1. Limited time only! Sign up for a free 30min personal 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!

Find Vc(t) given iC(t)

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-30_16-34-51.png

    iC(t) = 10cos(1000t+pi/4)
    Vc(0-) = 3v
    2. Relevant equations
    ic = C * d/dt*Vc(t)

    3. The attempt at a solution
    Vc(t) = 1/C * integral(Ic(t)dt) =
    1/10^-6 * 10sin(1000t+pi/4)/1000 = 10000sin(1000t+pi/4)

    What do I do with the initial condition?
     
  2. jcsd
  3. Oct 30, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    It's the integration constant.
     
  4. Oct 30, 2016 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    ##v_C(t)\ =\ V_C(0)\ +\ \frac 1C \displaystyle\int_{0}^{t} .... ##
     
  5. Oct 30, 2016 #4
    1/10^-6 * 10sin(1000t+pi/4)/1000 = 10000sin(1000t+pi/4) + K

    Vc(0-) = 3
    10000sin(0+pi/4) + K = 3
    7071.0678 + K = 3
    K = -7068.06781186547524400844362104
    ?
    That doesn't seem right...
     
  6. Oct 30, 2016 #5

    cnh1995

    User Avatar
    Homework Helper

    See NascentOxygen's post above. That's the formula you should use. Integration constant is nothing but Vc at t=0-.
     
  7. Oct 30, 2016 #6
    sin(1000t+pi/4)/100 [0 t]

    sin(1000t+pi/4)/100 + sin(pi/4)/100
    3 + 1000000(sin(1000t+pi/4)/100 + sqrt(2)*50)
    3 + 10000sin(1000t+pi/4) + 50000000sqrt(2)
     
  8. Oct 31, 2016 #7

    cnh1995

    User Avatar
    Homework Helper

    Well, actually this is right. I verified it using the above formula. I didn't look through your solution in detail earlier. But it looks correct to me now.
     
  9. Oct 31, 2016 #8
    Is there any way to simplify this so that there's no ugly constant?
     
  10. Oct 31, 2016 #9

    cnh1995

    User Avatar
    Homework Helper

    No. The constant can't be eliminated because of the initial condition. You can make it look better by writing it as
    Vc(t)=10sin(wt+pi/4)-7.068 kV.
     
  11. Oct 31, 2016 #10
    I asked my friend how he did it and he showed me this:
    10000sin(1000t+pi/4)
    Vc(0-) = 3
    10000sin(0+k+pi/4) = 3
    sin(k+pi/4) = 0.0003
    k + pi/4 = sin^-1(0.0003)
    k = (sin^-1(0.0003)-45degrees)/1000 = 0.045s
    Vc(t) = 10^4*sin(1000t)V
    Is this even remotely correct or did he just bs it?
    The reason I'm asking if it can be simplified further is because so far, most of the questions have had nice numbers.
     
  12. Oct 31, 2016 #11

    cnh1995

    User Avatar
    Homework Helper

    This is not where you add the integration constant.
    This does not give Vc=3V at t=0.

    Also, I don't understand what problem will having that constant in the equation cause. "Nice numbers" and "ugly constants" do not matter in an equation as long as it is mathematically and technically correct. Is it specifically asked in the question to remove the constant?
     
  13. Oct 31, 2016 #12

    gneill

    User Avatar

    Staff: Mentor

    Well, if you leave your calculator and root two alone the ugly constant is just ##5000 \sqrt{2} - 3##.
     
  14. Oct 31, 2016 #13

    NascentOxygen

    User Avatar

    Staff: Mentor

    That is a creative approach, but it's not correct.

    Over one full cycle (or period) the nett charge that the sinusoidal input current will have added to the capacitor is ZERO. This means that after each cycle of current, the capacitor voltage must back to where it started, i.e., it will have returned to VC(0). Your answer must show this.

    In contrast, if the capacitor voltage were (as your friend claims) a pure sinusoid with no DC offset, it would not show this periodic return to V(0) at times of t = n⋅T, where T is the period and n is an integer.
     
  15. Oct 31, 2016 #14

    NascentOxygen

    User Avatar

    Staff: Mentor

    Your friend's approach would be appropriate had the problem specified:
    iC(t) = 10cos(1000t + ##\theta## )
    Vc(0-) = 3v

    and so you'd be required to determine the ##\theta## that gives VC(0) as 3v. But this not what you are told about iC(t). You are not told it has some unknown phase; on the contrary, you are told it has some definite angle.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted