Integration mistake?

Scott92

Scott92

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Jun 11, 2022
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Have the below question but my working is inserted as picture but im getting a negative number and pretty sure ive made a mistake?
Assuming zero charge on the capacitor at t = 0, the current flowing may be quantified as.
i= e/r e -t/RC
Integrate this equation with respect to t and hence find the charge stored in the capacitor 0.5 seconds after the switch is closed.

1665593060774.png 1665593626281.png
 
blamocur said:
Do you mean (e/R) e^(-t/(RC)) ?
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Scott92 said:
Yes mate
Click to expand...
Actually (E/R)e^(-t/(RC)) ? I.e., 'E' and 'e' are not the same. It would save time to everybody if you were more careful with parenthesis and lower vs. upper case variables.

As for your computations, how did you get [imath]-EC\left[e^{-0.5} - e^0\right] = -EC[-0.393... - 484][/imath] ? I.e., why -0.393... ? And why 484 ?
 
Yeah I try type out but the maths inputs aren't great without proper fraction function.
The 0.5 and 0 are the time difference in the switch
The 484 is the due to R=10 C =22 see equation after since we know in the picture and the - 384 was what I got from calculation after we now write on the pic
 
Scott92 said:
Yeah I try type out but the maths inputs aren't great without proper fraction function.
The 0.5 and 0 are the time difference in the switch
The 484 is the due to R=10 C =22 see equation after since we know in the picture and the - 384 was what I got from calculation after we now write on the pic
Click to expand...

My point is that this last equation is wrong because [imath]e^{-0.5} - e^0 \neq -0.39346... - 484[/imath]!

Also: a) you have another error in the same line, and b) check your computation of [imath]RC[/imath] earlier in the text.
 
Last edited:
I still get on my calculator e^-05 - e^0 as -0.393 so what am I doing wrong?
 
Scott92 said:
I still get on my calculator e^-05 - e^0 as -0.393 so what am I doing wrong?
Click to expand...
I'm sorry but your work has a lot of mistakes.
1) As blamocur says, Mathematics is "case sensitive." e and E are not the same, neither are r and R, neither are c and C, neither are t and T, and neither are q and Q.

2) When you wrote your integral you included a term [imath]\int_0^{0.5}[/imath] all by itself. This is meaningless.

3) Where did the D come from? Was that supposed to be R?

4) You put the wrong numbers in when you calculated RC and you dropped the exponent of 10.

5) When you put the numbers into the result of your integration you didn't divide the time by RC.

6) As blamocur says, that -484 comes out of nowhere. It isn't a term in the equation!

7) You dropped a negative sign. Even if your 484 term was correct you have -EC(-484.39), which is positive.

8) [imath]\mu F = 10^{-6}[/imath] F, not [imath]10^{-4}[/imath].

9) And finally, your calculation is wrong. You don't have a [imath]\times 10^{-4}[/imath] at the end of that.

Look, we're here to help you find your mistakes. I hardly expect perfection. But did you even check your work before coming to us?

-Dan
 
Some of the capitals and none capitals is a PC error my bad ?
I knew I'd gone wrong from having a neg number as my final answer
Will start again as clearly more mistakes than I thought
Thanks for the help and patience
 
Hi sorry took my time to try again still not sure im correct here but think got rid most of the mistakes. Thanks 1672854838019.png
 
You have the correct equation for Q. Yay!

Then everything goes to Hades.

1. You said [imath]R = 10 k \Omega[/imath] and [imath]C = 22 \mu F[/imath]. So how can [imath]RC = 200 \Omega F[/imath]? Check your calculation.

2. Then you had -t/220 in the exponent, which mysteriously turning into just -t in the next step. What happened to the 220?

3. What did you do in the last line? C is not [imath]220 \times 10^{-6}[/imath] F.

4. Last line again. I don't know what you did to your powers of 10 but it's not [imath]\times 10^6[/imath].

5 Last line again. C'mon. How can [imath]0.0006925 \times 10^6[/imath] be the same as [imath]6.925 \times 10^{-10}[/imath]?

Please start checking your work more carefully. I'm not sure what class you are in, but if you are dealing with integrals you should be able to use your calculator better than this. This is just sloppy work.

-Dan
 
topsquark said:
You have the correct equation for Q. Yay!

Then everything goes to Hades.

1. You said [imath]R = 10 k \Omega[/imath] and [imath]C = 22 \mu F[/imath]. So how can [imath]RC = 200 \Omega F[/imath]? Check your calculation.

2. Then you had -t/220 in the exponent, which mysteriously turning into just -t in the next step. What happened to the 220?

3. What did you do in the last line? C is not [imath]220 \times 10^{-6}[/imath] F.

4. Last line again. I don't know what you did to your powers of 10 but it's not [imath]\times 10^6[/imath].

5 Last line again. C'mon. How can [imath]0.0006925 \times 10^6[/imath] be the same as [imath]6.925 \times 10^{-10}[/imath]?

Please start checking your work more carefully. I'm not sure what class you are in, but if you are dealing with integrals you should be able to use your calculator better than this. This is just sloppy work.

-Dan
Click to expand...
Im not in any class trying to do an online course but finding it difficult as work book isnt the best,
I thought the input for 10kΩ was 10^6 but is it 10^4 or am i confusing things more here
2.Yeah didnt notice that
3. looks like i did the RC not just c
Last line should be 2 different equations, shouldnt be an = symbol there should be spaced and separated.
Maths isnt a strong point for me but im trying..?‍♂️
 
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