How Does Charge in a Capacitor Satisfy the Given Differential Equation?

[alphysicist]

So:

e^{-\frac{t}{RC} + \frac{T}{RC}}} = \frac{Q}{Q_T}

is the same as:

e^{-\frac{t-T}{RC}} = \frac{Q}{Q_T}

and

Q_Te^{-\frac{t-T}{RC}} = Q

Q_T = VC(1-e^{-\frac{T}{RC}})

so:

VC(1-e^{-\frac{T}{RC}})e^{-\frac{t-T}{RC}} = Q

Which is the final product for t>T

TFM

That looks right to me.

 

[TFM]

Just to confirm, for t < T, the equation is:

Q = VC(1 - e^{-\frac{t}{RC}})

?

TFM

 

[alphysicist]

Just to confirm, for t < T, the equation is:

Q = VC(1 - e^{-\frac{t}{RC}})

?

TFM

That looks right to me. If you look in your book, you should find formulas that match these and you can verify the parts.

 

[TFM]

Thanks!

Final part now, which I have a feeling is practically the same as the last two parts.

Suppose instead that V = V_0sin\omega t, where V_0 and ω are constants. Find the general solution for Q(t). (You may assume that \int e^{t/\alpha} sin\omega t dt = \alpha e^{t/\alpha}\frac{(sin \omega t - \alpha \omega cos \omega t)}{1 + \alpha ^2\omega^2} + Constant )

So do I use the formula

V = R\frac{dQ}{dt} + \frac{Q}{C}

and replace the V, so that

V_0sin\omega t = R\frac{dQ}{dt} + \frac{Q}{C}

?

TFM

 

[TFM]

Assuming that this is the right equation,

CV_0sin\omega t = CR\frac{dQ}{dt} + Q

CV_0sin\omega t - Q = CR\frac{dQ}{dt}

CV_0sin\omega t dt - Q dt = CR dQ

Does this look correct?

Edit: I think I should actually of done:

CV_0sin\omega t - Q = CR\frac{dQ}{dt}

CV_0sin\omega t = CR + Q\frac{dQ}{dt}

Does this look better?

?

TFM

 

[alphysicist]

Thanks!

Final part now, which I have a feeling is practically the same as the last two parts.

Suppose instead that V = V_0sin\omega t, where V_0 and ω are constants. Find the general solution for Q(t). (You may assume that \int e^{t/\alpha} sin\omega t dt = \alpha e^{t/\alpha}\frac{(sin \omega t - \alpha \omega cos \omega t)}{1 + \alpha ^2\omega^2} + Constant )

So do I use the formula

V = R\frac{dQ}{dt} + \frac{Q}{C}

and replace the V, so that

V_0sin\omega t = R\frac{dQ}{dt} + \frac{Q}{C}

This equation is correct; I think the path you were going down in your next post after this was a wrong turn. (When the voltage was constant, we could take it to the charge integral; here that cannot be done.)

If you divide by R, you get and shuffle the terms, you get:

<br /> \frac{dQ}{dt} + \frac{Q}{RC} = \frac{V_0}{R} \sin\omega t<br />


How would you solve this differential equation? Note that there is a homongenous and non-homogenous part to the solution.

 

[TFM]

So:

V = R\frac{dQ}{dt} + \frac{Q}{C}

And Divide by R:

\frac{V}{R} = \frac{dQ}{dt} + \frac{Q}{RC}

V = V_0sin\omega t

So:

\frac{V_0sin\omega t}{R} = \frac{dQ}{dt} + \frac{Q}{RC}

\frac{V_0sin\omega t}{R} - \frac{Q}{RC} = \frac{dQ}{dt}

\frac{V_0sin\omega t}{R} - \frac{1}{RC} = \frac{1}{Q}\frac{dQ}{dt}

\frac{V_0sin\omega t}{R}dt - \frac{1}{RC}dt = \frac{1}{Q}dQ

Does this look right?

TFM

 

[alphysicist]

So:

V = R\frac{dQ}{dt} + \frac{Q}{C}

And Divide by R:

\frac{V}{R} = \frac{dQ}{dt} + \frac{Q}{RC}

V = V_0sin\omega t

So:

\frac{V_0sin\omega t}{R} = \frac{dQ}{dt} + \frac{Q}{RC}

\frac{V_0sin\omega t}{R} - \frac{Q}{RC} = \frac{dQ}{dt}

\frac{V_0sin\omega t}{R} - \frac{1}{RC} = \frac{1}{Q}\frac{dQ}{dt}

This line does not follow from the previous one; if you divide by Q, the first term will have Q in the denominator.

