Differential Equations and RQC

[TFM]

Homework Statement

A circuit consists of a voltage source, voltage V , a resistor, resistance R, and a capacitor,
capacitance C, in series.

(a) Show that the charge, Q, in the capacitor satisfies the equation RQ' + Q/C = V .

Homework Equations

$$R\frac{dQ}{dV} + \frac{Q}{C} = V$$

The Attempt at a Solution

I'm not quite sure what to do for this Question. In this Part you have to show how charge satisfies the given equation. Could anyone help show how to start this question, what it actually is asking you to do?

TFM

 

[alphysicist]

Hi TFM,

Homework Statement

A circuit consists of a voltage source, voltage V , a resistor, resistance R, and a capacitor,
capacitance C, in series.

(a) Show that the charge, Q, in the capacitor satisfies the equation RQ' + Q/C = V .

Homework Equations

$$R\frac{dQ}{dV} + \frac{Q}{C} = V$$

The Attempt at a Solution

I'm not quite sure what to do for this Question. In this Part you have to show how charge satisfies the given equation. Could anyone help show how to start this question, what it actually is asking you to do?

TFM

To start these types of questions you almost always use a Kirchoff voltage equation (or Kirchoff loop equation). What do you get when you write the loop equation for this circuit? Do you see how it reduces to the given equation?

 

[TFM]

So using Kirchoff,

Current in = Current Out

Sum of Voltage = Total Voltage

So, if you calculate the voltage, following from my attached diagram:

$$V_{input} = V_{Resister} + V_{Capacitor}$$

$$V_{resister} = IR$$

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

Inserting into Voltage equation above

$$V_{input} = IR + \frac{Q}{C}$$

but the equation doesn't have IR,

known equations featuring I:

Q=It - Time not in question
V=IR - would take ypu back to V

Not sure the above equations help, though?

TFM

 

[TFM]

I wonder if I have got my derivative correct. The next part of the question is to find an expression for Q(t)

Does this mean that RQ' + Q/C = V

should go to:

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

which would be

Q = It, I = Q/t

so:

$$V_{input} = IR + \frac{Q}{C}$$

$$V_{input} = \frac{Q}{t}R + \frac{Q}{C}$$

Although this can't be quite right because i have Q/t, not dQ/dt

?

TFM

 

[merryjman]

instead of using Q/t for current, you should be using dQ/dt. This gives you the differential equation you're looking for.

 

[TFM]

Thanks,

for the next part,

Suppose that R, C and V are constant and that Q is initially zero. Find an expression for Q(t). Sketch the solution.

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

R,C,V = Const.
Q(t=0) = 0

I have

$$Vdt = R + \frac{Q}{C} dQ$$

Which I get as

$$Vt = RQ + \frac{Q^2}{2C} + D$$

where d = integration constant

so inserting values

V(0) = R(0) + $$\frac{0^2}{2C} + D$$

D = 0

does this look right?

TFM

 

[merryjman]

I don't think you manipulated your differential equation correctly. Think about this rationally: as the capacitor continues to store energy of separated charge, it will eventually reach a limit to how much energy it can store. This means that Q(t) should start at zero (uncharged capacitor) and asymptotically approach a fixed value. What sort of functions behave this way?

The sort of functions that behave this way have a differential equation that looks like dU/U = K dt where U is any function and K is a constant. Get your equation into a form like that and integrate both sides.

 

[TFM]

Does this look right. So far I have:

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

$$\frac{V}{Q} = \frac{R}{Q}\frac{dQ}{dt} + \frac{1}{C}$$

$$\frac{V}{Q} - \frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C}$$

Factorise out:

$$\frac{1}{Q}(V - R)\frac{dQ}{dt} = \frac{1}{C}$$

$$\frac{1}{Q}\frac{dQ}{dt} = \frac{V - R}{C}$$

$$\frac{1}{Q} dQ = \frac{V - R}{C}dt$$

integrate gives:

$$lnQ = \frac{V - R}{C}t$$

does this look better?

Edit: No it doesn't

$$\frac{1}{Q}(V - R)\frac{dQ}{dt} = \frac{1}{C}$$

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

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

Integrates to:

$$lnQ = \frac{t}{VC - RC}$$

Does this look right now

TFM

 

[alphysicist]

Does this look right. So far I have:

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

$$\frac{V}{Q} = \frac{R}{Q}\frac{dQ}{dt} + \frac{1}{C}$$

$$\frac{V}{Q} - \frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C}$$

Factorise out:

$$\frac{1}{Q}(V - R)\frac{dQ}{dt} = \frac{1}{C}$$

This line is not correct. You have factored out a dQ/dt, but the first term did not have that in it.


