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Does charge in a capacitor containing circuit stop flowing when the potential of the capacitor becomes equal to the potential of the battery??

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  • $\begingroup$ Just to be clear, you are asking about one capacitor connected by wires to one battery, nothing else in the circuit? $\endgroup$ – DarenW Jun 17 '13 at 20:28
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The answer is just "yes, obviously, the voltage is zero". The answer below is unnecessarily computational, but I'm keeping it in case someone likes that.


Archived answer

I'll assume you are talking about an circuit with a capacitor and resistor inside. Then, let $Q$ be the charge, $t$ be the time, $C$ be the capacitance, $R$ be the resistance, $T$ be the time constant, and $V$ be the electromotive force. You must know of the differential equation: $$\frac{{{\text{d}}Q}}{{{\text{d}}t}} = \frac{{ - Q + CV}}{T}$$

Separating the equation and integrating, $$\frac{{{\text{d}}Q}}{{{\text{d}}t}} = \frac{V}{R}\exp \left( { - \frac{t}{T}} \right)$$

Since the current is the rate of change in the charge with respect to time, we can rewrite this equation in the following form:

$$I = \frac{V}{R}\exp \left( { - \frac{t}{T}} \right)$$

So, the potential of the battery being equal to the potential of the capacitor simply means that $V=0$, so $$I = \frac{0}{R}\exp \left( { - \frac{t}{T}} \right)=0$$

So yes, although it will take an infinite amount of time to reach this point.

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When you connect a DC-source to the capacitor, the capacitor will "instantly" have a charge which is equal to C*V (with V the voltage of the source, and C the capacitance of the capacitor). The current through a capacitor equals: I = dQ/dt = C*dV/dt

So when a constant voltage is applied, the current will be zero (since there is no change in voltage, dV/dt=0. Actually, dV/dt is a delta function when the voltage is applied (since the voltage is a step function), so there will be a short current pulse).

(When you have a circuit with a DC-source, a capacitor AND a resistor, the capacitor will load smoother with a damped exponential function. Eventually, the charge and voltage will be constant and there will be no current flowing.)

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You're question is ambiguous as other answers have shown.

(1) if you're thinking of a simple two element circuit with a battery and a capacitor, the capacitor voltage must, at all times, equal the battery voltage. Assuming the battery voltage is constant, the circuit current will be zero.

(2) if you're thinking of a simple series circuit that includes a battery, resistor, and a capacitor, the capacitor voltage asymptotically approaches the battery voltage.

(3) if you're thinking of something else, you'll simply have to be more specific.

The crucial consideration is this: the current through a capacitor is proportional to the rate of change of the voltage across the capacitor.

So, to answer the question in general, if the voltage across the capacitor is constant, the current through the capacitor is zero.

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It depends

Circuits: a capacitor connected in series or in parallel

Attached in series, then yes once charged, the capacitor represents an open circuit and current ceases to flow.

In parallel current can flow through the resistor which is does once the capacitor is charged (to any degree) reaching maximum current when the capacitor is fully charged.

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  • $\begingroup$ A very slight EE quibble: there's no time dependence in either drawn circuit. As drawn, the voltage across the capacitor is, has, and will be the battery voltage. $\endgroup$ – Alfred Centauri Jun 17 '13 at 22:14
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    $\begingroup$ @AlfredCentauri Well, there is that. And I though of it just after I uploaded this one. With real wire, of course, there is a little resistance on all paths; thought the time constants are pretty small unless you use a monster capacitor. $\endgroup$ – dmckee --- ex-moderator kitten Jun 17 '13 at 23:35
  • $\begingroup$ @AlfredCentauri Fixed. $\endgroup$ – dmckee --- ex-moderator kitten Jun 17 '13 at 23:48

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