I'm really confused. I'm studying about how capacitors are charged, and I learnt that there's a resistor attached in series with the capacitor when it is being charged. When the capacitor is fully charged, the current doesn't flow through the resistor anymore, as there's no potential difference across it. But, if that's the case, then how is circuit completed anymore, since resistor is in series with the capacitor?
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3$\begingroup$ A capacitor is a break in a circuit. The key thing to understand about capacitors is that electrical current can still flow through breaks, even when electrical charge cannot -- but what this means is that around the break, there is a charge buildup which then builds a voltage that opposes further current flow, until yes, that circuit is broken. I don't think this answers your question directly but maybe indirectly it gives some insight that maybe it is a mistake for our language to be so absolute about a "completed" circuit versus a "broken" one. $\endgroup$– CR DrostMay 15, 2019 at 14:42
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5 Answers
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I learnt that there's a resistor attached across the capacitor.
The situation you are describing is not one where there is a resistor across the capacitor. Across means one end of the resistor is connected to one end of the capacitor and the other end of the resistor is connected to the other end of the capacitor. That’s called being in parallel with the capacitor.
The situation you are describing is one where the resistor is in series with the capacitor and a source of electrical potential, say a battery. See the circuit diagram below which shows a battery, switch, resistor, and initially uncharged capacitor (no initial voltage on the capacitor).
When the switch S is open you can see that obviously no current will flow in the circuit. Things start at time t=0 when the switch S closes.
An ideal capacitor has the property that you can not change the voltage across it in zero time. So when the switch is first closed, the capacitor looks like a short circuit, i.e., zero resistance device. Imagine replacing the capacitor with a wire when the switch first closes. Now according to ohms law,
$$I=\frac{V}{R}$$
That means at time t=0 the current is simply the battery voltage divided by the resistor shown in the circuit. At time t=0 this is the maximum current that will flow in the circuit. After that, current decreases in time.
As current flows, however, charge is delivered to the capacitor and voltage across the capacitor increases in time. Put the capacitor back in the circuit. As long as the voltage across the capacitor, call it $V_{C}$, is still less then the battery voltage, $V$, current will continue to flow and will be
$$I=\frac{V-V_{C}}{R}$$
Eventually the build up of charge on the capacitor will result in $V_{C}=V$. From the previous equation that means $I=0$, that is, as you say “When the capacitor is fully charged, the current doesn't flow through the resistor anymore, as there's no potential difference across it”.
Since the current is now zero, the voltage across the resistor which equals $IR$ is zero.
Hope this helps.
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The voltage over the resistor is the difference between that over capacitor and the voltage source, that is, zero. The circuit is closed but the two voltage sources cancel out.
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Are we talking about a capacitor and a resistor in parallel? Then it is the other way around:
In the beginning, all current is directed towards the capacitor since that path has smallest (zero) effective resistance.
As the capacitor fills up with charge, further incoming charge is being resisted by repulsion more and more. The current is thus gradually diverted through the resistor instead.
When the capacitor is fully charged, it now acts as having infinitely large effective resistance, and all the current flows through the resistor.
At this filled stage, there is a potential difference (voltage) across the resistor just as there would be of the capacitor wasn't present. A filled capacitor has no influence on a DC circuit and can be considered a hole in the circuit (infinite resistance).
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The capacitor is two conducting plates seperated by an insulating material.When we connect the capacitor to a battery the cathode of the battery attracts electrons from one conducting plate.The anode of the battery repels ele ctrons to the second conducting plate.While doing so it willis create a potential difference across the two plates whixh will resist the flow of charges.ThisI will continue happening until the potential difference is big enough to stop the flow of charges.
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For a capacitor in series with a resistor in a closed DC circuit, consider the battery to be an electron "pump". The battery doesn't create any electrons, it just moves them from one place to another. This means that when the circuit is closed, the battery takes an electron off of one plate of the capacitor, moves it through the circuit, and deposits it on the other plate of the capacitor.
When the electron is moved to one plate of the capacitor, there is a net negative charge on that plate and a net positive charge of equal magnitude on the other plate, because that is where the electron came from. Since one electron has a very tiny charge, the battery can move other electrons to the negative plate, and it can keep doing this for a fairly large number of electrons. However, as the charge builds up on the negative plate, the electrons that are there repel any new electrons that the battery is adding to the plate, because like charges repel. This means that electrons are moved to the negative plate until the voltage drop across the capacitor is equal to the battery voltage. At that point, the battery can't add any more electrons to the negative capacitor plate, and current in the circuit stops. Thus, there is current flow in the circuit until the capacitor is fully charged, but as other posters have pointed out, there is no current flow across the plates of the capacitor because the capacitor plates are separated by a very tiny gap with an electrical insulator in between them.
answered May 15, 2019 at 18:50