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If an $emf$ leaves from point A, the concepts at play in the following diagram are straightforward:

  1. at any time, the $I_{R}$ = $I_{c}$

  2. as time increases, $\Delta V$ across $R$ decreases and it increases across $C$.

If I added another resistor, $R_{2}$, after capacitor $C$, what would the new resistor's current be?

My logic was that resistor two's current initially = 0, because charge first accumulates on the capacitor before it can travel onwards. $I_{R2}$ increases afterwards

My textbook claims, however, that $I_{R}$ = $I_{2R}$ = $I_{C}$. I disagree. How can $I_{2R}$ = $I_{C}$. If that were true, and the coulombs of charge passing through $C$ all continued to $R_{2}$, $C$ wouldn't accumulate any charge.

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If that were true, and the coulombs of charge passing through C all continued to R2, C wouldn't accumulate any charge.

The capacitor doesn't accumulate (electric) charge in order to charge. In a circuit, a capacitor is electrically neutral since the plates have (ideally) equal and opposite (electric) charge.

When we say that the capacitor is charged, we mean it in the same sense as when we say that a battery is charged. That is to say, we mean that the capacitor, like the battery, is 'charged' with energy.

So, in fact, the textbook is correct and the current entering the capacitor equals the current exiting the capacitor.

From another perspective, keep in mind that, since the circuit elements are linear, the order of the circuit elements in the circuit is irrelevant to the circuit solution.

This means that one can move R2 to be in-between R1 and C and get the same solution. But, by your reasoning, one would get two different solutions so it can't be correct.

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  • $\begingroup$ "When we say that the capacitor is charged, we mean it in the same sense as when we say that a battery is charged. That is to say, we mean that the capacitor, like the battery, is 'charged' with energy" In the equation CV = Q, what does Q physically represent then? Many Thanks $\endgroup$
    – Muno
    Commented Jan 11, 2015 at 23:17
  • $\begingroup$ @user21945, Q is the magnitude of the charge on either plate but it is a mistake to think of Q as the electric charge stored in the capacitor. Think of Q as the charge separated in the capacitor, i.e., Q charge has been transferred from one plate to the other leaving one plate with +Q charge and the other with -Q charge. Work must be done in separating this charge and that work equals the energy stored in the capacitor. $\endgroup$
    – Alfred Centauri
    Commented Jan 11, 2015 at 23:22
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If electrons pile up on one side of the capacitor at a certain rate, they leave the other side at the same rate. The capacitor is charged only in the sense that both sides have equal and opposite amount of charge imbalance. So the current does flow through both resistors the same amount, they are "different" electrons, but they travel through at the same rate of current.

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Current is the change in charge per unit time. Therefore as long as there is a movement of charge, current will flow, even whilst the charge is 'accumulating' in the capacitor.

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