Question

the charge on a capacitor in a circuit containing C,a resistance R,and a source of voltage E is given by q=CE(1-e^(-t/RC)).show that this satisfies the equation Rdq/dt+q/C=E

Answer 1

Hint:
If I compute the first derivative of the function \(\Large q(t)\), then I get:

\[\Large \frac{{dq}}{{dt}} = EC \cdot \frac{1}{{RC}}\exp \left( { - \frac{t}{{RC}}} \right)\]
please substitute that first derivative, together the same function \(\Large q(t)\) into your differential equation, you should get an identity

Answer 2

Is that exp the same as e?

Answer 3

e as in the opposite of a natural log?

Answer 4

yes

Answer 5

Thank you :)

Answer 6

no, thank "the maestro"

Answer 7

Thank both of you lol. You helped too :) I had no idea what exp meant. Shows how much I know about claculus :/

Answer 8

To find that derivative, did you first distribute the CE and then use d/dx e^x=e^x * x^i?

Answer 9

@IrishBoy123 @Michele_Laino

Answer 10

hey ruby, i think we started with:
\[\Large q(t)=CE(1-e^{\frac{-t}{RC}})\]

and we have to show that \(\Large R\dot q+\dfrac{q}{C}=E\) is true

Answer 11

Yes I'm just not sure how we got to

Answer 12

I'm sorry if I seem incompetent lol.

Answer 13

ruby, are you online now???

Answer 14

Yes

Answer 15

cool

did we get the first bit right: \(\large q(t)=CE(1-e^{\frac{-t}{RC}})\)

??

Answer 16

Yes, and you're correct in stating that we need to prove that \[q=CE(1-e ^{-t/RC}) \] satisfies the second equation

Answer 17

so we need : \(\large \dot q(t)=CE \dfrac{d}{dt}\left\{(1-e^{\frac{-t}{RC}})\right\}\)

Answer 18

Would you distribute the CE and then use the chain rule theorem?

Answer 19

the CE is a constant. keep it outside.

Answer 20

Alright, so just use chain rule on e^-t/RC?

Answer 21

\(\large \dfrac{d}{dt}(1-e^{\frac{-t}{RC}}) = \dfrac{d}{dt}(1)-\dfrac{d}{dt}e^{\frac{-t}{RC}}) \)

Answer 22

\(\large \dfrac{d}{dt}(e^{-\frac{t}{RC}}) = -\frac{1}{RC}~e^{\frac{-t}{RC}}\)

Answer 23

ruby

keep asking, if the answers are not helpful.

:p

Answer 24

And then I substitute that into the second equation? And try to get E=E?