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Why does the capacitor work like that?

enter image description here

In the left circuit the capacitor is discharging because there is a potential difference which was 5 V on one side and 4 V on the other.

Why doesn't it work the same in the second circuit? There is 5 V on one side and 2.5V on the other side

enter image description here

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    \$\begingroup\$ The voltage across the capacitor just isn't the same. Remember: voltage is ALWAYS measured between TWO points. In diagrams where they just list the voltage at one point, it is being referenced to some other node in the circuit which has been defined to be 0V (the reference node, aka labelled as ground) \$\endgroup\$
    – DKNguyen
    Aug 13, 2022 at 20:49
  • \$\begingroup\$ Have you tried to simulate it? \$\endgroup\$
    – winny
    Aug 13, 2022 at 20:51
  • \$\begingroup\$ @winny Yes. DKNguyen So why does it work like that ? Even thought there is a potential difference. \$\endgroup\$
    – user331990
    Aug 13, 2022 at 20:54
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    \$\begingroup\$ The voltage across the capacitor is simply not the full voltage from the source. That's all there is to it. Do you now what a resistor divider is? Don't even worry about the capacitor. Just look up how a resistor divider works. Sounds like you skipped too many steps. Not much different from how something dropped from the 10th floor will have less energy if falls on your head when you are standing on the 5th floor rather than the 1st floor. \$\endgroup\$
    – DKNguyen
    Aug 13, 2022 at 21:11
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    \$\begingroup\$ @user331990 Judging from your comments to other answers, I think you don't have the fundamentals right now to properly understand what is going on. Forget capacitors and just look at combination of only resistors for now. Then combinations of only capacitors. \$\endgroup\$
    – DKNguyen
    Aug 13, 2022 at 21:17

3 Answers 3

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The capacitor is a voltage source, somewhat like a cell (or battery of cells), except that unlike a cell, the potential difference across it changes with time, as current flows through it.

This means that at any instant in time, you could redraw your circuit replacing the capacitor with a cell of the same potential difference that the capacitor had at that instant. The currents in the circuit at that instant in time can be calculated as if the capacitor were a cell, and will therefore depend on the charge state (the voltage across the capacitor) at that instant.

Redrawing your right-hand circuit (the potential divider) three times, each time with the capacitor in a different charge state, we might get this:

schematic

simulate this circuit – Schematic created using CircuitLab

In each case, the potentials marked in red are the consequence of the two voltage sources, B1 and C, imposing their respective potential differences across various elements.

Resistor currents

In circuit A (left) the capacitor is completely discharged, and has no potential difference across it. For analysis we may consider it (for an instant in time) to be a zero volt source. Therefore the voltage across R1 is zero, and the current through it must also be zero. R2 must therefore have all the remaining 5V of B1's EMF across it, by Kirchhoff's voltage law (KVL), for a current of 5mA.

In circuit B (middle), the capacitor is initially charged to 5V, and those 5V are placed directly across R1, causing a current of 5mA through R1. By KVL again, R2 doesn't have any potential difference across it, so it's passing no current.

In circuit C (right), the capacitor is initially charged to 2.5V, which is imposed directly across R1, and so R1 passes 2.5mA. The remaining 2.5V of B1's EMF is across R2, also causing 2.5mA of current through R2.

Capacitor current

With all resistor currents known, we can work out what currents would flow through C, in order to obey Kirchhoff's current law (KCL). In circuit A, clearly all 5mA through R2 must be coming via C, since there's no current via R1.

In circuit B, for KCL to be obeyed, a current of 5mA must be flowing upwards through C, because none of the current flowing down through R1 is leaving via R2.

In circuit C, it's fairly obvious that no current can be flowing through C, since all of the 2.5mA flowing down through R1 is also leaving via R2.

Charging or discharging?

We can use current direction through C to determine if C is charging or discharging. In circuit A current flows downwards through C, so C is charging. That makes sense, because it has no initial charge.

For circuit B, the direction of current through C is in a direction that would act to discharge C. That is, the if C were a capacitor, then its potential difference would be reducing from 5V.

