Set up the differential equation showing the voltage V(t) for this RC circuit

  • #1
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Homework Statement:
A capacitor with Capacity C, a Battery with Voltage ## V_0 ## and Resistance R are connected.
C = 1mF, R = 500 Ohms, V_0 = 10V

a) Set up a differential equation showing the change in voltage over time.
b) draw a slope field of the equation
c) solve the equation
d) show that ## V(t) =V_0(1-e^(\frac{-t}{RC})) is a solution
e) draw the function
Relevant Equations:
C = Q/V, The voltage difference around a closed loop is zero.
Hi all.

I have another exam question that I am not so sure about. I've solved similar problems in textbooks but I have a feeling once again that the correct way to solve this problem is much simpler and eluding me.
Especially because my answer to a) is already the solution to c) and d) (I did my way before looking at c and d) and I can't figure out b with my equation.

So I set up ## V_0 -IR -\frac{Q}{C} = 0 ##
I seperated and got to ## Q = V_0C(1-e^(\frac{-t}{RC}))##
Then because Q/C = V I divided both sides by the Capacitance and got:
## V = V_0(1-e^(\frac{-t}{RC}))##
(Solution to c and d)

Now for b If I take the time derivative again I end up with ## \frac{dv}{dt} = -V_0(RC)(1-e^(\frac{-t}{RC})) ##
Which I guess I could draw but it seems to complicate to be correct?

If there is another way to start, a better way, could someone point me in that direction without giving the answer away?
Thank you all!
 
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Answers and Replies

  • #2
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So I set up ## V_0 -IR -\frac{Q}{C} = 0 ##

Using ##V(t) = \frac{Q}{C}##, you could have rewritten the differential equation in terms of V(t).

It looks like you should draw the slope field before solving the differential equation. You can use the rewritten version, the one in terms of V(t), then use your R, C, and ##V_0## values to make approximations to draw the slope field.
 
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  • #3
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Using ##V(t) = \frac{Q}{C}##, you could have rewritten the differential equation in terms of V(t).

It looks like you should draw the slope field before solving the differential equation. You can use the rewritten version, the one in terms of V(t), then use your R, C, and ##V_0## values to make approximations to draw the slope field.
Hi doggydan42,

thank you very much. I thought of this, but for some reason I thought it's not ok to just use V in there. But it makes sense. Let me try again, now I'm also curious if I'll get a similar derivative to the one I got doing my long and tedious way!

thanks so much!
 
  • #4
Chandra Prayaga
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Check your derivative dV/dt again.
 
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  • #5
DaveE
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One easy way to check your results is with dimensional analysis. If you find something like [meters]/[sec] = [meters], then you know something is wrong. Note that this doesn't tell you the answer is correct, it won't tell you about factors of -2 or π. But it's a quick way of spotting bad formulas.
https://en.wikipedia.org/wiki/Dimensional_analysis

For example, if you asked what sort of phenomena is measured in (bushells)/(fortnight-acre), then I can substitute the units:
bushell = volume = m3
fortnight = time = seconds
acre = area = m2
so, bushells/(fortnight-acre) = m3/(seconds*m2) = m/sec which is a velocity.
Note that I never cared about how many m3 are in a bushell. It's only about the type of dimension, not the conversion constants.
 
  • #6
vela
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Now for b If I take the time derivative again I end up with ## \frac{dv}{dt} = -V_0(RC)(1-e^(\frac{-t}{RC})) ##
Which I guess I could draw but it seems to complicate to be correct?
This expression for dV/dt isn't the one you want because it comes from a specific solution to the differential equation. What you want to do is follow @doggydan42's suggestion and write the original differential equation in terms of V. You'll end up with something of the form
$$\frac{dV}{dt} = f(t,V)$$ for some function ##f##. For various point on the tV-plane, you can calculate what dV/dt is equal to and draw a short line segment with that slope at that point.
 
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  • #7
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Hi all, thank you very much for your help.
At first I felt confident after@doggydan42's suggestion and tried last night, but I had problem identifying my ## \frac {dV}{dt} ## part.
I did a dimensional analysis as suggested by @DaveE of my solution and got Coloumbs * Ohms over seconds which seems to be correct at least.
I tried again this morning and I've been thinking where the Voltage changes.
The voltage over the resistor changes as the current is changing when the capacitors is discharging or charging.
I found in a book that R = dV/dI but R isn't changing in this case, right?
the only varying voltage part i'd have would be the capacitor.
But if I insert dV/dt there and just consider the current as the current I get ##V(t) =(V_0 -IR)*t##
:-(

EDIT: writing it out here made me see something, that's probably wrong, but may I ask:
Could I use ##R = \frac{dV}{dI}## and set I to ##\frac{dI}{dt}## and then I'd have ## V_0=\frac{dV}{dt} + V##
Feels completely wrong, but, I'm at my wit's end here. sorry..
 
  • #8
vela
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The ##I## in ##IR## is the current through the resistor. How is the current through the resistor related to the current through the capacitor? What is the relationship between the current through the capacitor and the voltage across the capacitor?
 
  • #9
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The ##I## in ##IR## is the current through the resistor. How is the current through the resistor related to the current through the capacitor? What is the relationship between the current through the capacitor and the voltage across the capacitor?
Hi Vela, thank you very much for helping.
I thought about your comment and tried a few ways earlier. Didn't quite seem to find a solution.
I came back after reviewing basics but I'm still stuck.

