I Voltage drop across a capacitor and resistor in series

  • Thread starter Tryhard314
  • Start date
10
4
Summary
The voltage drop across a resistor in series with a capacitor should be in my opinion only the potential difference between the plate of the capacitor connected to the same wire with the resistor and thep late
Cap.png
In this circuit a battery,Capacitor,and a resistance are in series.
For simplicity assume that there is a +4V in the positive terminal of the battery and -4V in the negative one and let A be the capacitor plate connected to the positive terminal and B the capacitor plate connected to the negative terminal.

During the first (milisecond) The capacitor is uncharged so all the voltage drop is across the resistor: 8V,
In my opinion it should be only 4V:

The red wire going out from the positive terminal,To the resistor and the capacitor is physically disconnected from the negative terminal.

And the only potential difference would be the difference between the capacitor's plate A potential and the positive terminal potential,During charging ( first miliseconds) it should be : +4V - 0V = 4V potential difference and not 8V

I know this is wrong but i don't know why.
 
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There are several ways to explain this - I'll attempt to do it in your terms:

Your assumption that the 'plate A' potential at t0 is '0V' is wrong. If 'plate B' is at '-4V' (which you seem to accept) and the capacitor is uncharged, then 'plate A' must be at the same voltage. Your assumptions imply a 'pre-charged' capacitor.
 

phyzguy

Science Advisor
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Another way to say what @Dullard said. Assume the capacitor is initially discharged. This means the potential on the top plate of the capacitor is the same as the potential on the bottom plate. If you take the potential of the bottom plate to be -4V, then the potential on the top plate is also -4V. So immediately after the switch is closed, the resistor has -4V on the right side and +4V on the left side, for a potential difference of 8V.
 
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Another way to say what @Dullard said. Assume the capacitor is initially discharged. This means the potential on the top plate of the capacitor is the same as the potential on the bottom plate. If you take the potential of the bottom plate to be -4V, then the potential on the top plate is also -4V. So immediately after the switch is closed, the resistor has -4V on the right side and +4V on the left side, for a potential difference of 8V.
That's a possible explanation but if the initially uncharged capacitor had a potential of -4V on the bottom plate,no charges will build up on this plate anymore because it's at the same potential with the negative terminal of the battery !Making the circuit open and the chimical reaction inside the battery halts
 
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The capacitor is uncharged so all the voltage drop is across the resistor: 8V,
In my opinion it should be only 4V:
When we say that the capacitor is uncharged, that means that V = 0 is the voltage difference across the capacitor, not the voltage relative to ground. In other words V=0 means that the two plates of the capacitor are the same voltage, it does not mean that either of the plates is at the same voltage as ground.

no charges will build up on this plate anymore because it's at the same potential with the negative terminal of the battery !Making the circuit open and the chimical reaction inside the battery halts
The charge built up on the plate is related to the voltage across the capacitor, not the voltage relative to ground. The capacitor initially has not charge on the plate so the voltage across the capacitor is 0. As current runs through the capacitor then charge builds up on the plate and the voltage increases accordingly.
 
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When we say that the capacitor is uncharged, that means that V = 0 is the voltage difference across the capacitor, not the voltage relative to ground. In other words V=0 means that the two plates of the capacitor are the same voltage, it does not mean that either of the plates is at the same voltage as ground.

The charge built up on the plate is related to the voltage across the capacitor, not the voltage relative to ground. The capacitor initially has not charge on the plate so the voltage across the capacitor is 0. As current runs through the capacitor then charge builds up on the plate and the voltage increases accordingly.
Let's start on correct bases From what i learned potential difference is what makes electrons flow in a circuit,So it's not only the potential difference across the capacitor that matter's it's the plates potential relative to the battery that matters too,

If you have a capacitor with both plates at -4V It's correct that there would be a 8 V potential difference across the resistor ,
+4V - (-4V) = 8V
But if you consider the negative terminal of the battery's it's only at -4V potential and the capacitor plate is already at -4V
-4 - (-4) = 0V
There would be no reasons for charge to further more accumulate on the capactor negative plate since it has already reached the negative terminal potential Further more, The battery needs a closed circuit to operate (So the redox reaction dosen't halt)and for this to happen electrons need to enter the battery positive terminal (It's true in this example) But also electrons need to leave the negative terminal (False in this example since the capacitor's negative plate and the negative terminal of the battery have the same voltage)

I am left with 2 possibilties:
1-You are incorrect (inlikely since this is a physics forum and I am a dumb student)
2-When the positive capacitor plate get's charged this makes somehow the negative plate of the capacitor potential lower.
 
