Nodal Analysis Impossibility!!! (Help)

Discussion in 'Homework Help' started by simon.r.potter, Nov 30, 2013.

  1. simon.r.potter

    Thread Starter New Member

    Nov 30, 2013
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    Hi Everyone,

    I've run into some serious problems with the question below.

    I've been researching for hours and have literally got nowhere!

    Any help at all would be greatly appreciated!!!

    Determine the node voltages V_1 and V_2 using Nodal Analysis and I_1 and I_2 using Mesh Analysis in fig-4?
    Task 2 - a.png

    No additional info other than the above!

    Thanks a bunch in advance!

    Kind Regards,
    Simon
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Step 1.
    Replace all inductors with their equivalent impedance.

    Step 2.
    Replace all capacitors with their equivalent impedance.

    Step 3.
    Treat capacitors and inductors as if they are resistors of corresponding impedance.
    Example:
    Impedance of inductor is XL=sL
    Impedance, XL, of 2 H inductor is XL=2s
    You are now have a purely resistive network. Apply Node-Voltage Method. Apply Mesh-Current Method.
     
  3. WBahn

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    Mar 31, 2012
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    My guess is that you don't know anything about impedances involving 's' yet.

    But hopefully you have learned about impedances involving jω. Have you?
     
  4. simon.r.potter

    Thread Starter New Member

    Nov 30, 2013
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    We have covered impedance's involving jω but not with s....could you direct me somewhere where I can find a bit of information?

    Kind Regards,

    Simon
     
  5. t06afre

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  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    s=jw
    So for inductor XL=jwL=sL
    For capacitor XC=1/(jwC)=1/(sC)

    I use s instead of jw because I am lazy to write jw all the time.

    Once you have the final formula, just place the values into it and get the answer.
     
  7. simon.r.potter

    Thread Starter New Member

    Nov 30, 2013
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    I can see that the use of s will greatly simplify the equations, but I don't have any values which plug into a final equation to get the values required in the question above. All I have is a value for the supply voltage, values of capacitance for the capacitors and inductance for the inductor and the resistance values for the resistors. Without the frequency I can't see how I can calculate impedance values for the capacitors and inductor...

    All of the values that I have to work with can be seen in the original post containing the question...
     
  8. shteii01

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    The expression for voltage source, 18cos(10t), gives you three pieces of information.

    The more exact expression for voltage source is actually 18cos(10t+0°)
    1) 18 is amplitude of the voltage in volts
    2) 10 is angular frequency w in rad/s
    3) 0° is phase angle in degrees
    See? Three pieces of information.

    Example.
    Remember the earlier 2 H inductor?
    XL=jwL=j(10 rad/s)(2 H)=j20 Ohm
    You can now replace the 2 H inductor with equivalent j20 Ohm impedance.
     
  9. simon.r.potter

    Thread Starter New Member

    Nov 30, 2013
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    My god thank you for this...

    Its starting to make sense!

    Thanks a BUNCH!

    Really cleared up my mind!

    p.s would you mind if I ask what you do for a living? In other words how do you know so much about electronics?

    Cheers,
    Simon
     
  10. shteii01

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    Feb 19, 2010
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    Funny you should ask ☺

    I graduated with BSEE from Pardue University this year. So these stuff is still fresh in my mind.
     
    Last edited: Dec 1, 2013
  11. WBahn

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    Have you had differential equations yet? If so, then this might help hang something in context. The 's' is the variable you end up with when you take the Laplace Transform of a differential equation.

    If you develop the solution for the circuit you have in the "time domain", you would have a pretty nasty differential equation stemming from the fact that, for each inductor, you have v(t) = L di(t)/dt and for every capacitor i(t) = C dv(t)/dt. But when you take the Laplace Transform, you end up with a set of algebraic equations in what is known as the "complex frequency" domain that are much easier to solve. Then you take the Inverse Laplace Transform of the solution to get an answer that is back in the time domain.

    In general, 's' is a complex value, usually expressed in rectangular form as s=σ+jω.

    If we set σ=0, so that s=jω, we end up with the Fourier Transform, which is just a special subset of the Laplace Transform.

    The Laplace Transform let's use capture both the transient response due to arbitrary changes in the signal or the circuit (i.e., closing switches and such) as well as the final steady state behavior, while the Fourier Transform gives us a way to greatly simplify things when all we are interested is the steady state behavior after all the transients have died out.

    Whether we use the Laplace Transform or the Fourier Transform, the great utility is that we can transform all of the components (resistors, inductors, and capacitors) into "impedances" that we can treat the same way we treated resistors when all we had were DC sources and resistors in our circuits. The big differences is that now our "impedances" are complex numbers and so the math to combine them, using the same series and parallel formulas we used in the DC case, is a bit more complicated. But that's a small price to pay for getting to avoid having to solve the differential equations directly!
     
  12. Marcin

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    Dec 1, 2013
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    Hello Mr.

    I study with Simon... he shared this link with me.
    I'm little confused about sign of XC. Shouldn't be neg. due to character of this component? I have no idea, just using lots of logic where possible.
    Thanks for reply in advance.
     
