1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential equations assignment T3

  1. Dec 2, 2013 #1
    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
     
  2. jcsd
  3. Dec 2, 2013 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Where did this come from?

    You didn't integrate correctly. R and C are constants.

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

     
  4. Dec 2, 2013 #3
    Sorry! It should be:
    "I want: (dV/dt)+PV=Q"

    This is just to remind me what form the equation should have.
     
  5. Dec 2, 2013 #4
    I'll try to get my head round this integration...
     
  6. Dec 2, 2013 #5
    I really struggle with this integration involving 1/CR.
    Is it just 1/(CR)*t and my IF will be e1/(CR)*t?
     
  7. Dec 2, 2013 #6
    ...et/(CR)
     
  8. Dec 2, 2013 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, that's right.
     
  9. Dec 2, 2013 #8
    I understand that because CR are constants, not functions I cannot have it as ln.
     
  10. Dec 2, 2013 #9
    Thnx! I'll do it again and post it asap.
     
  11. Dec 2, 2013 #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
     
  12. Dec 3, 2013 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  13. Dec 3, 2013 #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
     
  14. Dec 3, 2013 #13
    So if u=t/(CR), what is my dv/dt and v?
     
  15. Dec 3, 2013 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Why are you using integration by parts?
     
  16. Dec 3, 2013 #15
    Did you mean I should use a simple substitution?

    Veu=E/(CR)∫ eu dt
     
  17. Dec 4, 2013 #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
     
  18. Dec 4, 2013 #17

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Looks good. Keep going.
     
  19. Dec 12, 2013 #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?
     
  20. Dec 12, 2013 #19
    I'm getting that E=V+E/e(1/(CR))t

    This cannot be right can it?
     
  21. Dec 12, 2013 #20

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's correct. Why do you think anything's wrong?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Differential equations assignment T3
Loading...