Also, remember that after you get the integrals set up, these are definite integrals and so you have to worry about the limits.

 

[TFM]

Should it be then:

$$\frac{v}{Q}-\frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C}$$

$$\frac{1}{Q}(V - R \frac{dQ}{dt}) = \frac{1}{C}$$

$$\frac{1}{Q} = \frac{1}{C(V - R\frac{dQ}{dt})}$$

$$\frac{1}{Q} = \frac{1}{CV - CR\frac{dQ}{dt}}$$

This doesn't look right though...?

TFM

 

[alphysicist]

Should it be then:

$$\frac{v}{Q}-\frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C}$$

$$\frac{1}{Q}(V - R \frac{dQ}{dt}) = \frac{1}{C}$$

$$\frac{1}{Q} = \frac{1}{C(V - R\frac{dQ}{dt})}$$

$$\frac{1}{Q} = \frac{1}{CV - CR\frac{dQ}{dt}}$$

This doesn't look right though...?

TFM

I would suggest starting with the beginning equation and getting everything out of the denominator (even though later you'll get things in the denominator again).

Once you have that, the goal is to completely separate the equation so that q does not appear on one side of the equation, and t does not appear on the other. (This is where you'll have a fraction again.) At that point you can integrate each side separately. Do you get the answer?

 

[TFM]

So starting with:

$$\frac{v}{Q}-\frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C}$$

Multiply out the Q

$$V - R \frac{dQ}{dQ}=\frac{Q}{C}$$t

Multiply by C

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

split the derivative:

$$VC - RC dQ = Q dt$$

put the Q on the other side by dividing through by Q

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

Does this look better?

TFM

 

[alphysicist]

So starting with:

$$\frac{v}{Q}-\frac{R}{Q}\frac{dQ}{dt} = \frac{1}{C}$$

Multiply out the Q

$$V - R \frac{dQ}{dQ}=\frac{Q}{C}$$t

Multiply by C

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

This is correct up to here.

split the derivative:

$$VC - RC dQ = Q dt$$

The first term is missing a factor of dt.

Once you correct that, then move all terms with dt to one side, all terms with dq to the other, and factor out the differentials until you get:

(stuff) dt = (other stuff) dq

At that point you should be able to separate the variables.

 

[TFM]

So:

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

$$VCdt - RC \fracdQ = Qdt$$

$$-RC dQ = Qdt - VCdt$$

$$-RC dQ = (Q - VC)dt$$

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

Does this look okay now?

 

[alphysicist]

So:

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

$$VCdt - RC \fracdQ = Qdt$$

$$-RC dQ = Qdt - VCdt$$

$$-RC dQ = (Q - VC)dt$$

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

Does this look okay now?

That looks right to me.

 

[TFM]

So:

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

The right side will just go to t + C:

$$\frac{-RC}{Q - VC} dQ = t + C$$

$$\frac{-RC}{Q} + \frac{RC}{VC} dQ$$

$$\frac{-RC}{Q} + \frac{R}{V} dQ$$

So does this go to:

$$\frac{-RC}{ln(Q)} + \frac{RQ}{V} = t + C$$

?

TFM

 

[alphysicist]

So:

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

The right side will just go to t + C:

I would set limits on the t integral:

$$\int_0^t dt$$

That way the constant is evaluated.

Or if you want to be a bit more general, you could use

$$\int_{t_0}^{t} dt$$

but I don't think that is probably necessary here.

$$\frac{-RC}{Q - VC} dQ = t + C$$

This is okay, but your next line is not the thing to do. Just go ahead and integrate the left side.

$$\frac{-RC}{Q} + \frac{RC}{VC} dQ$$

$$\frac{-RC}{Q} + \frac{R}{V} dQ$$

So does this go to:

$$\frac{-RC}{ln(Q)} + \frac{RQ}{V} = t + C$$

?

TFM

 

[TFM]

Okay

$$\frac{-RC}{Q - VC} dQ = \int^t_0 dt$$

So

$$\frac{-RC}{Q - VC} dQ = t$$

But is the integral for the left:

$$\frac{-RC}{ln(Q - VC)} = t$$

?

 

[alphysicist]

Okay

$$\frac{-RC}{Q - VC} dQ = \int^t_0 dt$$

So

$$\frac{-RC}{Q - VC} dQ = t$$

But is the integral for the left:

$$\frac{-RC}{ln(Q - VC)} = t$$

?

No, the ln function goes in the numerator. Also, remember that you have to evaluate the limits of the integral.

 

[TFM]

Okay, so:

$$\int^Q_0 \frac{-RC}{Q - VC} dQ = t$$

$$\frac{ln(Q - VC)}{-RC} = t$$

?

