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Capacitance Discharge Question *Help*

Yo people I have a question to ask which has got me totally stumped.

There is a graph that shows the current in the resistor (y-axis), with time (x-axis) for part of the discharge. It shows an exponential curve, which decreases very rapidly, and then levels off. Io is worked out as being 5mA. The question that has me stumped however, is:

"Show that the shape of the graph is exponential." [2]

Time constant is known, and so is Io, and I imagine that this question has something to do with the exponential equation

I = Ioe-t/CR

But the problem is that I have no idea how to show why this is an exponential graph. Any help is good help.

Cheers
Anybody? :frown:
Reply 2
Avatar for Robob
Robob
17
Just show that from t = 0 to t = RC, that the current is now I/e. then from t = RC to t = 2RC that the current is now I/(2RC) etc. This will show it is exponential
Ahh right. Thanks a lot will do it now.
Reply 4
Avatar for Jonathan
Jonathan
12
Um, to show that it is exponential, the official way is algebraic not numerical, and easier IMO:

rearrange the equation like this

I = Io e-t/RC

ln I = ln (Io e-t/RC)

ln I = ln Io + ln e-t/RC

ln I = ln Io + -t/RC

ln I = -t/RC + ln Io

ln I = (-1/RC)*t + ln Io

which is now in the form
y = m x + c

where
y = ln I
m = -1/RC
x = t
c = ln Io

Now plot y against x, that is, ln I against t, to show this new relationship.

If the original relationship was really exponential, then this new relationship will be in form y=mx+c and hence a straight line.


This is the official way of doing it - there may be other ways that do it well though.
Reply 5
Avatar for Robob
Robob
17
silverjonny
Um, to show that it is exponential, the official way is algebraic not numerical, and easier IMO:

rearrange the equation like this

I = Io e-t/RC

ln I = ln (Io e-t/RC)

ln I = ln Io + ln e-t/RC

ln I = ln Io + -t/RC

ln I = -t/RC + ln Io

ln I = (-1/RC)*t + ln Io

which is now in the form
y = m x + c

where
y = ln I
m = -1/RC
x = t
c = ln Io

Now plot y against x, that is, ln I against t, to show this new relationship.

If the original relationship was really exponential, then this new relationship will be in form y=mx+c and hence a straight line.


This is the official way of doing it - there may be other ways that do it well though.


Yer i no that, but judging by her qu she only has the graph and it is only a 2mark question so i doubt she'll have 2 plot anotha graph, which is why i posted a simpler way
Reply 6
Avatar for Jonathan
Jonathan
12
In this situation then yes, good way, but in any other circumstances, that's the way to do it.
Reply 7
Avatar for Robob
Robob
17
silverjonny
In this situation then yes, good way, but in any other circumstances, that's the way to do it.


Yup i wud do it the same way as you suggested in other situations
Dude I'm not a her.. I'm a he.:|

Haha thanks for all the help anyways people. Much appreciated.

--------------

Dude I'm not a her.. I'm a he.:|

Haha thanks for all the help anyways people. Much appreciated.

Just goes to show how important it is to actually put down the number of marks available for the question
Reply 9
Avatar for Robob
Robob
17
angelfire1987
Dude I'm not a her.. I'm a he.:|

Haha thanks for all the help anyways people. Much appreciated.

Just goes to show how important it is to actually put down the number of marks available for the question


lol, soz jus saw the angel in the name and ...

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