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

[alphysicist]

So the first one, where t < T

$$Q = V_0C(1 - e^{-\frac{t}{RC}})$$

For the second, the Voltage is 0, so are we using the voltage stored in the Capacitor, or should the equation just be 0?

TFM

I cannot view the attachment yet, but if I'm reading it correctly, this will be essentially a discharging circuit. So since the capacitor already has charge on it, it will have a voltage.

As long as the capacitor has a charge, it will have a potential difference across it, and as long as the resistor has a current, it will have a potential difference across it. So the terms in the differential equation (which are just the volts across each part of the circuit) are not all zero until I=0 and Q=0.

So the first change for this differential equation (relative to the original one) is that V=0. There is one more; what would it be, and so what would the final diff. eq. be?

 

[TFM]

So will the equation be:

$$V_0 = R\frac{dQ}{dt} + V_{cap}$$

?

TFM

 

[alphysicist]

So will the equation be:

$$V_0 = R\frac{dQ}{dt} + V_{cap}$$

?

TFM

You know that V₀=0 here. What formula gives you the voltage across the capacitor? Also, be careful with your signs.

 

[TFM]

So

$$0 = R\frac{dQ}{dt} + V_{cap}$$

$$V_{cap} = \frac{Q}{C}$$

So:

$$-\frac{Q}{C} = R\frac{dQ}{dt} + V_{cap}$$

?

TFM

 

[alphysicist]

So

$$0 = R\frac{dQ}{dt} + V_{cap}$$

$$V_{cap} = \frac{Q}{C}$$

So:

$$-\frac{Q}{C} = R\frac{dQ}{dt} + V_{cap}$$

(You meant to eliminate the V_cap in this last equation right, since you replaced is with Q/C?)

The signs are not right. In first equation you have:

$$0 = R \frac{dQ}{dt} + V_{\rm cap}$$

This comes from Kirchoff's loop rule. How did you choose the signs of both of those to be positive?

 

[TFM]

I see - in the first part, we had a Voltage Source, so

$$V = R \frac{dQ}{dt} + V_{\rm cap}$$

In this part, we just have the capacitor, no independent Voltage Source, so

$$0 = R \frac{dQ}{dt} - V_{\rm cap}$$

$$V_{\rm cap} = R \frac{dQ}{dt}$$

and since

$$V_{cap} = \frac{Q}{C}$$

$$\frac{Q}{C} = R \frac{dQ}{dt}$$

?

TFM

 

[alphysicist]

I see - in the first part, we had a Voltage Source, so

$$V = R \frac{dQ}{dt} + V_{\rm cap}$$

In this part, we just have the capacitor, no independent Voltage Source, so

$$0 = R \frac{dQ}{dt} - V_{\rm cap}$$

$$V_{\rm cap} = R \frac{dQ}{dt}$$

and since

$$V_{cap} = \frac{Q}{C}$$

$$\frac{Q}{C} = R \frac{dQ}{dt}$$

?

TFM

TFM,

Sorry, I should have corrected one thing about the sign of the diff. eq. When doing Kirchoff's I was thinking of the terms for using the regular Ohm's law version (before plugging in the I=dQ/dt). In that case you would get:

0 = I R - Q C

So one term has a positive sign, and the other is negative--this is what Kirchoff's gives you.

Now the question is what do we put in for I? In this case, the current is related to the decrease in charge of the capacitor (it is discharging, so as time goes on the charge decreases). So we use:

$$I = - \frac{dQ}{dt}$$

Plugging this in leads to a minus sign in your last equation:

$$- \frac{Q}{C} = R \frac{dQ}{dt}$$

(I was still thinking in terms of the pure Kirchoff's relationships, not yet thinking about putting both terms in terms of Q. Sorry!)

With this last equation, you can separate variables, choose limits, and integrate to get the result for t > T.

 

[TFM]

So:


$$- \frac{Q}{C} = R \frac{dQ}{dt}$$

Multiply out the C:

$$-Q = RC \frac{dQ}{dt}$$

Divide by Q:

$$-1 = \frac{RC}{Q} \frac{dQ}{dt}$$

split the integral

$$-dt = \frac{RC}{Q} dQ$$

remove the RC from one side

$$- \frac{1}{RC} dt = \frac{1}{Q} dQ$$

set limits:

$$\int^t_T -\frac{1}{RC} dt = \int^Q_0 \frac{1}{Q} dQ$$

Does this look okay?

TFM

 

[alphysicist]

So:


$$- \frac{Q}{C} = R \frac{dQ}{dt}$$

Multiply out the C:

$$-Q = RC \frac{dQ}{dt}$$

Divide by Q:

$$-1 = \frac{RC}{Q} \frac{dQ}{dt}$$

split the integral

$$-dt = \frac{RC}{Q} dQ$$

remove the RC from one side

$$- \frac{1}{RC} dt = \frac{1}{Q} dQ$$

set limits:

$$\int^t_T -\frac{1}{RC} dt = \int^Q_0 \frac{1}{Q} dQ$$

Does this look okay?

TFM

Everything looks good except that the initial charge is not zero. Since the time integral starts at t=T, the charge integral must start with the charge that is on the capacitor at that time.

 

[TFM]

Would it start with charge Q, and go down to zero, so that:

$$\int^t_T -\frac{1}{RC} dt = \int^0_Q \frac{1}{Q} dQ$$

which would integrate to:

$$[-\frac{t}{RC}]^t_T = [lnQ]^0_Q$$

$$-\frac{t}{RC} + \frac{T}{RC} = -lnQ$$

?

TFM

 

[alphysicist]

Would it start with charge Q, and go down to zero, so that:

$$\int^t_T -\frac{1}{RC} dt = \int^0_Q \frac{1}{Q} dQ$$

No; if we wait a long enough time, the charge will go to zero. But we want to find an expression that will give us the charge at any time after t=T, and so you would not want to set the charge=0.


So call the lower limit Q_T or Q_i or something; that is the charge on the capacitor at time T, which you already know (you've already done the part that gives you the charge at any time up to T).

The final charge is Q which is the charge at some time t.

 

[TFM]

So:

$$\int^t_T -\frac{1}{RC} dt = \int^{Q_t}_Q \frac{1}{Q} dQ$$

$$[-\frac{t}{RC}]^t_T = [lnQ]^{Q_t}_Q$$

so:

$$-\frac{t}{RC} + \frac{T}{RC} = lnQ_T-lnQ$$

And this is for:

t > T

?

TFM

 

[alphysicist]

So:

$$\int^t_T -\frac{1}{RC} dt = \int^{Q_t}_Q \frac{1}{Q} dQ$$

Almost; Q_T is the lower limit.

Then integrate the same as before, and solve for Q. After that you can also replace Q_T with its real value.

 

[TFM]

OKay, so

$$\int^t_T -\frac{1}{RC} dt = \int_{Q_t}^Q \frac{1}{Q} dQ$$

$$\int^t_T -\frac{1}{RC} dt = \int_{Q_t}^Q \frac{1}{Q} dQ$$

This would give:

$$-\frac{t}{RC} + \frac{T}{RC} = lnQ - lnQ_T$$

Also, you say to replace Q_T with its true value, would that be It?

?

TFM

 

[alphysicist]

OKay, so

$$\int^t_T -\frac{1}{RC} dt = \int_{Q_t}^Q \frac{1}{Q} dQ$$

$$\int^t_T -\frac{1}{RC} dt = \int_{Q_t}^Q \frac{1}{Q} dQ$$

This would give:

$$-\frac{t}{RC} + \frac{T}{RC} = lnQ - lnQ_T$$

Also, you say to replace Q_T with its true value, would that be It?

?

TFM

I think you need to solve for Q, the same way as you did in the previous part: combine ln functions, etc.

Once you have Q by itself, you can then plug in the value for Q_T, which is the charge on the capacitor at time T. How can you determine what Q_T should be?

 

[TFM]

So:

$$-\frac{t}{RC} + \frac{T}{RC} = lnQ - lnQ_T$$

Goes to:

$$-\frac{t}{RC} + \frac{T}{RC} = ln\left(\frac{Q}{Q_T}\right)$$

Take exponentials:

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

$$Q_T\left(-frac{t}{RC} + e^{\frac{T}{RC}}\right) = Q$$

So now I have to find Q_T

Would it be:

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

?

TFM

 

[alphysicist]

So:

$$-\frac{t}{RC} + \frac{T}{RC} = lnQ - lnQ_T$$

Goes to:

$$-\frac{t}{RC} + \frac{T}{RC} = ln\left(\frac{Q}{Q_T}\right)$$

Take exponentials:

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

The left side is not correct. When you take the exponential of both sides, the entire left side goes in one exponential function, because:

$$e^{a+b} \neq e^a + e^b$$

$$Q_T\left(-frac{t}{RC} + e^{\frac{T}{RC}}\right) = Q$$

So now I have to find Q_T

Would it be:

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

That's right; you use this formula with t=T to find Q_T.

 

[TFM]

So:

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

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

and:

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

Finally giving:

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

?

TFM

 

[alphysicist]

So:

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

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

and:

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

Finally giving:

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

I think it's just a typo, but that T right before the e should not be there.

You can make your exponential function look nicer (if you like) by using:

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

It's not different, but it emphasizes that it is the time interval that's important.

 

[TFM]

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

 

[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:

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


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:

$$\frac{dx}{dt} + c x = f(t)$$

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

$$x' + ax = b --> x = Ce^{-at} + \frac{b}{a}$$

and linear:

$$x' + p(t)x = q(t) --> e^Fx = \int e^F q(t)dt + C, F = \int p(t) dt$$

?

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

$$x' + ax = b --> x = Ce^{-at} + \frac{b}{a}$$

and linear:

$$x' + p(t)x = q(t) --> e^Fx = \int e^F q(t)dt + C, F = \int p(t) dt$$

?

TFM

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