TFM

 

[alphysicist]

Okay, so:

$$\int^Q_0 \frac{-RC}{Q - VC} dQ = t$$

$$\frac{ln(Q - VC)}{-RC} = t$$

?

TFM

The (-RC) was correct in the numerator. (In fact, I would just go ahead and move the (-RC) to the right side denominator so it's not in the integral.)

You've evaluated the upper limit, but you still have to subtract off the value of the function at the lower limit.

 

[TFM]

So:

$$[ln(Q - VC)]^Q_0 -RC = t$$

$$[ln(Q - VC) - Ln(0 - VC)]-RC = t$$

$$[ln(Q - VC) - Ln(-VC)]-RC = t$$

Is this better?

TFM

 

[alphysicist]

So:

$$[ln(Q - VC)]^Q_0 -RC = t$$

$$[ln(Q - VC) - Ln(0 - VC)]-RC = t$$

$$[ln(Q - VC) - Ln(-VC)]-RC = t$$

Is this better?

TFM

No, the factor of (-RC) is still not in the right place. That's why I suggested to move it to the right side denominator. (If you do leave it to the left, it will be multiplyig the entire left side.)

So go back a few posts ago; after I suggested moving that factor, you would have had (after integrating the left side):

$$[ln(Q - VC)]^Q_0 = \frac{t}{-RC}$$

(This replaces the first line in your last post.) Then evaluating the limits like you did above would give:

$$[ln(Q - VC) - Ln(-VC)] = -\frac{t}{RC}$$

Now use what you know about logarithms to combine the two ln functions, and solve for Q.

 

[TFM]

Okay, so:

$$\frac{-RC}{Q - VC} dQ = t$$

divide both sides by -RC

$$\frac{1}{Q - VC} dQ = -\frac{t}{RC}$$

$$[ln(Q - VC) - Ln(0 - VC)] = -\frac{t}{RC}$$

This is the same as:

$$ln \left(ln frac/{Q - VC}{-VC}\right) = -\frac{t}{RC}$$

Take exponentials of both sides

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

$$Q - VC = -VCe^{-\frac{t}{RC}}$$

$$Q = VC - VCe^{-\frac{t}{RC}}$$

Is this correct now?

TFM

 

[alphysicist]

Okay, so:

$$\frac{-RC}{Q - VC} dQ = t$$

divide both sides by -RC

$$\frac{1}{Q - VC} dQ = -\frac{t}{RC}$$

$$[ln(Q - VC) - Ln(0 - VC)] = -\frac{t}{RC}$$

This is the same as:

$$ln \left(ln frac/{Q - VC}{-VC}\right) = -\frac{t}{RC}$$

Take exponentials of both sides

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

$$Q - VC = -VCe^{-\frac{t}{RC}}$$

$$Q = VC - VCe^{-\frac{t}{RC}}$$

Is this correct now?

TFM

That looks right to me. There's some more manipulation and defining you can do to get it into the standard textbook form, but I would think that what you have is what they are probably looking for.

 

[TFM]

That looks right to me. There's some more manipulation and defining you can do to get it into the standard textbook form, but I would think that what you have is what they are probably looking for.

I suppose you could factorise down to:

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

Thanks

The next part of the question then goes on to ask:

Suppose instead that V = V0 for t < T and V = 0 for t > T, where V0 is constant and Q is again initially zero. Find the new expression for Q(t), both for t < T and t > T.
(Hint: both solutions must match at t = T.)

I assume that T is another time piece and not Temperature

Would solving this be similar to the previous part?

TFM

 

[alphysicist]

I suppose you could factorise down to:

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

Thanks

The next part of the question then goes on to ask:

Suppose instead that V = V0 for t < T and V = 0 for t > T, where V0 is constant and Q is again initially zero. Find the new expression for Q(t), both for t < T and t > T.
(Hint: both solutions must match at t = T.)

I assume that T is another time piece and not Temperature

Would solving this be similar to the previous part?

It's a similar process. You already have the expression for up to time T. On repeating the process for times later than T, one big difference is that the lower limits on both integrals will not be zero. And of course you will want to think about what that does to your differential equation that you get from the Kirchoof loop equation.

 

[TFM]

So for the t > T one, it is just

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

And for t < T, we start of with

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

go as far as:

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

and just change the limits?

TFM

 

[alphysicist]

So for the t > T one, it is just

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

Well, that's the one for t<T (and they want you to set $V\to V_0$).

And for t < T, we start of with

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

For the later times (t > T), the differential equation will be different in two ways. Draw the circuit and use the loop rule; what is the new diff. eqn?

 

[TFM]

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

 

[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?