A2: Exponential Discharge Equation

V at a certain discharge period = (V initial) x (e to the power of -t/RC)

When capacitor is discharging.

I'm completely happy with this equation, but I've got a problem re-arranging it.

I'm doing a homework on it, and the information given is everything apart from the 't' in the equation.

It's a question where it wants to know how much longer the capacitor will take to discharge once it reaches a certain value of V.

In the question I've got, it asks how long the security light will working once the V reaches 2.5V.

Basically, how do you re-arrange to equation to read

-t = ....

How do you work it out? :frown:

Reply 1

sionww

You need to rearrange the formula by taking natural logarithms. :frown:

I don't do A-level Mathematics. How can they expect me to be able to do this? :frown:

V = V0 e –t/RC

Þ V / V0 = e –t/RC

Þ loge V - loge Vo = -t/RC [When you divide two numbers, you subtract their logs]

Þ 0.693 – 2.485 = - t/10

Þ -t/10 = -1.792

Þ +t/10 = +1.792

Þ t = 1.792 × 10 = 17.9 s

Reply 2

Ribbons

loge V - loge Vo = -t/RC

there! uve got it!.....see, to start off with, take e to one side, so u have:

V/V(initial) = e^-t/RC
Ln both sides, (ln is the same as log_e )

ln V/V(initial) = -t/RC

RC x [ln V- ln V(initial)]= -t

i believe thats wat u wanted?

Reply 3

Ribbons

and i havent really dont electricity in physics yet.......so im not really sure how ud work it out....but, ya , wat uve done, looks okay to me!

Reply 4

sionww

PSdilemma

My second reply was just something I found on a website. It does what I want to do, but I don't understand the Maths behind it.

When I was doing my homework, I got to:

V/Vinitial = e-t/RC

I was stuck then.

I don't do A-level Maths, so I haven't done logs.

Reply 5

sionww

I can answer the question now, but what annoys me is that I don't understand the Maths behind it with the logs.

Reply 6

Ribbons

i see...
well basically, in maths, for logs, there are certain rules u have to accept......

1) log_e e^x is the same as x.....
and log_ex can also be written as lnx....i use ln because it just looks a little bit less horrible.

2) ln x/y can also be written as lnx-lny
(or log x/y=logx-logy)

so... ur equation is

V=V_initial x e^-t/RC

take the Vs to one side, so u have,

V/V_initial=e^-t/RC

now log_e both sides, or Ln, same thing:

ln [V/V_initial]=lne^-t/RC

from rule one, the e goes, and ur left with e^-t/RC

so u have

ln V/Vin= -t/RC

u can leave it at that, or say

lnv-lnV(initial)= -t/RC

RC(lnv-lnV_initial)=-t

hope i made SOME sense! :biggrin:

Reply 7

Ribbons

I can answer the question now, but what annoys me is that I don't understand the Maths behind it with the logs.

yes, it IS unfair on the physics students who dont do maths.....

Reply 8

sionww

PSdilemma

yes, it IS unfair on the physics students who dont do maths.....

You made perfect sense.

I just don't understand how to 'ln' the numbers. What do I press on my calculator?

I don't understand how to loge the numbers.

Reply 9

pal_sch

The button that says ln. Often second function of the log button or the e button.

It works by taking the power e has to be raised by to get the result after it. For example;

ln e^5 = 5
ln e^44 = 44
ln 1 = 0 (e^0 = 1)

Reply 10

Ribbons

sionww

You made perfect sense.

I just don't understand how to 'ln' the numbers. What do I press on my calculator?

I don't understand how to loge the numbers.

there should be a "ln" button on ur calculator...somewhere near the "log" button......and since log_e is the same as ln....u can use ln for log_e

Reply 11

sionww

PSdilemma

there should be a "ln" button on ur calculator...somewhere near the "log" button......and since log_e is the same as ln....u can use ln for log_e

I found it! My calculator is old and the writing on some of the buttons have worn.

I've found an 'ln' button, and above it is the e_x which I can use with the help of shift.

Lets see if I can get an answer for this question now.

Reply 12

sionww

I found it! My calculator is old and the writing on some of the buttons have worn.

I've found an 'ln' button, and above it is the ex

which I can use with the help of shift.

Lets see if I can get an answer for this question now.

yay, I did it! My answer agreed with the answer at the back of the book!

Thank you for all your help!!!!! :smile: