Why current through all the resistors in series is considered to be same? [duplicate]

1. Consider the circuit shown, where $R_{1} > R_{2}$. I meant to say that resistance offered by the resistor 1 is greater than the resistance offered by the resistor 2. Keep $t$ constant i,e calculate the charge transferred through each resistor over the time $t$. All the way it means that charge transferred through the resistor 1 ($Q_{1}$) is slightly less than charge transferred through the resistor 2 ($Q_{2}$) over the same time $t$.Because, resistor 1 offers high resistance than resistor 2, amount of charge flowed through resistor 1 will be less than the amount of charge flowed through resistor 2 over the same time t.

2. As it is well known that basic equation for current ($I$) can be given as $I = \frac{Q}{t}$, where $Q$ is the electric charge transferred through the resistor over a time $t$.

3. If you agree with the previous points, current through each resistor can be given as: $I_1 = \frac{Q_1}{t}$ and $I_2 = \frac{Q_2}{t}$

Thus, if you have agreed all the three points, you will conclude that "current through each resistor is different."

But, the current through each resistor is considered to be same through all the resistors in case of series connection.So,here comes the unanswered question at the end "How can this contradiction be explained?"

NOTE 1: I have noticed many saying that, if current (motion of charge per unit time) "I" had pass through the conductor, then same current (motion of charge per unit time) should come out of the conductor, taking the analogy that same water (not considered rate of water) which had flown into the pipe should come out.It is not so,if they have considered rate of motion of charge at first they must consider rate of flow of water latter.You can see Nabeshin, explaining the problem with analogy of flowing water.click here.It is also said that, if current through each resistor is different, electrons would pile up.Yes, there will be slight piling up, that is what results in heating of resistors.But considering this any one can't say current could not be different through different resistors.

NOTE 2: I have noticed many others proving current through each resistor to be same, by keeping the ammeter in between the resistor's and else any where, and the value comes out to be same!!It doesn't mean current through each resistor is same, they have calculated current in between the resistor and else any where, but not through the resistor.This is the reason why we can't use ohm's law equation.I think we must find any other way to find current through the resistor, because you can't place an ammeter through a resistor.Well, you might ask "why current is coming out to be same in between the resistors and else any where?", this is because charges get compensated to form a steady flow after coming pass through resistors, which actually causes heating up of resistors.We can't expect the same current through both of the resistors in series, it depends on the resistance value they offer.Due to compensated motion of the charge the current difference that flowed through either of the resistors might be tending to zero, so we would have considered the current to be same.

NOTE 3: If you still want to consider the analogy of water.Consider the following picture, where you can visualize flat surfaces as resistors and down falling movement of water as other part of conductor.You can see flow of current is slow at the flat surface (resistor), once again speed will be compensated and it moves with the actual speed it had before.Thus, speed will be different at flat surface (resistors), and not same through out the motion of water (circuit).

Anyway I want anyone from here onward's to give answer either saying current is same in all the resistors "proving my explanation false" or obeying my explanation.Please don't opt other way for proving current to be same in all the resistors, just say where am I going wrong in the explanation, it has still remained as "doubt" for me and others.
"If you do agree with my answer, please report a answer specifying reasons why you obey it, other wise it would mean as if there was no acceptance on my explanation."
I want David Z take off the sign of duplicate, as I have reasoned out how I am not satisfied with previous answers.

I hope my doubt will be cleared at least now!

[Statements made above are up to my view.Any correction advisory is welcome.]

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marked as duplicate by David Z♦Oct 16 '13 at 18:18

How do you get $R = Q/T$? That's current, not resistance. –  Javier Oct 16 '13 at 15:40
@CURIE, could you please put a space after your commas and periods? –  Hasan Oct 16 '13 at 16:16
I don't see how your "note" addresses that. Also, you should link to the earlier question or answers in your own question, for example something like "I already looked at this question but the answers there don't explain [concept]...." –  David Z Oct 16 '13 at 18:37
I think this has been said many times, but if you insist: you went wrong in assuming different amounts of charge flow through the resistors. Experience tells us that's not true, there's not a lot of room for discussion. –  Javier Oct 17 '13 at 15:18
@CURIE, Your assumption that different charges flow through each resistor is wrong. This has been explained to you multiple times. The reason for this has also been explained. It is your problem if you refuse to accept everyone's explanation, without providing a coherent, sensible reason for rejecting said explanations. –  Pranav Hosangadi Oct 18 '13 at 5:40

Your problem is assuming that the charge transferred through the resistors is different. I don't know where you got that from, so I don't really know how to refute it other than by saying that since the currents must be the same, so must be the charge transfer in a given time.

Edit in response to your comment: What you said is plainly not true. Experimentally, the same current passes through both resistors. What happens is that the voltage drop is different, according to Ohm's law: $V = IR$. When the resistance is higher, the potential difference must be higher so the same current can pass through.

I know that if you think about it, it looks like there would be less current through the bigger resistor, but it's not like that and no thought experiment will help you. Think about what would happen: if less charge passed through the first resistor, a charge density would start to appear, and it would create electric fields that would interfere with the current. If you get a couple of resistors, cables, a battery and a voltmeter, you can easily see this is not the case.

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To demonstrate that the (steady-state) current is the same on both resistors, simply look at the charge state of a spot chosen between them. If the current were not the same charge would accumulate there changing the potential seen by each resistor in such a way as to undo the imbalance. –  dmckee Oct 17 '13 at 14:29

All the way it means that charge transferred through the resistor 1(Q1) is less than charge transferred through the resistor 2(Q2) over the same time T.