In circuit C, there's no current through C, so C is neither charging or discharging.

Conclusions

Circuit C demonstrates what would happen when the voltage across a capacitor equals the voltage that would be between those same two nodes if the capacitor weren't there at all. If you removed component C, there would be 2.5V across R1. If you then place a 2.5V source (a charged capacitor, a battery, a photovoltaic cell, literally any voltage source of 2.5V) across R1, nothing changes!

When C is a capacitor, if it has no initial charge, clearly it will begin to charge and that charging will continue until the potential difference across it equals the potential difference that would be across R1 if C weren't there, which is 2.5V

If C has an initial charge of 5V, it will begin to discharge, and will continue to do so until the potential difference across it equals the potential difference that would be across R1 if C weren't there, which is 2.5V.

In fact, regardless of the current state of charge of C, it will always charge or discharge until the voltage across it becomes equal to the voltage that would be present if C weren't there at all.

When capacitor C's charge arrives at 2.5V, charging/discharging halts, because in that condition, no current flows via C any more.

Thevenin equivalent

This behaviour might be easier to understand when B1, R1 and R2 are replaced by their Thevenin equivalent circuit. That would look like this, in the blue box, right:

schematic

simulate this circuit

I'll leave it up to you to study Thevenin equivalency, but it's clearer in the right-hand circuit that the capacitor will eventually settle at a charge of 2.5V, regardless of its initial state. Also, by reducing the network of R1 and R2 to a single resistance Rt (500Ω in this case), the time constant of the capacitor's charge/discharge curve becomes obvious.

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  • \$\begingroup\$ Your schematics may be confusing by using a battery symbol for a capacitor. \$\endgroup\$
    – PStechPaul
    Aug 14, 2022 at 4:13
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A capacitor's charging current is proportional to the rate of change of voltage across it, not the actual voltage. In the right-hand diagram, the current through the two resistors is equal as are the voltages across them. So there is no reason why any current should flow through the capacitor. The circuit is stable with no changing voltages. dv/dt=0 and so Icharge also equals 0.

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  • \$\begingroup\$ But on the left one it is discharging because of voltage differential. So on the right one it doesn't work ? \$\endgroup\$
    – user331990
    Aug 13, 2022 at 21:11
  • \$\begingroup\$ On the left one there is a voltage across the resistor forcing a charging current. On the right one, all of the current in the top resistor flows through the bottom resistor because they have equal voltages across them, When the current in the two resistors is equal there can be no current through the capacitor by Kirchoff's current law. \$\endgroup\$
    – user173271
    Aug 13, 2022 at 21:16
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Why doesn't it work the same in the second circuit? There is 5 V on one side and 2.5V on the other side

Do you understand the concept of steady state?
The second circuit appears to be in steady state - if you measure voltages and/or currents at one moment, they will be the same at another moment. For this to be true, the capacitor must remain charged at 2.5V.

One might ask, "Well, how did the capacitor get charged?".
At some point during the circuit's construction, current flows...the capacitor attains its voltage. In the second circuit, current continues to flow through the two resistors, but it is constant - the circuit has reached its steady state.
At first construction, current flow from the 5V source is higher than steady state current - this extra current charges up the capacitor.

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  • \$\begingroup\$ Can we move it into chat ? \$\endgroup\$
    – user331990
    Aug 13, 2022 at 22:16
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – user331990
    Aug 13, 2022 at 22:16
  • \$\begingroup\$ @DKNguyen in chat it will be easier for me to follow what are you trying to tell me. \$\endgroup\$
    – user331990
    Aug 13, 2022 at 22:35
  • \$\begingroup\$ @All - After so many comments, chat is now the only option on this answer, as comments on Stack Exchange should not be used for extended discussion. Since the system has created a chatroom for this answer already (link above) we will leave that in place, rather than the usual process of creating a new chatroom. The comments which were here (and which are now in the chatroom) have been deleted from here. Keep it in chat now, please. Any further comments posted here may deleted without notice. Thanks. \$\endgroup\$
    – SamGibson
    Aug 13, 2022 at 22:42

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