You can ignore the spanish, but you can imagine what's going on

The thing is, they use this direction for current flow to derive the equation

$-iR-\frac{q}{c}=0$ and then derive the equation $q(t)=Q_o e^\frac{-t}{RC}$ from the differential equation $\frac{dq}{dt}=\frac{-q}{RC}.$

This seems pretty artifical to me, because if I use the other direction for current, which seems more natural to me, I don't get the same equation for $q(t)$. Any ideas?

Another thing which seems artificial is that their equation produces a "negative current" which is really just there to satisfy the =0 equality. And which means, and wrongly, that the resistor is giving energy to the system and not the exterior. (That's what I think, I may be wrong).

  • $\begingroup$ Which direction do they use the one with violet? $\endgroup$ – soumyadeep Oct 8 '14 at 16:57
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    $\begingroup$ If you choose the opposite direction you get $iR + q/c = 0$. It's the same. $\endgroup$ – David Hammen Oct 8 '14 at 16:57
  • $\begingroup$ Their is nothing artificial as such.As David Hammen said the other direction gives the same equation.The thing is sometimes you will face circuits where you will be forced to move against the current and there it would be negative.May be your text bok wants to make students familiar with this concept. $\endgroup$ – soumyadeep Oct 8 '14 at 17:03
  • $\begingroup$ Why wouldn't it be $q/c-iR=0$? as I'm interpreting it, there is a voltage drop inside the resistor and voltage increase due to the capacitor $\endgroup$ – DLV Oct 8 '14 at 17:04
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    $\begingroup$ Hello so you understood it?I did not see the video as i Suddenly understood everything as suddenly as i forgot-strange .... $\endgroup$ – soumyadeep Oct 8 '14 at 17:40

It may seem that the current direction is artificial, but not really. It matches the direction of the electron flow in the wires. Their choice however is arbitrary because one MUST get the same answer regardless of direction choice.

Initially you should write an equation for the potential changes (aka voltages) without any reference to current. Including the current too early is simply confusing the issue.

$$ V_{ab} + V_{bc}+V_{ca}=0,$$ or ``the voltage of 'a' relative to 'b' plus the voltage of 'b' relative to 'c' is zero'' (because 'a' and 'c' are at the same potential.) One could just as easily write $$ V_{ba} + V_{cb}=0.$$ Now, expand the voltages, carefully. $V_{ab}$ is an ohmic voltage drop, so $V_{ab}=iR$ because of the sign convention of Ohm's Law. Notice that $V_{ab}=-V_{ba}$.

$V_{bc} = \frac{q(t)}{C}$ because this is a voltage across a capacitor. This gives us $$ iR+\frac{q}{C}=0 $$ which is the same equality as their result.

Relationship between the current $i$ through the resistor and the charge $q$ on the capacitor must be studied very carefully, paying attention to signs. We do this by realizing that the capacitor and resistor are in series, so they will have the same effective current, $i_R=i_C$.

Paying attention to the polarity of charge on the capacitor, the effective capacitor current is $i_C=\frac{dq}{dt}$, flowing in the direction from positive plate to negative plate. That direction matches the current arrow, $i$ in the diagram, so we are okay to use this derivative as a direct substitution (no need to change the sign, which is an IMPORTANT discovery).

Combining all this gives the d.e. you listed: $$\frac{dq}{dt}=\frac{-q}{RC}$$ which yields the decreasing exponential solution.

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I think you have a problem with Kirchoff's laws.Now it states that the potential drop across a closed loop is 0.

Your txt seems to have taken the direction of I as the direction of electron flow.Now suppose you move along the wire anticlockwise(The txt has moved clockwise.) Across the resistor ,potential drops but you travel along the original direction of current.So the potential drop becomes positive so $iR$.Across the capacitor you move from the positive to negative plate.Obviously the potential drop .So the potential drop along this direction is positive so $+Q/C$.

For what your txt says ,they move clockwise.Along the resistor, as you move opposite to the direction of current the potential rises along the direction you move.So the potential drop is negative.Similar is the case for the capacitor.

Note:May be you are confused by the direction of current,but thats just a convention.Your book seems to have used another convention.

2.Now coming to the video of Youtube.I suppose you have read the first part of the answer as it explains the equations and contains the most important part potential drop.

Well in the video it says,$V_c+V_r=0$.Now think one thing-What is $V_c$-it is the potential difference!So is $V_r$.What are they- positive right?So it says positive +positive=0!!!!kidding...

Actually it should be $V_c=V_r$.Now their blunder lies here... What you should do- just forget these potential differences and stuff and their equating.You have Kirchoff's laws for the same thing.The laws say that Potential dropis zero-why?As the electrostatic field is conservative-so the poential drop (or rise) is 0.But the blunder the videomaker makes is he takes both potential rise and drop together...

I hope now you understand the importance of potential rise or drop.Yes that makes it mechanical,artificial but what to do?Suppose you have a circuit like this- enter image description hereIn the right loop hoe would you do it by physicsl sense?You have to apply the laws blindly.Okay even you pull it off here there exists more complex circuits.

Okay,now after this as they made this blunder it leads to their next blunderOut of the middle of nowhere the introduce a negative sign-why?to make the current positive!!!But maths equations in physics have the fame of giving you the correct signs at last.So why manipulate signs midway?

In the last part even if you do the correct way using potential drop,you will get i as negative.That is because we have and your text book has chosen the wrong direction of current.You might be wondering why i say so-i just said its different convention.BUT in the writing of equations somewhere a fault has crept in.Wanna know where??It is in the potential drop of the capacitor. You took the potential to decrease from the positive plate to the negative plate.But that happens only whan there is positive charges moving.

Aha..So the equations speak.They are saying -look man you chose the current to be clockwise...But my $I$ comes as negative So its in the opposite direction

Thats the significance of the minus.But IF you manipulate the signs midway like the video did you are actually making the equations dumb!!!It is now you controlling the direction of current.But you dont want to do that do you-there will be places where you will be helpless only equations and signs will assist you in the direction of current.

But unfortunately the video seems to care about none of these.They just exploited the simplicity of the situation using DEAD WRONG and MISLEADING methods.

So i hope you understand my points and the mistakes of the video.If i can think of some more counterarguments i will update it.

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  • $\begingroup$ I doubt you'll need to see more than 5 mins of the video I posted to understand what I meant by it having to do with dq/dt. Maybe you did see it and you disagree with it, in which case i'd like to know, so that i'm not happy with a wrong interpretation. $\endgroup$ – DLV Oct 8 '14 at 18:34
  • $\begingroup$ @David Hello i edited my answer pls see it. $\endgroup$ – soumyadeep Oct 9 '14 at 7:30
  • $\begingroup$ @David Are you still unsatisfied with the answer? $\endgroup$ – soumyadeep Oct 9 '14 at 16:25
  • $\begingroup$ Sorry, I just saw this. I'll have to study your answer later. Thanks though. $\endgroup$ – DLV Oct 9 '14 at 17:03

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