I don't understand how this follows from the preceding two statements.

If R1>R2,resistance offered by resistor 1 is greater than resistor 2,so the amount of charge transferred through the resistor 1 will be greater than the amount of charge transferred through resistor 2 in the same time interval T.

This is just plain wrong. If the resistors were parallel connected such that they have identical voltage across, then the above is correct (if you change greater to less).

But the resistors are not parallel connected.

Instead, they are series connected which, by definition, means that they have identical currents through and, thus, the voltage across R1 is greater than across R2.

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.If R1>R2,resistance offered by resistor 1 is greater than resistor 2,so the amount of charge transferred through the resistor 1 will be greater than the amount of charge transferred through resistor 2 in the same time interval T.For example,if we have lot of hurdles(R1) in one path and a small amount of hurdles in another path(R2).If we keep test of how many vehicles(Q) can pass through two different path in the same time interval,vehicles(Q) passed through the path where there were lot of hurdles(R1) will be less and greater vehicles(Q) will be passed through another path(R2) –  Feynman Oct 16 '13 at 16:20
@CURIE, that's just plain wrong. –  Alfred Centauri Oct 16 '13 at 16:46
@CURIE, see the update to my answer –  Alfred Centauri Oct 16 '13 at 16:53
I had edited incorrectly saying "greater than" instead of "lesser than".But,it's good you noticed it.Well apart from it,I think you haven't read the question properly which says that if you had agreed all the points, current through each resistor is different contradicting to the definition what you mentioned.Well that is what the question is about explaining this contradiction.If you don't agree any of the point mentioned, comment on it.I am here to answer you. –  Feynman Oct 16 '13 at 17:10
@CURIE It sounds like Alfred understood your question perfectly well. –  Dmitry Brant Oct 16 '13 at 17:29

Quite simply:

Since current is the amount of charge that flows through a resistor in a given time, if the same current flows through two resistors, the same charge must also flow through them (of course, given that the time is the same for both).

I do not see how you concluded that the charge through both resistors is different, since any charge that passes through $R_1$ must also pass through $R_2$. Since charges and currents are related, this statement goes for currents too. Current through $R_1$ is equal to current through $R_2$, since it has nowhere to go. This is, in fact, the statement of Kirchhoff's Current Law

To make it simpler, consider charge (or current) flowing through the circuit as a car on a one-lane street. It can not turn back, only travel from B to A. Once it has gone through $R_1$ it has no option but to go through $R_2$. On the other hand, if there was a parallel connection, or for the purpose of this analogy, a fork in the road, the car would have the option of going through either one of the roads. In the case of electric charge / current, we see it as splitting across both paths in the proportion $Q_1 : Q_2 = R_2 : R_1$

EDIT

Here's another way to look at it:

The charge passing through a resistor depends on the resistance as well as the potential difference across the resistor. Now consider two resistors, connected such that the potential difference across them is the same. In this case, your points would be correct, i.e. the charge and hence the current passing through each resistor is different. However, this case, by definition, is the case of parallel connection.

In case of a series connection, the potential difference across the individual resistors is not equal, so you can not claim what you do in your question. Rather, $\frac{V_1}{V_2} = \frac{R_1}{R_2}$, which ensures that the charge and current passing through both resistors is equal.

If you insist for a mathematical proof:

$$I = \frac{V}{R} \implies I_1 = \frac{V_1}{R_1} and \ I_2 = \frac{V_2}{R_2}$$ $$\therefore \frac{I_1}{I_2} = \frac{V_1}{R_1} \times \frac{R_2}{V_2}\\ \therefore \frac{I_1}{I_2} = \frac{V_1}{V_2} \times \frac{R_2}{R_1}\\ \therefore \frac{I_1}{I_2} = \frac{R_1}{R_2} \times \frac{R_2}{R_1}\\ \therefore \frac{I_1}{I_2} = 1\\ \therefore I_1 = I_2\\ \therefore Q_1 = Q_2$$

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Thank you for asking the question.Now I could make out where you are going wrong.I hope you have misunderstood the statement"charge flowing through different resistor is different".I didn't mean that, different "charge" is going through different resistor, instead I meant "different amount of charge" is going through the different resistors,but they are the same charges which had gone pass through R1.It will be great full if you can even go through NOTE given in the question. –  Feynman Oct 16 '13 at 17:27
@CURIE, that is exactly what I mean. I have understood your question perfectly. Consider that car to be a charge of 1 Coulomb, you should see that the charge of 1C has no option but to pass through $R_2$ once it has passed through $R_1$. In other words, all charge that passes through $R_1$ MUST pass through $R_2$. Is this clear now? –  Pranav Hosangadi Oct 16 '13 at 17:31
Hossangadi.Yes,all the charges once pass through R1 must pass through R2.But the amount charges passed through the resistor 1 and resistor 2 vary in the specified time interval.Well would you add on any thing in support or against the question now. –  Feynman Oct 16 '13 at 17:41
Alright, then please explain where you think the supposed difference in the charge flowing through both resistors is coming from. –  Pranav Hosangadi Oct 16 '13 at 17:46
I think you have not understood what I said in the first comment.Go through it and then edit once again.CURIE:) –  Feynman Oct 16 '13 at 17:56