electrons traveling negative to positive potentials. while most higher institution use the direction of current from positive to negative or conventional current. We have to live with it, for myself I treat current as the flow of electrons.
lol.. i think what answer she is looknig for is the definition of current is given by the rate of flow of charges, and the direction of the current is taken as direction of POSITIVE CHARGES FLOWING. since electron flow in one direction is equivalent of thinking of positive charges flow in the opposite direction, we choose the convention that way!
I guess that would help in the understanding of PNP devices, the motion of positive charge carriers.
I was almost going to bring up holes. That mimics the flow of positive charge! And is opposite the flow of electrons.
@Mashy , would you agree that you can have negative current, and that is in the direction of the electron flow? Just because current is directional?
I think it was conventionally chosen to be this. It might have chosen something else.
\[\Delta Q?\] I think you are mixing! Delta Q just can mean that the average of negative/positive charges crossing the cross section A during the time Delta t.
I was thinking we chose the proton and electron signs arbitrarily, and the signs continued to work from that. So current of is the flow of positive charge? On the topic of \(\Delta Q=Q-Q_0\), the amount of charge is countable. I will let \(\bar I\) be the average current, just like \(\bar v=\dfrac{d}{t}\). If there is a change in velocity, then that equation holds while \(v=\dfrac{d}{t}\) does not necessarily hold. And so, if \(5.00\ [C]\) has gone by at \(t=15.0\ [s]\), and \(10.00\ [C]\) has gone by when \(t=19.0\ [x]\), then \(\bar I=\dfrac{\Delta Q}{\Delta t}=\dfrac{Q-Q_0}{t-t_o}\\=\dfrac{10.00\ [C]-5.00\ [C]}{19.00\ [s]-15.00\ [s]}\\=\dfrac{5.00\ [C]}{4.00\ [s]}\\=1.25\ \left[\dfrac{[C]}{[s]}\right]=\bar I\) I might be wrong, seeing as neither charge nor time have direction. Charge has a sign, though. And if we're talking about the charge that moved in a certain direction (across an area), then that gives it direction. And it's historically sensible and conceptually suitable to say that positive current should be the motion of the positive charge. With that logic in mind, the direction must be assigned in the equation. I have to go. On that much, I'd like to see what others have to say! :)
@theEric Well said. Hopefully current will go in the desired direction, if it doesn't then you have a wiring problem.
Thanks, @radar ! Haha, and thankfully ammeters read \(-\) and \(+\)! :)
So the direction of current comes from which way you set to be positive or negative. So, if we say the positive current direction is the direction a positive test charge would move, then current goes from positive or negative. That work well when using the electric field. But then, maybe you can say it's the other way around. Current goes from negative to positive. That is the flow of electrons. I guess you can switch the sign of the proton and electron charges to achieve the different sign, which is not changing the prediction of any events but just changing how we write them. And, even if there is no sign swap, if we change the equations for electric field and others, can we just choose the directions to be different? Just like \(g\approx -9.81\ [m/s]\) because we just commonly consider "up" to be "positive," we can say \(g=9.8\ [m/s^2]\) when we consider positive to be towards the center of the Earth. However, that is another field force system, and my mind might still be confused here. The sign is towards or away from the Earth's center of mass - not a simple rectangular coordinate system..
Wow, I didn't notice this question is from 9 months ago!! Haha!
Ok, what all you mentioned was practically true, though. But it is different when we choose an upward or inward direction in order to calculate the problem of falling bodies. But here is something different. I remember that the physicists had picked out the desired direction so as to get rid of some unwanted problems might happen during the calculations. Even so, you have immensely put well expositions. Thanks...
Thanks!
Thankx awlz 4 clearing my doubts :) :)
Haha, you're very welcome! Thanks! :)
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