Differential equations assignment T3

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Homework Help Overview

The discussion revolves around a differential equation related to the charging of a capacitor through a resistor. The equation is given as CR(dV/dt) + V = E, where V is the voltage across the capacitor, E is the steady voltage applied, C is the capacitance, and R is the resistance. Participants are exploring the solution to this equation and evaluating voltage under specific conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to solve the differential equation for E and evaluate V under given parameters. There are discussions about the correct integration techniques and the use of integrating factors. Some participants question the integration steps and the treatment of constants.

Discussion Status

The discussion is active, with participants providing feedback on each other's approaches and clarifying misunderstandings. Some guidance has been offered regarding integration techniques, and there is a recognition of the need for careful handling of constants in the equations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on understanding the integration process and the application of differential equations in this context.

mathi85
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Hi!

I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.
Thank you in advance for your time.

Task 3:

A capacitor C is charged by applying a steady voltage E through a resistance R. The p.d. between the plates, V, is given by the differential equation:

CR(dV/dt)+V=E

a) Solve the equation for E given that when time t=0, V=0.
b) Evaluate voltage V when E=50V , C=10μF, R=200kΩ and t=1.2s

Solution:

a)
I want: (dV/dt)PV=Q

CR(dV/dt)+V=E /:(CR)
(dV/dt)+(1/CR)V=E/(CR)

P=1/(CR) and Q=E/(CR)

Integrating factor =e∫ P dt
∫ P dt=∫ 1/(CR) dt = ln(CR)t

IF=eln(CR)t=CRt

y(IF)=∫ (IF)*Q dx
∴V(IF)=∫ (IF)*Q dt
VCRt=∫ CRt*(E/(CR)) dt
VCRt=∫ tE dt
VCRt=(Et2)/2+c

General Solution:

VCRt=(1/2)Et2+c

t=0 when V=0

∴0=0+c
c=0

Particular Solution:

VCRt=(1/2)Et2 /*2
2VCRt=Et2 /:t2
E=(2VCRt)/(t2)

b)

VCRt=(1/2)Et2 /:CRt
∴V=(1/2)[(Et2)/(CRt)]

V=(1/2)[(50*1.22)/(10*10-6*200*103*1.2)]
V=30volts
 
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mathi85 said:
Hi!

I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.
Thank you in advance for your time.

Task 3:

A capacitor C is charged by applying a steady voltage E through a resistance R. The p.d. between the plates, V, is given by the differential equation:

CR(dV/dt)+V=E

a) Solve the equation for E given that when time t=0, V=0.
b) Evaluate voltage V when E=50V , C=10μF, R=200kΩ and t=1.2s

Solution:

a)
I want: (dV/dt)PV=Q
Where did this come from?

CR(dV/dt)+V=E /:(CR)
(dV/dt)+(1/CR)V=E/(CR)

P=1/(CR) and Q=E/(CR)

Integrating factor =e∫ P dt
∫ P dt=∫ 1/(CR) dt = ln(CR)t
You didn't integrate correctly. R and C are constants.

IF=eln(CR)t=CRt
This is wrong too. It's true that eln a=a, but that's not what you have here because of t.

y(IF)=∫ (IF)*Q dx
∴V(IF)=∫ (IF)*Q dt
VCRt=∫ CRt*(E/(CR)) dt
VCRt=∫ tE dt
VCRt=(Et2)/2+c

General Solution:

VCRt=(1/2)Et2+c

t=0 when V=0

∴0=0+c
c=0

Particular Solution:

VCRt=(1/2)Et2 /*2
2VCRt=Et2 /:t2
E=(2VCRt)/(t2)

b)

VCRt=(1/2)Et2 /:CRt
∴V=(1/2)[(Et2)/(CRt)]

V=(1/2)[(50*1.22)/(10*10-6*200*103*1.2)]
V=30volts
 
Sorry! It should be:
"I want: (dV/dt)+PV=Q"

This is just to remind me what form the equation should have.
 
I'll try to get my head round this integration...
 
I really struggle with this integration involving 1/CR.
Is it just 1/(CR)*t and my IF will be e1/(CR)*t?
 
...et/(CR)
 
Yes, that's right.
 
I understand that because CR are constants, not functions I cannot have it as ln.
 
Thnx! I'll do it again and post it asap.
 
  • #10
Now it makes even less sense to me:

∫ P dt = ∫ 1/(CR) dt = t/(CR)
∴IF=et/(CR)

Vet/(CR)=∫ et/(CR)*E/(CR) dt

I'm not sure about the integration. If I treat E/CR as number then I can go:

Vet/(CR)= (CR)/E∫ et/(CR) dt
 
  • #11
This is fine:
$$Ve^{t/RC} = \int \frac{E}{RC}e^{t/RC}\,dt.$$ You're correct that E/RC is a constant so that you can pull it out of the integral; however, this isn't correct:
$$Ve^{t/RC} = \frac{RC}{E}\int e^{t/RC}\,dt.$$ Why did you flip the fraction over when you pulled it out of the integral? It should be
$$Ve^{t/RC} = \frac{E}{RC}\int e^{t/RC}\,dt.$$ Now use the substitution u = t/RC.
 
  • #12
Sorry. I was tired last night and started making stupid mistakes.

I understand that now I have to use integration by parts method however in examples I've done so far I always had two functions of x, i.e. ∫ x*e2x dx
In this case I know that:

u=x
du/dx=1
dv/dx=e2x ∴ v=(1/2)e2x

and:
∫ u(dv/dx) dx = u*v-∫ v*(du/dx) dx
 
  • #13
So if u=t/(CR), what is my dv/dt and v?
 
  • #14
Why are you using integration by parts?
 
  • #15
Did you mean I should use a simple substitution?

Veu=E/(CR)∫ eu dt
 
  • #16
I keep trying... This is what I have now:
Vet/(CR)=E/(CR)∫ e[1/(CR)]t dt
Vet/(CR)=E/(CR)*CRet/(CR)+c
 
  • #17
Looks good. Keep going.
 
  • #18
Right I come to the point where:

Ve(1/(CR))t=Ee(1/(CR))t+c /:e(1/(CR))t
V=E+c/(e(1/(CR))t)

if I apply t=0 when V=0

0=E+c
c=-E

and Part. Sol.

V=E+[-E/e(1/(CR))t]

Where am I going wrong?
 
  • #19
I'm getting that E=V+E/e(1/(CR))t

This cannot be right can it?
 
  • #20
That's correct. Why do you think anything's wrong?
 
  • #21
It just didn't look right to me.
Thank you very much for your help!
 

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