How Do You Invert the Function Q(t) in a Camera Flash Capacitor Model?

Comael321
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Homework Statement



When a camera flashes, the batteries begin recharging the flash capacitor which stores the charge Q according to the function Q(t) = Q* (1-e-t/a) where t is the elapsed time in seconds since the camera flash and Q* and a are non-zero

(a) What does Q* represent?
(b) Find the inverse of this function
(c) How long does it take to recharge the capacitor to 90% if a = 5?

Homework Equations


The Attempt at a Solution



I said that Q* was the constant of proportionality, but I'm not sure how right that is.

My attempt at the inverse came out as t*a/ln(Q*) = Q

I'm wondering if my answer for (a) is at all correct, and where to go from here, I have no idea how to attempt (c) and I think my answer to (b) is wrong.
 
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Comael321 said:

Homework Statement



When a camera flashes, the batteries begin recharging the flash capacitor which stores the charge Q according to the function Q(t) = Q* (1-e-t/a) where t is the elapsed time in seconds since the camera flash and Q* and a are non-zero

(a) What does Q* represent?
(b) Find the inverse of this function
(c) How long does it take to recharge the capacitor to 90% if a = 5?

Homework Equations





The Attempt at a Solution



I said that Q* was the constant of proportionality, but I'm not sure how right that is.

My attempt at the inverse came out as t*a/ln(Q*) = Q

I'm wondering if my answer for (a) is at all correct, and where to go from here, I have no idea how to attempt (c) and I think my answer to (b) is wrong.

Try plotting Q(t) for a couple of different Q* values (but for the same a---just pick some value). What do you get?

RGV
 
We're not allowed to use a graphing calculator in this course, otherwise I would've played around with the graphs a bit, but I don't want to become dependent on my calculator to solve these, cause it'll bit me in the arse in the exam
 
a. I believe that Q* should be some type of initial constant, and a is proportionality constant. I'm not entirely sure of this, but that's what I'm thinking just off of the top of my head. If it helps you, I'm thinking of this as a differential equation; I can explain more to you if you would like.

b. The inverse should be:
Q = Q(t)

t = Q*(1-e^{\frac{-Q}{a}})

1-\frac{t}{Q*}=e^{\frac{-Q}{a}}

ln(1-\frac{t}{Q*})=-Q*a

Q=\frac{-ln(1-\frac{t}{Q})}{a}

c. You plug in .9 for the value of Q(t), and 5 as the value for a. The value for Q* is 1 I believe, but I am not absolutely certain on that one. With these values, solve the equation for t.
 
Comael321 said:
We're not allowed to use a graphing calculator in this course, otherwise I would've played around with the graphs a bit, but I don't want to become dependent on my calculator to solve these, cause it'll bit me in the arse in the exam

You are certainly allowed to draw graphs on pieces of scrap paper and throw them away later, after you have used them to gain insight into some issues. The point is that you should not need to draw a graph at all, but since you are missing a key insight, drawing a graph would be helpful to you---more helpful in the long run than being told by someone what is the correct answer.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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