s3a said:
(1) I just realized the "R" and "11" were subscripts to the delta.
The deltas are just a common notation for determinant.
(2) I don't know what a mesh matrix equation is.
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 I
1, I
2, ..., I
n 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 I
1, I
2, I
3.
Loop 1 has a 60V source in it, a resistance of 7 ohms with current I
1 flowing through it and a resistance of 12 ohms with net current (I
1-I
2) 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 I
1 +12(I
1-I
2)
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
<br />
\left[ \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right]<br />
\left[ \begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \end{array} \right]<br />
=<br />
\left[ \begin{array}{c} 60 \\ 0 \\ 0 \end{array} \right]<br />
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 I
1 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 I
1 passing through it. The second entry, -12, is the negative of the total resistance in loop 1 with I
2 passing through it. It is negative because I
2 flows through this the resistance in the opposite direction of I
1, so in the KVL equation around loop 1 where the direction of I
1 represents voltage drops across resistors, the direction of I
2 will represent a voltage rise. The last entry in the top row is 0. I
3 does not pass through any resistances in loop 1 so I
3 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 I
m will be positive as I
m 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 I
m 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 I
2.
So the second row of
Z = [ -12 18 -6 ] because in the second loop, I
1 flows through a total resistance of 12 ohms, I
2 flows through a total resistance of 12+6=18 ohms, and I3 flows through a total resistance of 6 ohms. I
2 is the reference direction for KVL in loop 2 so its resistance will be positive representing a voltage drop of 18*I
2. The other resistances are negative because I
1 and I
3 flow in the opposite direction of I
2 in loop 2. There are no voltage sources in loop 2 so KVL around the loop will = 0 and V
2 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.
(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.
The deltas are determinants. The determinant of matrix
Z is denoted Δ
Z or in this case because all impedances are resistances your solution's notation is using Δ
R for the determinant of the
Z matrix:
<br />
\Delta_{Z} = \Delta_{R} = \left| \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right|<br />
From the matrix equation above, you can solve for individual loop current I
n using Cramer's Rule:
<br />
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}}<br />
where Δ
1 is Δ
Z with the first column replaced by
V and Δ
11 is Δ
Z with the first row and first column removed and is made possible by cofactor expansion along the first column of Δ
1.
You can write similar equations for I
2 and I
3 but we are interested in finding R
1, the input resistance V
1/I
1. To find input resistance we would zero out all sources (how would this change vector
V?), apply a test voltage V
1 (how would this change vector
V) and divide by the current I
1 passing through the test source. We don't need to zero out anything as there is only one voltage source V
1=60 that we will also use as the test voltage.
So
<br />
R_{1} = \frac{V_{1}}{I_{1}} = \frac{V_{1}}{V_{1} \Delta_{11} / \Delta_{Z}} = \frac{\Delta_{Z}}{\Delta_{11}}<br />
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.