The current decreases as the Voltage (Electric PE) increases in the capacitor.
So I thought that maybe ## dI = -\frac {dV}{dt}## I put that in to IR and I ended up with just V_0 = - V in the end.
It's nonsense but interestingly one of my first results was V = V_0 but I can't find the page where I did that anymore. I think there and then I just replaced IR with dV/dt.

edit: I have another question to your comment also @vela, I thought there is no current through the capacitor?
 
  • #10
vela
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In your original attempt, you went from
$$V_0 - IR - \frac QC = 0$$ to
$$Q=CV_0(1-e^{-t/RC}).$$ How did you deal with ##I## there to get everything in terms of ##Q##? And what was your reasoning?
 
  • #11
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In your original attempt, you went from
$$V_0 - IR - \frac QC = 0$$ to
$$Q=CV_0(1-e^{-t/RC}).$$ How did you deal with ##I## there to get everything in terms of ##Q##? And what was your reasoning?
Hi @vela thank you very much.

the current was more intuitive, I knew it's dq/dt so it just came.
I did some algebra (Q to the dq/dt side) and then I integrated and then raise to the power of e to get Q out of the natural log to get the Result.
dq/dt is changing due to the amount of charge on the capacitor.
 
  • #12
vela
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##I## in the term ##IR## represents the current flowing through the resistor. ##dQ/dt## is the current flowing into the capacitor. How do you know they are equal?

You have the relation ##Q=CV## for the capacitor. Try using that.
 
  • #13
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Hi vela,
thank you very much. It seems like your earlier comment put me on the right track but I messed up somehow. I tried replacing I in IR through and the Resistance canceled and I ended up with the V_0 = V expression.

I will try to think about it some more and see if I can't come up with it.
thank you for your patience!!
 
  • #14
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##I## in the term ##IR## represents the current flowing through the resistor. ##dQ/dt## is the current flowing into the capacitor. How do you know they are equal?

You have the relation ##Q=CV## for the capacitor. Try using that.
I think I might be making an integration mistake.

Is ## V_0 = CR*\frac{dV}{dt} +V ## the correct Differential equation?
edit: I am definitely making integration mistakes. Sorry! I thought I had to use the integrating factor. And then used u sub. But with some algebra now, I could do without and I almost got the correct solution.
I ended up with ##V_0 - V_0 e^(\frac{t}{RC})## instead of with e raised to the negative (t/RC).
I think I'll better review some calculus and come back soon!

I still end up with ##V_0 = V## or ## V_0 = -V ## depending on which way I believe the current goes.
 
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  • #15
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Is ## V_0 = CR*\frac{dV}{dt} +V ## the correct Differential equation?
edit: I am definitely making integration mistakes. Sorry! I thought I had to use the integrating factor. And then

You have the right differential equation. You could use an integration factor to solve it, or you could use separation of variables. Try solving it again, and pay careful attention to the sign in the exponent. Please show all your work in solving the differential equation next time.

I still end up with ##V_0 = V## or ## V_0 = -V ## depending on which way I believe the current goes.
How are you getting this?
 
  • #16
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Hi @doggydan42 thank you.
I tried to make a clean version of what happened. I didn't upload the work previously because once i get confused it becomes a mess that only I can read.
First is the integrating factor.

I was able to solve the latter for the right solution, though if I use -dV/dt for the current in the loop.
instead of dV/dt in the equation, you mentioned was correct.
I feel really silly, I know I'm making a sign mistake, but I just seem unable to locate where.
thank you for helping!
 

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  • #17
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I tried to make a clean version of what happened. I didn't upload the work previously because once i get confused it becomes a mess that only I can read.
You're missing a constant that you need to solve for using an initial or boundary condition.

Edit:
It looks like you are incorporating the initial condition into the integral as one of the bounds. But you have to use that as a bound instead of ignore it.

Edit 2:
In your latter solution, watch what you are integrating over, and look at the signs. Try a u-substitution to make it clear.
 
  • #18
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You're missing a constant that you need to solve for using an initial or boundary condition.

Edit:
It looks like you are incorporating the initial condition into the integral as one of the bounds. But you have to use that as a bound instead of ignore it.

Edit 2:
In your latter solution, watch what you are integrating over, and look at the signs. Try a u-substitution to make it clear.

Hi doggydan42, could it be that the differential equation should be : ## V_0 = -\frac{dV}{dt}RC + V ## ?
 
  • #19
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Hi doggydan42, could it be that the differential equation should be : ## V_0 = -\frac{dV}{dt}RC + V ## ?
I don't believe so. Solve the earlier differential equation using your first method again. It looks like your mistake is setting ##e^0=0## when evaluating the integral over the bounds.
 
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  • #20
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Omg! You're right doggydan42, I evaluated e^0 as 0 instead of 1. I made this mistake each time I solved it. It's so frustrating, these simple things seem to still happen to me so often. (This morning I was working on another problem where I kept 2*1 = 3 and couldn't figure out what I was doing wrong.)

Thank you very much. I'm so sorry, I really worked through it about 10 times and I seem to have always automatically discarded e^0 as 0.
 

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