But I was saying that this problem is different from the previous one; the procedure you followed for the previous problem is not the procedure to solve this one.

In my last post, I had that the diff. eq. was:

\frac{dQ}{dt} + \frac{Q}{RC} = \frac{V_0}{R} \sin\omega t

This has the form:

<br /> \frac{dx}{dt} + c x = f(t)<br />

with c being a positive constant here. When you have a diff. eq. in this form, how can you solve it? (Have you studied how to solve differential equations yet?) It has to do with the fact that the solution will have a homogenous and inhomogenous part.

 

[TFM]

So:

\frac{V_0sin\omega t}{R} - \frac{Q}{RC} = \frac{dQ}{dt}

\frac{dQ}{dt} + \frac{Q}{RC} = \frac{V_0sin\omega t}{R}

It seems to be a mixture of Linear with Constant Coefficients

x&#039; + ax = b --&gt; x = Ce^{-at} + \frac{b}{a}

and linear:

x&#039; + p(t)x = q(t) --&gt; e^Fx = \int e^F q(t)dt + C, F = \int p(t) dt

?

TFM

 

[alphysicist]

So:

\frac{V_0sin\omega t}{R} - \frac{Q}{RC} = \frac{dQ}{dt}

\frac{dQ}{dt} + \frac{Q}{RC} = \frac{V_0sin\omega t}{R}

It seems to be a mixture of Linear with Constant Coefficients

x&#039; + ax = b --&gt; x = Ce^{-at} + \frac{b}{a}

and linear:

x&#039; + p(t)x = q(t) --&gt; e^Fx = \int e^F q(t)dt + C, F = \int p(t) dt

?

TFM

Looks good; what do you get when you apply that last one to the problem?

 

[TFM]

Well,

p(t) = \frac{1}{RC}

q(t) = \frac{V_0}{R} sin\omega t

So:

F = \int \frac{1}{RC}dt

F = \frac{t}{RC}

so:

e^{\frac{t}{RC}}x = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C

TFM

 

[alphysicist]

In the last line you want a Q instead of an x.

 

[TFM]

Oops...

e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C

So do I integrate the right side now, with what was given in the question?

TFM

 

[alphysicist]

Oops...

e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C

So do I integrate the right side now, with what was given in the question?

TFM

Yes.

 

[TFM]

So:

e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C

e^{\frac{t}{RC}}Q = \frac{V_0}{R}\int e^{\frac{t}{RC}} sin\omega t dt + C

Using the replacement formula:

\int e^{t/\alpha} sin\omega t dt = \alpha e^{t/\alpha}\frac{(sin \omega t - \alpha \omega cos \omega t)}{1 + \alpha ^2\omega^2} + Constant

With \alpha = RC

Gives:

e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}sin\omega t-RC\omega cos\omega t}{1+(RC)^2\omega^2} + Constant

TFM

 

[alphysicist]

So:

e^{\frac{t}{RC}}Q = \int e^{\frac{t}{RC}}\frac{V_0}{R} sin\omega t dt + C

e^{\frac{t}{RC}}Q = \frac{V_0}{R}\int e^{\frac{t}{RC}} sin\omega t dt + C

Using the replacement formula:

\int e^{t/\alpha} sin\omega t dt = \alpha e^{t/\alpha}\frac{(sin \omega t - \alpha \omega cos \omega t)}{1 + \alpha ^2\omega^2} + Constant

With \alpha = RC

Gives:

e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}sin\omega t-RC\omega cos\omega t}{1+(RC)^2\omega^2} + Constant

TFM

Looks good so far (except keep the parenthesis that are in the formula around the (\sin\omega t-\alpha\omega \cos\omega t)).

 

[TFM]

So:

e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant

So should I rearrange to get q:

q = \frac{\frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant}{e^{t/RC}}

??

TFM

 

[alphysicist]

So:

e^{t/RC}q = \frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant

So should I rearrange to get q:

q = \frac{\frac{V_0}{R} \frac{RCe^{t/RC}(sin\omega t-RC\omega cos\omega t)}{1+(RC)^2\omega^2} + Constant}{e^{t/RC}}

??

TFM

Okay; now I would suggest you write the right hand side as three terms (one with a sine, one with a cosine, and one with the constant). Once you have that, you can determine the constant.