Mesh analysis

Discussion in 'General Electronics Chat' started by Ndjs, Mar 1, 2013.

  1. Ndjs

    Thread Starter New Member

    Sep 26, 2012
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    0
    00216.png

    Loop 1

    loop1.png

    Loop 2

    loop2.png

    My question is relation to (I1+I2) why is it positive or negative in each equation.

    I am unable to find a good way to consistently know whether it is it a negative -R(I1+I2) or a positive +R(I1+I2)


    I also find it hard to label the polarity of the resistors any help would be great I have taken the pictures from. http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

    I am still not able to understand completely.
     
    Last edited: Mar 1, 2013
  2. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    This is not a pure math question, so I have moved it to a forum where it is a better fit.
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,748
    4,796
    You can randomly assign the polarities of your voltages and currents however you want, but then the signs that you use in your equations have to be consistent with the polarities you've assigned.

    If a current is fllowing through a resistor from the positive terminal to the negative terminal, then the voltage of the positive terminal relative to the negative terminal will be positive.

    The "passive sign convention" says that, for a passive component, positive current flows through the device from the positive terminal to the negative terminal. For a source, the opposite is true. This is so that the power for both will turn out to be positive when they are behaving the way you expect them to, namely sources delivering power and passive components absorbing it. Sometimes you don't know whether a device, particularly a source, is acting as a source or a load. So you guess and declare it to be one or the other. If the power turns out to be negative, then you guessed wrong, but you don't have to redo the analysis.

    in your diagram you have R1, R2 and R3. I'm going to use V1, V2, and V3 as the voltages across them using the polarities you've established. I'm going to use I1, I2, and I3 as the currents through each of the resistors with the polarity per the passive sign convention. What you call I1 and I2 I am going to call Ia and Ib.

    So first perform KVL around the two loops:

    Loop A: B1 - V1 - V2 = 0
    Loop B: V2 + V3 - B2 = 0

    In both cases I went around the loop clockwise totaling the voltage increases. But I could have gone the other way in one or the other and I could have totaled votlage drops in one or the other. As long as I am consistent within each equation, the only effect is that both sides of the equation get multiplied by -1, which doesn't change anything.

    Now I can replace the voltages across the resistors with the currents through them via Ohm's Law. I don't even have to glance at the diagram because the passive sign convention guarantees that Vx = Ix*Rx.

    Loop A: B1 - I1*R1 - I2*R2 = 0
    Loop B: I2*R2 + I3*R3 - B2 = 0

    Now I look at Ia and Ib, as you have them defined (your I1 and I2, respectively) and I see that both loop currents are always in the opposite direction of the device currents. So I have:

    I1 = -Ia
    I2 = -Ia + -Ib = -(Ia+Ib)
    I3 = -Ib

    Now I substitute these in to the prior set of equations:

    Loop A: B1 - (-Ia)*R1 - (-(Ia+Ib))*R2 = 0
    Loop B: (-(Ia+Ib))*R2 + (-Ib)*R3 - B2 = 0

    A bit of simplifying:

    Ia*(R1+R2) + Ib*R2 = -B1
    Ia*R2 + Ib*(R2+R3) = -B2

    With a bit of practice (and not very much, either), you will be able to write these equations down by inspection as your starting point.

    From there it is straight forward algebra. Notice that I have not used the numerical values for anything. Instead, I have done it symbolically. Don't plug numbers in until the end. That way, if you make a mistake, you can track down the mistake much more easily and patch the work instead of having to start all over. Also, track your units throughout your work. As soon as you put a numerical value into an expression, that value has units that are a fundamental part of it. Leaving them off is like only putting in some of the digits. This is important because most mistakes you make will screw up the units. Frequently, you will see the units getting screwed up, like trying to add voltage to current, right when it happens. But other times you won't see it until later. But if you track the units and, at some point, notice that they aren't correct then you can confidently stop right there because you KNOW the answer is going to be wrong. So instead of wasting time continuing to work meaningless drivel, you can go back and track down the error (which the units will greatly facilitate) and fix the problem and get moving along the correct path much quicker.
     
  4. Ndjs

    Thread Starter New Member

    Sep 26, 2012
    18
    0
    Thank you my biggest issue was not putting down current direction assumed through the resistors in respect to the polarity I chose for these.

    I will do a few questions and see how I go your explanation is excellent and easy to follow. I think a few questions with the correct current direction through the resistors should sort me out.

    I can do the maths side of it fine just not the setting up of the equations when it comes to 2 loop currents being involved.
     
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