Graph of capacitor voltage

[TsAmE]

An earthed 2µF capacitor C is charged from a +9V supply rail via a 1MΩ resistor. At time t = 0, C is uncharged.

Sketch a graph showing how the capacitor voltage vc varies with time. Show the tangent line and asymptote, and the critical point a t = RC.

(Correct answer attached).

There are a couple of things I am not sure of:

*Why is it that the graph approaches an asymptote of vc, and not +9V?
*How do you know which point to draw the tangent line from?
*When using the formula t = -RCln(1 - v/v) to calculate the time for the capacitor to reach a certain voltage, what are the standard units that you must convert R and C to, for the calculation?

 

[wassabi]

*Why is it that the graph approaches an asymptote of vc, and not +9V?

Because the current passing through R results in a voltage drop across R.
9V - VdropR = Vcap


t in seconds
C in uF
R in ohms
V in volts
R*C is in seconds

For the rest, I think this page will be of help to you.
http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/capchg.html#c2

 

[LvW]

*Why is it that the graph approaches an asymptote of vc, and not +9V?

Because the current passing through R results in a voltage drop across R.
9V - VdropR = Vcap

Hi wassabi,

are you really sure about that?
______________________________
Hi TsAmE,
the tangent always starts at the origin.

 

[wassabi]

Rephrasing: the current passing through R will decrease as C approaches 100% charge. The Vdrop across R then approaches but never actually hits zero.

 

[stefannm]

1. The asymptote should be 9V. I think the graph is off by a factor of 10.

2. I don't know. By the definition of tangent line, you could draw it anywhere on the graph. Perhaps it can be argued that the origin is the most interesting because it will give an indication of the maximum current because i = C dv/dt where dv/dt is the slope of the tangent line.

3. Farads = A*s/V and Ohms = V/A. So when you multiply them you get seconds.

 

[LvW]

Rephrasing: the current passing through R will decrease as C approaches 100% charge. The Vdrop across R then approaches but never actually hits zero.

OK, that means that the capacitor voltage will rise as high as 9 volts (for t approachung infinite), right?

Added after 3 minutes:

2. I don't know. By the definition of tangent line, you could draw it anywhere on the graph. Perhaps it can be argued that the origin is the most interesting because it will give an indication of the maximum current because i = C dv/dt where dv/dt is the slope of the tangent line.

Of course, you can draw a tangent where you want.
However, only the tangent through the origin makes sense because it gives you an information about the time constant: The tangent through 0/0 crosses the maximum horizontal voltage line at a time t=T which is identical to T=RC.

 

[karesz]

Hi,
2,
Yes; Stefan has right!
Tangent means per definition: is a line , what is in 90 degree to drow to the "micro part" of the graph ...
K.

 

[stefannm]

Thanks LvW. I never realized that before.

 

[LvW]

Thanks LvW. I never realized that before.

It's easy to prove with the differential quotient (derivative) of the loading function at t=0.

 

[TsAmE]

Rephrasing: the current passing through R will decrease as C approaches 100% charge. The Vdrop across R then approaches but never actually hits zero.

Why does the current decrease and Vdrop across R approach 0?

1. The asymptote should be 9V. I think the graph is off by a factor of 10.

The graph is right, it the memo I got, but it shows the asymptote as the capacitor voltage, which I dont get.

 

[LvW]

The graph is right, it the memo I got, .......

Ahh - I see. OK, a memo can't be wrong.
In this case, we have to rewrite all of the textbooks.

 

[TsAmE]

The graph is right, it the memo I got, .......

Ahh - I see. OK, a memo can't be wrong.
In this case, we have to rewrite all of the textbooks.

Sorry but I am confused that is why I am asking, cause I see a vc label and not V for the asymptote.

 

[matbob]

Hi TsAmE,

What stefannm told is right: "The asymptote should be 9V. I think the graph is off by a factor of 10."

There is a scaling factor of 10 for the y axis.

You may verify this using the equation for capacitor voltage:

Vcap= Vfinal + (Vinitial - Vfinal) e^(-t / (RC)), where R is in ohms and C in farads

and Vfinal is the voltage to which the capacitor is charged, Vinitial is the initial voltage of the capacitor.

so, for R=1M , C=2uF , Vfinal=9V and Vinitial=0V :

at t=0s, Vcap=0V
at t=1s, Vcap=3.54V
at t=2s, Vcap=5.69V
.
at t=5s, Vcap=8.26V
.
at t=8s, Vcap=8.83V


and so on.

You would get same values if you multiply the voltage along y axis by 10.

 

[ahmed277]

time constant : the time the output reaches 63% of its final value or: the time the output reaches its final value assuming that the initial rate remains constant

 

[TsAmE]

Oh ok I see makes a bit more sense now. I have another doubt, shouldnt the asymptote of vc be slightly less than 9V since when the capacitor charges up to +9V, it has to go through the 1MΩ?

 

[matbob]

Hi TsAmE,

No, the asymptote would be at 9V. Let the capacitor charge to 9V which happens at t = infinity. Now what the resistor sees is 9V on both its terminals. So there won't be any current through the resistor which means no drop across it.

 

[TsAmE]

No, the asymptote would be at 9V. Let the capacitor charge to 9V which happens at t = infinity. Now what the resistor sees is 9V on both its terminals. So there won't be any current through the resistor which means no drop across it.

Sorry but I dont really understand. Why wouldnt there be a potential difference between the terminals of the resistor? Since when the current travels from the positive rail to the capacitor, it loses some voltage when it goes through the resistor?

 

[matbob]

Hi TsAmE,

When the capacitor is fully charged, ie, when it is at 9V, there won't be any current flow (provided, the capacitor is loseless). So there won't be a pot difference across the resistor.

 

[TsAmE]

When the capacitor is fully charged, ie, when it is at 9V, there won't be any current flow (provided, the capacitor is loseless). So there won't be a pot difference across the resistor.

So a capacitor will always charge up to the positive rail voltage, despite there being any resistances in the way? And the circuit line between the end terminal of the resistor and top terminal of capacitor would = 9V?

 

[matbob]

Hi TsAmE,

Yes, but it happens only in an ideal capacitor after infinite time.The capacitor charges near to the applied voltage after few time constants (time constant = R*C ) , say six. You may put t=6RC in the equation I mentioned and you can find that Vcap=8.977V.