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From what i learned potential difference is what makes electrons flow in a circuit,
This is not correct in general. This statement correctly describes the behavior of a resistor, but not other circuit elements. In an inductor electrons will flow even with no potential difference. In a capacitor a potential difference is often associated with no electron flow. In a battery the potential difference is fixed regardless of the electron flow. So in general, particularly in circuits that are not purely resistive, this is not a good rule to base your understanding on.
 

phyzguy

Science Advisor
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You need to understand that only potential difference matters. The actual value of the potential is arbitrary. There is no difference in the behavior of your circuit between the folowing three cases:
(1) The bottom of the battery is at -4V and the top is at +4V.
(2) The bottom of the battery is at 0V and the top of the battery is at 8V.
(3) The bottom of the battery is at 1000V and the top of the battery is at 1008V.
 
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You need to understand that only potential difference matters. The actual value of the potential is arbitrary. There is no difference in the behavior of your circuit between the folowing three cases:
(1) The bottom of the battery is at -4V and the top is at +4V.
(2) The bottom of the battery is at 0V and the top of the battery is at 8V.
(3) The bottom of the battery is at 1000V and the top of the battery is at 1008V.
Ok then you can say if the bottom of the battery is a -4V and the bottom of the capacitor is at -4V there is 0V potential difference and there is no reason for electrons to flow to the capacitor Is this wrong ?
 
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Ok then you can say if the bottom of the battery is a -4V and the bottom of the capacitor is at -4V there is 0V potential difference and there is no reason for electrons to flow to the capacitor Is this wrong ?
Yes, this is wrong. That is what I already mentioned above. Potential difference is not what makes electrons flow in general. That statement is only correct for resistors, not for anything else. In particular, the -4 V terminal of the capacitor and the -4 V terminal of the battery are connected by a wire, and in an ideal wire electrical current flows freely with no voltage.
 

phyzguy

Science Advisor
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Ok then you can say if the bottom of the battery is a -4V and the bottom of the capacitor is at -4V there is 0V potential difference and there is no reason for electrons to flow to the capacitor Is this wrong ?
No, this is not wrong. But look what hapens when you close the switch. You now have +4V on the left side of the resistor, so an 8V potential difference across the resistor. So current begins to flow through the resistor. This current raises the potential on the right side of the resistor. Now there is a potential difference between the top and bottom of the capacitor, so current begins to flow into the capacitor.
 
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Yes, this is wrong. That is what I already mentioned above. Potential difference is not what makes electrons flow in general. That statement is only correct for resistors, not for anything else. In particular, the -4 V terminal of the capacitor and the -4 V terminal of the battery are connected by a wire, and in an ideal wire electrical current flows freely with no voltage.
Wait are you the same dale from physics stack exchange (i am the op on physics stack exchange too)xDD
Assume the wire have a bit of resistance so we don't start an other debate .
 
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There would be no reasons for charge to further more accumulate on the capactor negative plate since it has already reached the negative terminal potential
This is where you go wrong. When the switch is open, there will be almost no charge on the negative plate to keep it at -4 V. (this amount of charge is usually ignored, because it is so small)
When you open the switch you will get a positive charge on the other plate, this will attract more negative charges. The equation for how much charge you can get is Q =U/C, where U is the voltage difference between the plates. The amount of charge only depends on the difference between the potentials, not on the potential of the plate.
If you ignore the resistance of the wire between the battery and the negative plate they will be at the same potential always. You don't need a potential difference to get a current through a 0 ohm resitor.
If you don't ignore the resistance of the wire, the potential of the negative plate will become slightly less negative after the switch is opened and while there is still current through the capacitor. This is insignificant if the wiring resistance is much smaller than R.
 
