Making Capacitance the subject of discharge equation

[Daniel2244]

Homework Statement

I have to rearrange the equation V=lnVo -t/RC to calculate C. The gradient from graph = -1/RC

Homework Equations

The Attempt at a Solution

I got R/C=ln(Vo-V)/t
but the answer I get is wrong

 

[kuruman]

The discharge equation is V = V₀e^-t/RC from which you get -t/(RC) = ln(V/V₀). So what is RC (not R/C) ?

 

[Daniel2244]

The discharge equation is V = V₀e^-t/RC from which you get -t/(RC) = ln(V/V₀). So what is RC (not R/C) ?

Wouldn't it just be RC=ln(V/V₀)/t

 

[kuruman]

It would not be that. Please try again doing the algebra steps very carefully.

 

[Daniel2244]

It would not be that. Please try again doing the algebra steps very carefully.

I have no idea. I've never used ln before and I don't know how to rearrange using it. However, what I got this time is RC=t/ln(V/V₀)

 

[Daniel2244]

I have no idea. I've never used ln before and I don't know how to rearrange using it. However, what I got this time is RC=t/ln(V/V₀)

but if the gradient of the graph gives -t/RC would it just be C=t/R (but it gives me the wrong answer)

 

[kuruman]

but if the gradient of the graph gives -t/RC would it just be C=t/R (but it gives me the wrong answer)

The gradient of which graph? What did you plot against what?

 

[Daniel2244]

The gradient of which graph? What did you plot against what?

Ln(V/V₀) against time
gives me a gradient -1/RC

 

[kuruman]

No. The gradient is -1/(RC). There is no t. What gradient do you get from your plot? How will you (did you) use it to find the time constant RC?

 

[Daniel2244]

No. The gradient is -1/(RC). There is no t. What gradient do you get from your plot? How will you (did you) use it to find the time constant RC?

Yes, sorry, I was meant to put -1/RC. The gradient I got was 0.015 so would I then do e^0.015 which gives me 1.01 which is 1% dicharge at 10 seconds. but how does this help me find the capacitance?

 

[kuruman]

The gradient I got was 0.015 so would I then do e^0.015 ...

No, you would not do that. You missed the point of the exercise. You plotted ln(/V₀) against time. That's good. The equation is $$\ln(V/V_0)=-\frac{t}{RC}$$ and the plot gives you a straight line of gradient 0.015. OK so far? Now define $\ln(V/V_0) = y$ and $\frac{1}{RC}=m$. The equation becomes $y=-mx.$ This is the equation of a straight line of gradient m = 0.015 which, by definition, is also equal to $1/(RC)$. Given the previous sentence, what is $RC?$

 

[Daniel2244]

No, you would not do that. You missed the point of the exercise. You plotted ln(/V₀) against time. That's good. The equation is $$\ln(V/V_0)=-\frac{t}{RC}$$ and the plot gives you a straight line of gradient 0.015. OK so far? Now define $\ln(V/V_0) = y$ and $\frac{1}{RC}=m$. The equation becomes $y=-mx.$ This is the equation of a straight line of gradient m = 0.015 which, by definition, is also equal to $1/(RC)$. Given the previous sentence, what is $RC?$

Ok, so would I do 1/0.015=66.6 then 66.6/99400=6.7069E^-4 (99400 being resistance) then after converting to microF i get 670.69microF.

 

[kuruman]

Yes, except that I would not carry that many significant figures.

 

[Daniel2244]

Yes, except that I would not carry that many significant figures.

Thanks man, your help was much appreciated!