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Obtain the input resistance

  1. Jan 10, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem I would like help with is 4.4 which is in the file 4.4.jpg. (The problem requires problem 4.2 which is why I included it as 4.2.jpg.) The solution is included however, I do not understand it since it is very brief and, I don't think I've covered this before.

    2. Relevant equations
    V = IR
    R_(input,1) = ΔR/Δ11 (This equation makes no sense to me.)

    3. The attempt at a solution
    I can see how I_1 = V_(input,1)/R_(input,1) = 60/R_(input,1) = 60/10 = 6 A but, the first equation with the deltas confuses me.

    Could someone please explain to me the logic behind that equation?

    Any help would be greatly appreciated!
     

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  2. jcsd
  3. Jan 10, 2013 #2
    That is Cramer's Rule used to solve for an unknown in a matrix equation.
     
  4. Jan 10, 2013 #3
    Sorry I was pulled away and it was too late to edit.

    It's an application of Cramer's Rule, maybe processed enough that it's not obvious.

    Your mesh matrix equation is V=ZI where the entries of the matrix V are the voltage sources in each loop. To find the input resistance at the left of the 7 ohm resistor (let's call it node 1), you'd set all the other independent sources to 0 and apply a test voltage V1 (=60 V, for example). Your matrix equation is already set up like that since the only independent source is V1=60.

    So

    [itex]R_{in,1}=\frac{V_{1}}{I_{1}}=\frac{V_{1}}{\Delta_{1} / \Delta}=\frac{V_{1} \Delta}{V_1 \left| \stackrel{18 \ -6}{-6 \ 18} \right|}=\frac{\Delta_{R}}{\Delta_{11}}[/itex]

    where Cramer's Rule was used first to substitute for I1 and then a cofactor expansion of Δ1 around V1=60 led to the final form.

    ΔR is the determinant of the impedance matrix Z and Δ11 is the determinant of the same matrix when row 1 and column 1 have been removed. This is a result of doing a cofactor expansion along the column containing V1 at position (1,1) in the matrix with all other voltage sources set to 0.
     
    Last edited: Jan 10, 2013
  5. Jan 10, 2013 #4

    s3a

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    Actually, I need to take a few steps back even from that.

    (1) I just realized the "R" and "11" were subscripts to the delta.

    (2) I don't know what a mesh matrix equation is.

    (3) I'm having trouble filling in the values for the matrices which, if I'm understand correctly, are represented by the deltas with their subscripts.

    Could you help me from here please?
     
  6. Jan 11, 2013 #5
    The deltas are just a common notation for determinant.

    It's the system of equations you get from mesh analysis put in matrix form.

    Problem 4.2 that you posted gives a completely worked out example of this. In mesh analysis, you assume currents I1, I2, ..., In flow around closed loops in the circuit. KVL is then written for each loop and you wind up with a system of equations.

    In 4.2, there are three loops with associated currents I1, I2, I3.

    Loop 1 has a 60V source in it, a resistance of 7 ohms with current I1 flowing through it and a resistance of 12 ohms with net current (I1-I2) flowing through it. The currents in the 12 ohm resistor subtract because they flow in opposite directions. The KVL equation for loop 1 is then:

    60 = 7 I1 +12(I1-I2)

    The worked out problem 4.2 shows the other two KVL equations and puts the system of three equations into matrix form:

    Z I = V
    [itex]
    \left[ \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right]
    \left[ \begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \end{array} \right]
    =
    \left[ \begin{array}{c} 60 \\ 0 \\ 0 \end{array} \right]
    [/itex]

    There is a shortcut in writing this matrix equation which you can see if you collect like terms in the KVL equations written in your solution to 4.2. If you look at the top row of the matrix equation above, this is KVL for the first loop with I1 in it. The voltage drop around the loop equals the net voltage increase supplied by sources (in this case 60V). The first entry in the first row of matrix Z, 19, is the total resistance in loop 1 with I1 passing through it. The second entry, -12, is the negative of the total resistance in loop 1 with I2 passing through it. It is negative because I2 flows through this the resistance in the opposite direction of I1, so in the KVL equation around loop 1 where the direction of I1 represents voltage drops across resistors, the direction of I2 will represent a voltage rise. The last entry in the top row is 0. I3 does not pass through any resistances in loop 1 so I3 does not appear in the KVL equation around loop 1. These resistances are multiplied by the current vector to get the total voltage drop around a loop. The currents in mesh analysis are always assumed to flow in a clockwise direction -- this will ensure that the net current flowing through common branches will be the difference between two mesh currents and therefore the signs of the resistances in matrix Z will always follow the pattern described -- in row m, corresponding to KVL for loop m, the entry multiplying Im will be positive as Im is the reference direction for current flow in that loop. All other resistances in that row of Z will be negative as all other mesh currents will be flowing in the opposite direction of the reference direction Im in that loop.

    Wow, hard to write down in words. Hopefully you already know this well enough to follow.

    Just to confirm some understanding, the second row of the matrix equation represents KVL around the second loop corresponding to current I2.

    So the second row of Z = [ -12 18 -6 ] because in the second loop, I1 flows through a total resistance of 12 ohms, I2 flows through a total resistance of 12+6=18 ohms, and I3 flows through a total resistance of 6 ohms. I2 is the reference direction for KVL in loop 2 so its resistance will be positive representing a voltage drop of 18*I2. The other resistances are negative because I1 and I3 flow in the opposite direction of I2 in loop 2. There are no voltage sources in loop 2 so KVL around the loop will = 0 and V2 is therefore 0 in the matrix equation.

    Knowing this allows you to write the Z and V matrices by inspection rather than painstakingly writing out the system of equations and going to matrix form from there. The other important thing to take away is that non-zero entries in the V vector represent the total voltage in the loop supplied by voltage sources.

    The deltas are determinants. The determinant of matrix Z is denoted ΔZ or in this case because all impedances are resistances your solution's notation is using ΔR for the determinant of the Z matrix:

    [itex]
    \Delta_{Z} = \Delta_{R} = \left| \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right|
    [/itex]

    From the matrix equation above, you can solve for individual loop current In using Cramer's Rule:

    [itex]
    I_{1} = \frac{\Delta_{1}}{\Delta_{Z}} = \frac{\left| \begin{array}{ccc} 60 & -12 & 0 \\ 0 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right|}{\left| \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right|} = \frac{60 \left| \begin{array}{cc} 18 & -6 \\ -6 & 18 \end{array} \right|}{\left| \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right|} = \frac{60 \Delta_{11}}{\Delta_{Z}}
    [/itex]

    where Δ1 is ΔZ with the first column replaced by V and Δ11 is ΔZ with the first row and first column removed and is made possible by cofactor expansion along the first column of Δ1.

    You can write similar equations for I2 and I3 but we are interested in finding R1, the input resistance V1/I1. To find input resistance we would zero out all sources (how would this change vector V?), apply a test voltage V1 (how would this change vector V) and divide by the current I1 passing through the test source. We don't need to zero out anything as there is only one voltage source V1=60 that we will also use as the test voltage.

    So

    [itex]
    R_{1} = \frac{V_{1}}{I_{1}} = \frac{V_{1}}{V_{1} \Delta_{11} / \Delta_{Z}} = \frac{\Delta_{Z}}{\Delta_{11}}
    [/itex]

    which is what you were asking about originally. The next question is can you go straight to this formula as the written solution did? You probably can if you pay close attention to how vector V is modified by zeroing all sources and applying a test voltage in the loop in question. You would be measuring the resistance seen by that voltage source by computing V/I with I=the current through the source. There would be a little more complication if the test source was in a branch shared by two mesh currents.
     
    Last edited: Jan 11, 2013
  7. May 2, 2013 #6

    s3a

    User Avatar

    Hello, aralbrec.

    Sorry for the late response; it's because I was dealing with school. Now that school is done, I can continue with my own no-panic learning. :)

    Thank you very much. :) Thanks to you, I now understand how to do this problem (which was my main goal) but, since you asked the following two questions, I would like to answer them:

    1)
    V would be a 0 column vector.

    and

    2)
    Are you saying, for this thought experiment, that the 60 V power supply would be replaced by a test voltage ##V_1##? If so, then, I believe you would just replace the 60 with the variable ##V_1##.

    (Ultimately, I don't think it matters if I didn't grasp that since I now understand what I wanted to overcome but, I'm still a little curious.)
     
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