Algebra Manipulation Help

[Electric1]

I am trying to solve for C (capacitance) for both of the following formulas, but I keep tripping up on the equation conversion.

VC = V - Ve(-t/RC) to:
C=-tLn/R*(V-VC/V)

I = I0 e(-t/RC) to:
C=-tLn/R*(I0-I/I0)

E is the natural log. I have values, but I know my current formulas are wrong using them, as VC does not equal the known amount using my incorrect formulas.

 

[MrAl]

I'll do the first one for you:

t=RC*ln(V/(V-Vc))

When trying to solve these equations first simplify as much as possible so that you are left with 'e' up to some power on the right side, then take the natural log of both sides.

 

[Electric1]

Thanks for the reply. But I'm trying to solve for C, not the known variable of t.
But going off that....

t=RC*ln(V/(V-Vc))

and change it to solve for C

C=t/R*ln(V/(V-Vc))

But that still doesn't satisfy the known values I have for R,V,VC, and Ve.
Maybe Im just inputing the values wrong.

Help again?

 

[Pommie]

t=RC*ln(V/(V-Vc))

C=t/(R*ln(V/(V-Vc)))

Mike.

 

[carbonzit]

Butbutbut ... all you did was put parentheses around R*ln(V ...). How does that change anything?

 

[Pommie]

Butbutbut ... all you did was put parentheses around R*ln(V ...). How does that change anything?

It makes it correct.

Edit, put it in a calculator with and without the parentheses and you will get two very different answers.

Mike.

 

[Electric1]

Ah, I see where I went wrong.
Guess Ill need some help with this from a friend or something, keep messing up this algebra.

Thanks a lot Mike, Al and Pommie

 

[carbonzit]

OK, I see. But when I saw

C=t/R*ln(V/(V-Vc))

I saw it as meaning "t over (all the rest of that stuff)", so no need for extra parens.

(That's the problem with text-based math ... we need tEx.)

 

[Pommie]

If you write it like,

[LATEX]C=\frac{T}{R*ln(V/(V-Vc))}[/LATEX]

Then parentheses aren't required.

The above was done by entering [LATEX]C=\frac{T}{R*ln(V/(V-Vc))}[/LATEX].

Mike.

 

[MrAl]

OK, I see. But when I saw

C=t/R*ln(V/(V-Vc))

I saw it as meaning "t over (all the rest of that stuff)", so no need for extra parens.

(That's the problem with text-based math ... we need tEx.)

Hi,

Parens are used to eliminate the ambiguity. For example, if i write:
K1=t/R*ln(V/(V-Vc))
and:
K2=t/R*ln(V/(V-Vc))

which one do i want to have everything after the 't' in the denominator, and which one do i want only the 'R' in the denominator?
What if one has it and one doesnt, how would we know which one has it and which one doesnt?
Following the rules of algebra, both of those only have 'R' in the denominator, so we really need parens to clear it up:

K1=t/R*ln(V/(V-Vc))
and:
K2=t/(R*ln(V/(V-Vc)))

Now we know that K1 only has 'R' while K2 has the whole part after the 't' in the denominator.

We really must use parens whenever there is a doubt or else the reader wont be able to evaluate it in the correct order. For example:
y=x^2*3/8*a/7+b/sqrt(2)/9/8/7/6/5/4/3/2
What is the true denominator there?

Another thing to watch out for is sometimes some authors still use the trig convention:
cos x
instead of the more proper:
cos(x)

Using "cos x" causes problems when we get something like:
cos x*2*pi
Is this "cos(x)*2*pi", or is it "cos(x*2)*pi", or is it "cos(x*2*pi)"? Anybodies guess
Note that if we write instead:
cos(x*2*pi)
there is no problem understanding how to evaluate this expression.
It gets even weirder than that though as with:
cos 2*pi*x+ph
which is meant to mean:
cos(2*pi*x+ph)
Note again the parens eliminate any doubt as to how to evaluate the expression.

The simple rule then is: "Use parens when there is any possibility of an alternate order of evaluation".

 

[MrAl]

Thanks for the reply. But I'm trying to solve for C, not the known variable of t.
But going off that....

t=RC*ln(V/(V-Vc))

and change it to solve for C

C=t/R*ln(V/(V-Vc))

But that still doesn't satisfy the known values I have for R,V,VC, and Ve.
Maybe Im just inputing the values wrong.

Help again?

Hi again,

When you want to solve for a given variable you try to get everything else to the other side of the equation.
For example, in the equation:
t=R*C*ln(V/(V-Vc))
lets first lump V/(V-Vc) into A:
t=R*C*ln(A)
Now if we want to solve for C, we just divide by everything else. We'll start by dividing by R first, and we get:
t/R=C*ln(A)
and we can see that we still dont have C alone yet, so we have to divide by ln(A) next, and we get:
t/(R*ln(A))=C
[note ln(A) was divided so it too goes in the denominator of the left side]
and flipping this around we get:
C=t/(R*ln(A))
and replacing A with it's original value we get:
C=t/(R*ln(V/(V-Vc)))
and note that we use parens to make sure the reader knows the correct order of evaluation here.