  13. shteii01

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    Impedance of capacitor (XC).
    XC=1/(jwC)
    One of the elements that you have here is 1/j. However. There is another way to represent 1/j. 1/j is the same as -j.
    So. XC=1/(jwC)=(1/j)(1/(wC))=-j(1/(wC))=-j/(wC)
    That is how you get the negative sign.

    This is from wiki, it shows how 1/j and -j is the same thing:
    1/j=(1/j)(j/j)=j/(j^2)=j/(-1)=-j
    http://en.wikipedia.org/wiki/Imaginary_unit#Multiplication_and_division
     
  14. WBahn

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    Somewhat surprisingly, this isn't a simple question. That's because there are two schools of thought regarding reactance.

    The explanation by shteii01 pretty well covers one school of thought, though I want to make some terminology clarifications.

    You have "impedance", usually symbolized by Z, and it has a real component, which we call the "resistance", usually symbolized by R, and an imaginary component, which we call the "reactance", usually symbolized by X.

    Somewhat counterintuitively (at least to me), the "imaginary component" is a real number -- specifically it is the real coefficient of the imaginary unit, j, in a complext number.

    Bring this all together, we have

    Z = R + jX

    Now we can start with the impedance of our three main components:

    Z_R = R
    Z_L = jωL
    Z_C = 1/(jωC) = (j/j)*(1/(jωC)) = j/(-ωC) = j(-1/(ωC))

    Doing match up, we have

    R_R = R
    X_R = 0

    R_L = 0
    X_L = ωL

    R_C = 0
    X_C = -1/(ωC)

    This is one school of thought and in this school of thought you have resistance and reactance. Resistance is (almost) always positive while reactance can be positive or negative depending on if it is inductive or capacitive.

    The other school of thought centers around formulas and geometric interpretations of them and does not have complex numbers. So in this school of thought you have two types of reactance, both of which are positive. You have inductive reactance X_L = 2∏fL and you have capacitive reactance, X_C = 1/(2∏fC). Note that this school of thought also almost invariably uses hertzian frequencies whereas the prior school of thought, while it uses both, tends to focus more on radian frequency. You then have in all your formulas (X_L-X_C). So the negative sign is turned into a subtraction operator in a formula.

    The bottom line is that you will see widely both schools of thought represented. So when you see the definition (or computation) of the reactance of a capacitor and it doesn't have a negative sign, it is almost certainly just a case of being in this second school of thought. It's important to recognize that, though, because that will let you properly interpret subtraction operators, or lack thereof, in various formulas. It's even more important to be sure that the formulas you use match the school of thought in which the definition of reactance is coming from.
     
  15. tbinder3

    Member

    Jun 30, 2013
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    I just follow it easier, and more of an understanding this way.

    Xc=1/2∏(f)( C) and XL= 2∏(f)(L)

    f= frequency
    C= capacitance
    L= henrys or inductance

    In order to add or multiply these to a get a final impedance value to later get your current, you must understand Polar and Rectangular Notation. I made this post the other day on another thread. Just realized this follows up on WBahn's post, after understanding the true concept of it other than going straight to the algebra, also he expressed his reactance formulas the same lol

    "Polar Notation is a complex number, a Magnitude and Vector angle.
    Example: 12v<0° The 12v being your magnitude, and 0° being your vector angle.

    When solving for total current, you must divide two complex numbers. I=V/Z
    Hoping that you know how to solve for impedance.

    In my example i'm using the voltage source as a reference, at 0°
    Continuing my example: Giving us 12v<0° / 12Ω<30°

    Now, when dividing vectors, you divide the magnitudes (12v / 12Ω =1A) AND subtract the angles (0° - 30° =-30°). Giving us I total= 1A<-30°.

    Just want to point out that, when subtracting angles, it is common to subtract by a negative number, example: (0° - -30°) resulting in a positive angle.

    If you were to add or subtract a vector, you would have to put it in Rectangular Notation first. Rectangular notation has 2 parts as well, we will use the terms "Real" and "Imaginary" . It will look something like this (-1 + J6) The "-1" being the "real" part, and "J6" being the so-called "imaginary" part. Also want to make note here, that a resistor example in rectangular notation would like similar to 100ohm + J0. And reactive components like a cap or inductor would have one similar to 0 - J100. When adding rectangular numbers as such, it would be "Real" +- "Imaginary" giving us for our example: 100 - J100

    When converting a Polar Notation number to Rectangular Notation, I memorized this example: C(magnitude)<θ°(angle) (polar notation) C<θ°=polar notation
    The Real part= C * cos(θ) or Magnitude * cos(Angle)
    Imaginary= C * sin(θ) or Magnitude * sin(Angle)

    Example: 1A<-30°
    Real = 1 * cos(-30) = .866
    Imaginary = 1 * sin(-30) = -.5

    Answer= .866<-.5°

    Of course there is a lot going on, and a lot more to understand, and i'm sure i missed basic concepts, but this was put together quickly as an example."

    lol
     
    Last edited: Dec 3, 2013
  16. Marcin

    New Member

    Dec 1, 2013
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    Thank You very much for blowing out my uncertainty.
    Now I'm sure what to do.
    Xc= -j(1/(ωC))
    all XL are positive and all Xc are neg in Ω's
    Thanks again
     
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