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Wait are you the same dale from physics stack exchange (i am the op on physics stack exchange too)xDD
Assume the wire have a bit of resistance so we don't start an other debate .
Yes, that is me :-)
 
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This is where you go wrong. When the switch is open, there will be almost no charge on the negative plate to keep it at -4 V. (this amount of charge is usually ignored, because it is so small)
When you open the switch you will get a positive charge on the other plate, this will attract more negative charges. The equation for how much charge you can get is Q =U/C, where U is the voltage difference between the plates. The amount of charge only depends on the difference between the potentials, not on the potential of the plate.
If you ignore the resistance of the wire between the battery and the negative plate they will be at the same potential always. You don't need a potential difference to get a current through a 0 ohm resitor.
If you don't ignore the resistance of the wire, the potential of the negative plate will become slightly less negative after the switch is opened and while there is still current through the capacitor. This is insignificant if the wiring resistance is much smaller than R.
Ahhhhhhh Thanks now i understand so there is like kinda of two type of things happening here first there is like the same charges that act on the wire to increase it's potential(I am talking about surface charges and it's like really really small amount of charge) and i got confused between Surface charges and like the Actual capacitance (now i understand why i didn't really get why dielectic premitivity or distance would matter because i thinked of them as two seperate plates !)

Thanks @willem2 , @Dale , @phyzguy and @Dullard for your time !
Do you mind if i quote some of your posts on Physics Stack exchange (Ofc a bit later after i am sure i really got it ) (I will give a link to this discussion)
 
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After rereading what dale said it might have been my bad english that failed me xD
 
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Another way to say what @Dullard said. Assume the capacitor is initially discharged. This means the potential on the top plate of the capacitor is the same as the potential on the bottom plate. If you take the potential of the bottom plate to be -4V, then the potential on the top plate is also -4V. So immediately after the switch is closed, the resistor has -4V on the right side and +4V on the left side, for a potential difference of 8V.
Just one last question If both capacitor's plates were at 0V at the start,then we place them in the circuit,(switch open), The Plate b will reach a potential of -4V (but with close to no capacitance) How does the other plate(A) get it's potential to -4V too since the capacitor is in uncharged state and both the plates should be at the same potential ?
Can the plates like interact with each other through the insulating metariel placed between them ?
 
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How does the other plate(A) get it's potential to -4V too since the capacitor is in uncharged state and both the plates should be at the same potential ?
Since the two plates are so close together they will naturally be at the same potential if they are uncharged.
 

Baluncore

Science Advisor
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I am surprised no one defined some zero potential reference node, and labelled the nodes in the circuit. Grounding the bottom or mid-point of the battery would have simplified discussion of the analysis. Floating circuits are like castles in the air.
 
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And the only potential difference would be the difference between the capacitor's plate A potential and the positive terminal potential,During charging ( first miliseconds) it should be : +4V - 0V = 4V potential difference and not 8V
I know this is wrong but i don't know why.
I tried hard to understand the question you presented because my English is not good.

Anyway I want to say that the behavior of capacitor can basically be described by the following formula.

VC=Q, where V is the potential difference between plate A and plate B of the capacitor, C is capacitance, Q is charge

From VC=Q we can derived the two equations below : -

V = (1/C) int (I) dt, ie., voltage across the capacitor is equal to 1/C times the time integral of current flows through it.

I = C (dV/dt), ie., current through a capacitor is equal to C times the derivative of the voltage across the capacitor with respect to time.

Therefore, you can apply a current source to obtain the corresponding voltage change, or you can apply a voltage source to obtain the corresponding current change of the capacitor.

In addition, you can assign any voltage to the bottom plate of the capacitor, for example -4V. If the initial charge is zero, then V =Q/C=0, ie., the voltage across the capacitor is zero, so the voltage of the top plate of the capacitor is -4V + 0 = -4V.

Hoping this is helpful for you.
 
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