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Math Help - RC Circuit Differential Equation

  1. #1
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    RC Circuit Differential Equation

    I was trying to do this practice problem, but the constants of the equation are killing me!
    The question is:

    -The differential equation for the voltage vc across the capacitor is dvc/dt=(V(t)-vc)/RC.
    Find the formula for the general solution of the RC circuit equation above if the voltage source is contant for all time, i.e. V(t)=K for all t.
    -Find the solution for the voltage vc(t) with initial-value vc(0)=1 in the RC circuit with constant voltage source V(t)=K

    The first question I get stuck because once I replace V(t)=K, it becomes dvc/dt=(K-vc)/RC, now I could isolate RC by dvc/dt=1/RC(K-vc). The problem is after that I'm having problems differentiating.
    The second question is easy. Once the formula for the general solution of the RC circuit is found all I have to do is plug it in vc(t)=1 and solve for contant right?
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  2. #2
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    Your first question is separable. That is, you have

    \dfrac{dv_{c}}{K-v_{c}}=\dfrac{dt}{RC}.

    Now integrate.

    Your intuition about the second part is correct, just make sure to do v_{c}(0)=1, not v_{c}(t)=1.

    Make sense?
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  3. #3
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    dvc/(k-vc)= dt/RC,
    Then I integrate both sides but first taking 1/RC out of the integral, since its constant. then
    integral of (dvc/(k-vc))= (1/RC) integral of dt

    that will equal to

    ln|K-vc|= (1/RC)t+c,

    solving for vc I have to exponentiate both sides which will equal to

    e^(ln|K-vc|)= e^(t/RC+c), in which is the same as K-vc=e^(t/RC)e^(c),

    solving for vc the equation will be,

    vc=-e^(t/RC)e^(c)+K

    Is that correct? Are the signs correct (-,+)?
    Now do I have to replace e^(c) with another letter. My professor usually subs with k, but I feel like I can't since there is a K on the equation already. Does it matter?

    As you said, my intuition on the second question is right, however when I plug 0 for all ts and equal the equation to 1
    i get e^c=K-1

    what I did was since vc(0)=1, so
    vc(0)=-e^(t/RC)e^(c)+K
    1=-e^(0/RC)e^(c)+K, aince e^(0)=1 then
    1=(-1)e^(c)+K , solving for e^(c)
    1-K=(-1)e^(c)
    K-1=e^(c)

    Now that i know that K-1=e^(c), I can go back to my first equation and plug (K-1) to e^(c)
    by doing that i get
    vc=-e^(t/RC)e^(c)+K
    vc=-e^(t/RC)(K+1)+K

    Would these be the correct answer or they are asking something else? I don't know the values for R nor C, neither the values for V(t)= K. Did the answers I gave satisfy what both question are asking? I feel like I did something is wrong and the solution is supposed to be something simpler.
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  4. #4
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    Quote Originally Posted by hansbahia View Post
    dvc/(k-vc)= dt/RC,
    Then I integrate both sides but first taking 1/RC out of the integral, since its constant. then
    integral of (dvc/(k-vc))= (1/RC) integral of dt

    that will equal to

    ln|K-vc|= (1/RC)t+c,
    Two comments:

    1. The integration is incorrect. It should be

    -\ln|K-v_{c}|=\dfrac{t}{RC}+B.

    2. As you have already seen, I used a symbol other than C for my constant of integration. You don't want to confuse the constant of integration with the capacitance!

    If you carry these corrections through, what does that give you?
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  5. #5
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    Oh yes! I forgot to put the "-" sign when using substitution, with that being fixed all i have to change is

    -ln|K-vc|= (1/RC)t+B
    vc=+e^(t/RC)e^(B)+K

    and the second part solving for e^(B)

    1-K=(1)e^(c)
    e^(c)=1-K

    Now that i know that e^(c)=1-K, I can go back to my first equation and plug (1-K) to e^(c)
    by doing that i get

    vc=e^(t/RC)e^(c)+K

    vc=e^(t/RC)(1-K)+K

    Now everything should be right. However, does the these two answer satisfy what its being asked? Should I change K to V(t)?


    vc(t)=e^(t/RC)(1-K)+K
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  6. #6
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    The cool thing about differential equations is that it's easy to check your own answer: differentiation is easier than integration. Just differentiate your solution, plug it into the DE and the initial conditions, and see if they're all satisfied. Are they?
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