As time continues, does I (current) through a capacitor INCREASE or DECREASE?

[ipodtouch]

I'm having a lot of trouble with this concept.

I know that when time continues until q in capacitor = Voltage, The current through the capacitor is 0.

V=IR+(q/C)
I=0
V=(q/C)

Alright, considering the formula,

I=(V/R) - (q/RC)

when I=0, V/R= (q/RC) because R and V obviously do not change.


So then as time progresses the Current I through a capacitor obviously increases.

But considering the formula for the current through a capacitor over time:

I= I x (e^(1t/RC))

It seems that current through a capacitor decreases as time continues?



I was wondering if anyone could explain this contradiction?

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[MedPR]

I'm having a lot of trouble with this concept.

I know that when time continues until q in capacitor = Voltage, The current through the capacitor is 0.



Alright, considering the formula,


when I=0, V/R= (q/RC) because R and V obviously do not change.


So then as time progresses the Current I through a capacitor obviously increases.

But considering the formula for the current through a capacitor over time:

It seems that current through a capacitor decreases as time continues?



I was wondering if anyone could explain this contradiction?

Are you saying that current I through a capacitor increases? Or did you quote that from a book? It seems to me like


V=IR

C=Q/V, C=Q/IR

 

[johnwandering]

The current through a capacitor is
I=(q/RC)

as time progresses, the q value increases--> increasing Current


hmmmm. But the time equation is saying that it decreases....
Perhaps the variables are attributed to different circuit components?

 

[ipodtouch]

I have it written in my notes, sorry I can't seem to find it in the book.

 

[MedPR]

The current through a capacitor is
I=(q/RC)

as time progresses, the q value increases--> increasing Current


hmmmm. But the time equation is saying that it decreases....
Perhaps the variables are attributed to different circuit components?

Why would increasing Q mean increased current?

 

[411309]

As time increases, charge on a capacitor Q increases. C= Q/V V=IR C= Q/IR IRC=Q

Also another equation to through out there- Q=It . So im pretty sure as time increases current increases.


My only concern is if the capacitor is charging or discharging. If the capacitor is charging Q is increasing with time, if it's discharging then Q is decreasing with time which affects the previous two equations.

Is the capacitor discharging?

 

[MedPR]

As time increases, charge on a capacitor Q increases. C= Q/V V=IR C= Q/IR IRC=Q

Also another equation to through out there- Q=It . So im pretty sure as time increases current increases.


My only concern is if the capacitor is charging or discharging. If the capacitor is charging Q is increasing with time, if it's discharging then Q is decreasing with time which affects the previous two equations.

Is the capacitor discharging?

Does it really matter if its charging or discharging? If you are looking at Q=It, if C is discharging, then there is a negative change. Negative Q and positive t = negative I, so current is decreasing. If C is charging, then Q is positive, t is positive, so I must also be positive.

Where did you get the Q=It equation from?

 

[411309]

Does it really matter if its charging or discharging? If you are looking at Q=It, if C is discharging, then there is a negative change. Negative Q and positive t = negative I, so current is decreasing. If C is charging, then Q is positive, t is positive, so I must also be positive.

Where did you get the Q=It equation from?

Q would be decreasing that doesn't necessarily mean negative and the formula isnt delta Q/V.

Q=it is a nifty little equation I found in a TBR problem. Whenever I find random equations like these I write them on a whiteboard just for reference that I keep above my computer, lol.

 

[ipodtouch]

Why would increasing Q mean increased current?

I posted the equation:
I= (q/RC)


charge and current are directly proportional. Unless someone can explain otherwise...

 

[johnwandering]

let's get this straight:


I= (q/RC)
Clearly shows that increasing charge (increasing time) means increasing current.


I= I x (e^(-t/RC))
Clearly shows that increasing time means decreasing current.



Something is not attributed correctly between these two equations. Perhaps one is Current Through the Capacitor, and the other is Current Through the Circuit?

 

[STxHorn]

Already done with the MCAT so I'm a bit rusty, but conceptually, I think you're off here. A capacitor stores charge. Over time as q increases in the capacitor it charges more slowly, that is it takes more time to charge the last half of the charge in the capacitor than it does the first half. That's why the equation for q is an exponential function (q = Q(1 - e^(-t/tau)). At first the charge flows readily into the capacitor, later as charge increases it takes more work to get the same amount of charge into the capacitor. This all occurs over very, very small amounts of time.

Taking all that into consideration it seems to me that current will be higher at first eventually dropping to almost nothing as the capacitor becomes (almost) fully charged. (The calculus tells us it's never completely fully charged except at infinite t.). I hope that makes some sense.


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[hellocubed]

Already done with the MCAT so I'm a bit rusty, but conceptually, I think you're off here. A capacitor stores charge. Over time as q increases in the capacitor it charges more slowly, that is it takes more time to charge the last half of the charge in the capacitor than it does the first half. That's why the equation for q is an exponential function (q = Q(1 - e^(-t/tau)). At first the charge flows readily into the capacitor, later as charge increases it takes more work to get the same amount of charge into the capacitor. This all occurs over very, very small amounts of time.

Taking all that into consideration it seems to me that current will be higher at first eventually dropping to almost nothing as the capacitor becomes (almost) fully charged. (The calculus tells us it's never completely fully charged except at infinite t.). I hope that makes some sense.


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You are correct, when a capacitor and a resistor are in series, the current will flow Entirely through the capacitor when q=0, meaning the current through a capacitor is highest at t=0ish.



However, there is the equation that is continually brought up.

C=(q/V)
V=(q/C)
IR=(q/C)
I=(q/RC)


This is a legitimate equation that is formulated in the BR.
What is this referring to?
When time continues, q definitely increases -----> Current increases

What current is this referring to?

 

[MedPR]

You are correct, when a capacitor and a resistor are in series, the current will flow Entirely through the capacitor when q=0, meaning the current through a capacitor is highest at t=0ish.



However, there is the equation that is continually brought up.

C=(q/V)
V=(q/C)
IR=(q/C)
I=(q/RC)


This is a legitimate equation that is formulated in the BR.
What is this referring to?
When time continues, q definitely increases -----> Current increases

What current is this referring to?

What is TBR's explanation of that equation?

If C=Q/V, then decreasing C while holding Q constant means you must be increasing Voltage. If you increase voltage while holding resistance constant, you increase current. To me it seems like the I=q/RC equation relates current through a circuit as a charging/discharging capacitor.

 

[milski]

There is no such formula as I=q/RC. You derived that by using Ohm's law for a capacitor and that cannot be done. You cannot replace any occurrence of V with IR - that's applicable only for linear resistors.

The exponential decay is the correct formula for the current of a charging capacitor - it decreases asymptotically towards 0.

 

[hellocubed]

If C=Q/V, then decreasing C while holding Q constant means you must be increasing Voltage. If you increase voltage while holding resistance constant, you increase current. To me it seems like the I=q/RC equation relates current through a circuit as a charging/discharging capacitor.

I believe you are correct. Sorry i don't have the BR book with me.


The formula definitely exists.
http://webhw.unca.edu/bennett_phys222/content/labs/rc/rc_principles.asp

It is the Capacitor component of
I=V/R - Q/RC

 

[milski]

Ok, I can concede on that. You can use it when the capacitor is creating the potential - in other words when it is discharging and there are no other sources of emf. You still cannot use it for the case of a capacitor being charged.

 

[ipodtouch]

If C=Q/V, then decreasing C while holding Q constant means you must be increasing Voltage. If you increase voltage while holding resistance constant, you increase current. To me it seems like the I=q/RC equation relates current through a circuit as a charging/discharging capacitor.

I was under the notion that the C value of a capacitor was a constant that cannot be changed (aside from a dielectric)?


Also, what does this mean?
What prohibits us from using the

I=V/R-q/RC

equation when the capacitor is charging? Because, it seems to me that q just increases and there don't seem to be any violations.

 

[MedPR]

I was under the notion that the C value of a capacitor was a constant that cannot be changed (aside from a dielectric)?


Also, what does this mean?
What prohibits us from using the

I=V/R-q/RC

equation when the capacitor is charging? Because, it seems to me that q just increases and there don't seem to be any violations.

Yea, I'm confused about that too. Why can't we derive equations using V=IR?

 

[hellocubed]

Why can't we derive equations using V=IR

I believe because R value is supposed to be independent of C?

I am also curious why it cannot be used when the capacitor is charging.

 

[milski]

You can use I=V/R-q/RC but both V and q are changing so you cannot infer anything about what happens to I.

My major problem was plugging any random V in a formula with IR - you can do that only in the case where V is a voltage drop over a resistor. For example if Vc was the voltage drop over a capacitor, you cannot just write Vc=IcRc for a lot of reasons, including the fact that there is no Rc associated with the capacitor.

For a simple circuit where you have a resistor and a capacitor, you know that the currents are the same, so you can use V=IR for the resistor and use I from there. Adding any more elements to the circuit will change that, since you have to take into account the currents through each element.

Last, the V over a capacitor is continuously decreasing when it is being discharged. Without knowing what it is, I=V/R-q/RC is not very useful.

PS: I'm a bit short on time right now, so if that's not too clear, we can continue.