Understanding Input Resistance

[s3a]

Homework Statement

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.

Homework Equations

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

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!

 

[aralbrec]

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

That is Cramer's Rule used to solve for an unknown in a matrix equation.

 

[aralbrec]

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

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

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.

Δ_R 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.

 

[s3a]

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?

 

[aralbrec]

(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 I₁, I₂, ..., 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 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
$\left[ \begin{array}{ccc} 19 & -12 & 0 \\ -12 & 18 & -6 \\ 0 & -6 & 18 \end{array} \right] \left[ \begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \end{array} \right] = \left[ \begin{array}{c} 60 \\ 0 \\ 0 \end{array} \right]$

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 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 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.

(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:

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

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

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

where Δ₁ is Δ_Z with the first column replaced by V and Δ₁₁ is Δ_Z 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

$R_{1} = \frac{V_{1}}{I_{1}} = \frac{V_{1}}{V_{1} \Delta_{11} / \Delta_{Z}} = \frac{\Delta_{Z}}{\Delta_{11}}$

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.

 

[s3a]

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)

zero out all sources (how would this change vector*V?)

V would be a 0 column vector.

and

2)

apply a test voltage V1*(how would this change vector*V)

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.)