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Intergrate

hi, i've been working on the following question below and would like any help whatsoever….for the circuit shown below, 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.

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sounds like isaac physics?
Original post by Palmer17
hi, i've been working on the following question below and would like any help whatsoever….for the circuit shown below, 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.

Do you know how to integrate an exponential? If so, what are you having problems with?
Reply 3
No not intergrated exponential and wasn’t sure how to start my working out
Original post by Palmer17
No not intergrated exponential and wasn’t sure how to start my working out


Do you know you to differentiate an expoential, can you "reverse" it?
Reply 5
I have done some work on this but I don’t think it’s correct
Original post by Palmer17
I have done some work on this but I don’t think it’s correct


Post what youve tried. The question asks you to integrate the current to get the charge, so you are expected to know how to integrate an exponential
(edited 1 year ago)
Reply 7
It won’t let me upload my image
Reply 8
I can differentiate but never tried “reverse it”. I’m just working from a work book which is the only info I have. Hence I’m stuck
Reply 9
(edited 1 year ago)
Original post by Palmer17
I can differentiate but never tried “reverse it”. I’m just working from a work book which is the only info I have. Hence I’m stuck


A quick google gave many hits so things like
https://www.youtube.com/watch?v=MTL1yTVtNMQ&ab_channel=ExamSolutions
In your case, the b in the exponent is zero and the a=...
Reply 11
Ok so where would you start with my equation? I over think everything and tie myself in knots



Original post by mqb2766
A quick google gave many hits so things like
https://www.youtube.com/watch?v=MTL1yTVtNMQ&ab_channel=ExamSolutions
In your case, the b in the exponent is zero and the a=...
Original post by Palmer17
Ok so where would you start with my equation? I over think everything and tie myself in knots


From the video you should be able to integrate things like
e^(-t)
E e^(-t)
(E/R) e^(-t)
(E/R) e^(-t/(RC))
so try and do simpler ones first and check by differentiating your answer and make sure you get the original function. Post what youve done for these, or similar ones that you choose.
Reply 13
e^(-t) dt = e^(-t) + C



Original post by mqb2766
From the video you should be able to integrate things like
e^(-t)
E e^(-t)
(E/R) e^(-t)
(E/R) e^(-t/(RC))
so try and do simpler ones first and check by differentiating your answer and make sure you get the original function. Post what youve done for these, or similar ones that you choose.
Original post by Palmer17
e^(-t) dt = e^(-t) + C


No. If you differentiate the right hand side you do not get e^(-t).
Reply 15
-e^(-t)+C
Original post by Palmer17
-e^(-t)+C


Agreed as when you differentiate
-e^(-t)
You differentiate the exponent -t to get -1 and that multiplies the original exponential so
-1*(-e^(-t)) = e^(-t)
which is what you started with. So keep going with the others (or your own similar ones).
(edited 1 year ago)
Reply 17
E(-e^-t)+C
Original post by Palmer17
E(-e^-t)+C


Agreed, though youd write it normally as
-Ee^(-t) + C

So the third one should be similar?
Reply 19
-(E/R)e^(-t)+C

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