1st order circuit problem help

Discussion in 'Homework Help' started by inkosi, Sep 26, 2014.

  1. inkosi

    Thread Starter Member

    Dec 10, 2013
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    [​IMG]

    I need help with this question. At t = 0-, V = 0. I don't know how to find V at t = 0+. Here is the solution manual's answer:

    [​IMG]

    What is Vcn and Vcf? Natural and steady state response? I know how to find Vcn but what have they done in that step I circled? I don't get what's going on there..

    Help would REALLY be appreciated. Thanks.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Vcn(t) can be interpreted as the natural or unforced response, and Vcf(t) the forced response.
    Since Vcf(t)=K the first derivative is zero, so they solve for K.
    Once you get K, you can then add the two responses and solve for A as in:
    Vc=K+A*e^(-a*t)

    knowing that Vc=0 at t=0.

    BTW, Vc=0 at t=0+ because of the physical nature of the capacitor. It pays to know that the voltage across a capacitor can not change in zero time.
    This nature is true of the inductor too, where the current can not change in zero time.
    Knowing these two facts helps with these circuits.
    Looking at the definition of the capacitor:
    dv/dt=i/C

    and rearranging a little we get:
    dv=i*dt/C

    so you can see right away that if the increment in time (dt) is zero then the increment in the voltage (dv) must also be zero, regardless of the value of the current or capacitance (provided we dont have infinite current which is usually not allowed).
     
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  3. inkosi

    Thread Starter Member

    Dec 10, 2013
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    That's interesting. Thanks for the response. I found this question a little tricky because it's not how I learnt how to solve these type of questions. We would find V(t = 0-), V (t = 0+), V(t = infinity), Rth (infinity) and then use the same general equation K1 + K2e^-t/Tow where K1 was v(infinity) and K2 v(t = 0+) - v(t = infinity).

    Using that method I got the same answer.. but it like 4-5 steps only so I don't really see the purpose of doing all of those "complicated" steps with the method the solution manual used..
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
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    Knowing the general form of a solution and the working out the associated boundary conditions is definitely quicker.
    However the method outlined is more comprehensive to the extent that it provides insight as to how a solution might be obtained starting from first principles. If you weren't told about the general form and had no knowledge of the likely solution how would you solve the problem?
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    Yes, we learn different methods for doing these circuits and this is just one way to do it. This way however has a mathematical basis for approaching it in this way though. That is, the circuit solution can be expressed as the sum of natural and forced responses while the (linear) differential equation for the solution can be expressed as the sum of the complementary solution and the particular solution. So if you solve the differential equations using an integrating factor you'll see the two parts show up in the equation as being added together to form the solution, one part being the natural response and the other the forced response.

    This i think is the most basic method taught. A more modern method i think is the coupled ODE method where we simply set up the equation(s) as a set of ODE's and then solve them, but in various coursework you'll need to do it the way the professor wants you to do it. You can however use other methods to check your work, which you should always do anyway.
     
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