Linear Algebra/Topic: Analyzing Networks

Problem

[] Current Law. For any point in a network, the flow in

Problem

[] Voltage Law. Around any circuit the total drop equals

the total rise. \end{itemize} In the above network there is only one voltage rise, at the battery, but some networks have more than one.

For a start we can consider the network below. It has a battery that provides the potential to flow and three resistors (resistors are drawn as zig-zags). When components are wired one after another, as here, they are said to be in series.^[Index]

By Kirchhoff's Voltage Law, because the voltage rise is 20~volts, the total voltage drop must also be 20~volts. Since the resistance from start to finish is 10~ohms (the resistance of the wires is negligible), we get that the current is (20/10)=2~amperes. Now, by Kirchhoff's Current Law, there are 2~amperes through each resistor. (And therefore the voltage drops are: 4~volts across the 2~oh m resistor, 10~volts across the 5~ohm resistor, and 6~volts across the 3~ohm resistor.)

The prior network is so simple that we didn't use a linear system, but the next network is more complicated. In this one, the resistors are in parallel.^[Index] This network is more like the car lighting diagram shown earlier.

We begin by labeling the branches, shown below. Let the current through the left branch of the parallel portion be i_1 and that through the right branch be i_2, and also let the current through the battery be i_0. (We are following Kirchoff's Current Law; for instance, all points in the right branch have the same current, which we call i_2. Note that we don't need to know the actual direction of flow— if current flows in the direction opposite to our arrow then we will simply get a negative number in the solution.)

The Current Law, applied to the point in the upper right where the flow i_0 meets i_1 and i_2, gives that i_0=i_1+i_2. Applied to the lower right it gives i_1+i_2=i_0. In the circuit that loops out of the top of the battery, down the left branch of the parallel portion, and back into the bottom of the battery, the voltage rise is 20 while the voltage drop is i_1\cdot 12, so the Voltage Law gives that 12i_1=20. Similarly, the circuit from the battery to the right branch and back to the battery gives that 8i_2=20. And, in the circuit that simply loops around in the left and right branches of the parallel portion (arbitrarily taken clockwise), there is a voltage rise of 0 and a voltage drop of 8i_2-12i_1 so the Voltage Law gives that 8i_2-12i_1=0. \begin{displaymath} \end{displaymath} The solution is i_0=25/6, i_1=5/3, and i_2=5/2, all in amperes. (Incidentally, this illustrates that redundant equations do indeed arise in practice.)

Kirchhoff's laws can be used to establish the electrical properties of networks of great complexity. The next diagram shows five resistors, wired in a series-parallel^[Index] way.

This network is a Wheatstone bridge^[Index] (see \nearbyexercise{exer:WheatstoneBr}). To analyze it, we can place the arrows in this way.

Kirchoff's Current Law, applied to the top node, the left node, the right node, and the bottom node gives these. Failed to parse (syntax error): i_0 = i_1+i_2 \\ i_1 = i_3+i_5 \\ i_2+i_5 = i_4 \\ i_3+i_4 = i_0

Kirchhoff's Voltage Law, applied to the inside loop (the i_0 to~i_1 to~i_3 to~i_0 loop), the outside loop, and the upper loop not involving the battery, gives these. Failed to parse (syntax error): 5i_1+10i_3 = 10 \\ 2i_2+4i_4 = 10 \\ 5i_1+50i_5-2i_2 = 0

Those suffice to determine the solution i_0=7/3, i_1=2/3, i_2=5/3, i_3=2/3, i_4=5/3, and i_5=0.

Networks of other kinds, not just electrical ones, can also be analyzed in this way.

Problem

[] Many of the systems for these problems

Problem

Calculate the amperages in each part of each network.

1. This is a simple network.

1. Compare this one with the parallel case discussed above.

1. This is a reasonably complicated network.

Answer

1. The total resistance is 7~ohms.

With a 9~volt potential, the flow will be 9/7~amperes. Incidentally, the voltage drops will then be:~27/7~volts across the 3~ohm resistor, and 18/7~volts across each of the two 2~ohm resistors.

1. One way to do this network is to note that the 2~ohm

resistor on the left has a voltage drop across it of 9~volts (and hence the flow through it is 9/2~amperes), and the remaining portion on the right also has a voltage drop of 9~volts, and so is analyzed as in the prior item.

We can also use linear systems.

Using the variables from the diagram we get a linear system </math> which yields the unique solution i_1=81/14, i_1=9/2, i_2=9/7, and i_3=81/14.

Of course, the first and second paragraphs yield the same answer. Esentially, in the first paragraph we solved the linear system by a method less systematic than Gauss' method, solving for some of the variables and then substituting.

Using these variables

one linear system that suffices to yield a unique solution is this. Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): <math>\begin{bmatrix} i_0 &- &i_1 &- &i_2 & & & & & & & & = &0 \\ & & & &i_2 &- &i_3 &- &i_4 & & & & = &0 \\ & & & & & &i_3 &+ &i_4 &-&i_5 & & = &0 \\ & &i_1 & & & & & & &+&i_5 &-&i_6 = &0 \\ & &3i_1 & & & & & & & & & & = &9 \\ & & & &3i_2 & & &+ &2i_4&+&2i_5& & = &9 \\ & & & &3i_2 &+ &9i_3& & &+&2i_5& & = &9 \end{bmatrix}

</math> (The last three equations come from the circuit involving i_0-i_1-i_6, the circuit involving i_0-i_2-i_4-i_5-i_6, and the circuit with i_0-i_2-i_3-i_5-i_6.) Octave gives i_0=4.35616, i_1=3.00000, i_2=1.35616,

Problem

In the first network that we analyzed, with the three resistors in series, we just added to get that they acted together like a single resistor of 10~ohms. We can do a similar thing for parallel circuits. In the second circuit analyzed,

the electric current through the battery is 25/6~amperes. Thus, the parallel portion is equivalent^[Index] to a single resistor of 20/(25/6)=4.8~ohms.

1. What is the equivalent resistance if we change

the 12~ohm resistor to 5~ohms?

1. What is the equivalent resistance if the two are each

8~ohms?

1. Find the formula for the equivalent resistance if

the two resistors in parallel are r_1~ohms and r_2~ohms.

Answer

Using the variables from the earlier analysis, \begin{displaymath} \end{displaymath} The current flowing in each branch is then is i_2=20/8=2.5, i_1=20/5=4, and i_0=13/2=6.5, all in amperes. Thus the parallel portion is acting like a single resistor of size 20/(13/2)\approx 3.08~ohms.

A similar analysis gives that is i_2=i_1=20/8=4 and i_0=40/8=5~ amperes. The equivalent resistance is 20/5=4~ohms.

Another analysis like the prior ones gives is i_2=20/r_2, i_1=20/r_1, and i_0=20(r_1+r_2)/(r_1r_2), all in amperes. So the parallel portion is acting like a single resistor of size 20/i_1=r_1r_2/(r_1+r_2)~ohms. (This equation is often stated as:~the equivalent

Problem

For the car dashboard example that opens this Topic, solve for these amperages (assume that all resistances are 2~ohms).

1. If the driver is stepping on the brakes, so the

brake lights are on, and no other circuit is closed.

1. If the hi-beam headlights and the brake lights are on.

Answer

1. The circuit looks like this.

1. The circuit looks like this.

Problem

Show that, in this Wheatstone Bridge,

r_2/r_1 equals r_4/r_3 if and only if the current flowing through r_g is zero. (The way that this device is used in practice is that an unknown resistance at r_4 is compared to the other three r_1, r_2, and r_3. At r_g is placed a meter that shows the current. The three resistances r_1, r_2, and r_3 are varied— typically they each have a calibrated knob— until the current in the middle reads 0, and then the above equation gives the value of r_4.)

Answer

Problem

[]There are networks other than electrical ones, and

we can ask how well Kirchoff's laws apply to them. The remaining questions consider an extension to

Problem

Consider this traffic circle.

This is the traffic volume, in units of cars per five minutes.

We can set up equations to model how the traffic flows.

Adapt Kirchoff's Current Law to this circumstance. Is it a reasonable modelling assumption?

Label the three between-road arcs in the circle with a variable. Using the (adapted) Current Law, for each of the three in-out intersections state an equation describing the traffic flow at that node.

Solve that system.

Interpret your solution.

Restate the Voltage Law for this circumstance. How reasonable is it?

Answer

An adaptation is:~in any intersection the flow in equals the flow out. It does seem reasonable in this case, unless cars are stuck at an intersection for a long time.

We can label the flow in this way.

Because 50 cars leave via Main while 25~cars enter, i_1-25=i_2. Similarly Pier's in/out balance means that i_2=i_3 and North gives i_3+25=i_1. We have this system. </math>

The row operations \rho_1+\rho_2 and rho_2+\rho_3 lead to the conclusion that there are infinitely many solutions. With i_3 as the parameter, of course, since the problem is stated in number of cars, we might restrict i_3 to be a natural number.

If we picture an initially-empty circle with the given input/output behavior, we can superimpose a z_3-many cars circling endlessly to get a new solution.

A suitable restatement might be:~the number of cars entering the circle must equal the number of cars leaving. The reasonableness of this one is not as clear. Over the five minute time period it could easily work out that a half dozen more cars entered than left, although the into/out of table in the problem statement does have that this property is satisfied. In any event it is of no help in getting a unique solution since for that we need to know the number of cars circling

Problem

This is a network of streets.

The hourly flow of cars into this network's entrances, and out of its exits can be observed.

(Note that to reach Jay a car must enter the network via some other road first, which is why there is no `into Jay' entry in the table. Note also that over a long period of time, the total in must approximately equal the total out, which is why both rows add to 235~cars.) Once inside the network, the traffic may flow in different ways, perhaps filling Willow and leaving Jay mostly empty, or perhaps flowing in some other way. Kirchhoff's Laws give the limits on that freedom.

1. Determine the restrictions on the flow inside this network

of streets by setting up a variable for each block, establishing the equations, and solving them. Notice that some streets are one-way only. (Hint:~this will not yield a unique solution, since traffic can flow through this network in various ways; you should get at least one free variable.)

1. Suppose that some construction is proposed for

Winooski Avenue East between Willow and Jay, so traffic on that block will be reduced. What is the least amount of traffic flow that can be allowed on that block without disrupting the hourly flow into and out of the network?

Answer

1. Here is a variable for each unknown block; each known

block has the flow shown.

We apply Kirchoff's principle that the flow into the intersection of Willow and Shelburne must equal the flow out to get i_1+25=i_2+125. Doing the intersections from right to left and top to bottom gives these equations. Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): <math>\begin{bmatrix} i_1 &- &i_2 & & & & & & & & & & = &10 \\ -i_1 & & &+ &i_3 & & & & & & & & = &15 \\ & &i_2 & & &+ &i_4 & & & & & & = &5 \\ & & & &-i_3&- &i_4 & & &+ &i_6 & & = &-50 \\ & & & & & & & &i_5 & & &- &i_7 = &-10 \\ & & & & & & & & & &-i_6&+ &i_7 = &30 \end{bmatrix}

</math> The row operation \rho_1+\rho_2 followed by \rho_2+\rho_3 then \rho_3+\rho_4 and \rho_4+\rho_5 and finally \rho_5+\rho_6 result in this system. Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): <math>\begin{bmatrix} i_1 &- &i_2 & & & & & & & & & & = &10 \\ & &-i_2&+ &i_3 & & & & & & & & = &25 \\ & & & &i_3 &+ &i_4 &- &i_5 & & & & = &30 \\ & & & & & & & &-i_5&+ &i_6 & & = &-20 \\ & & & & & & & & & &-i_6&+ &i_7 = &-30 \\ & & & & & & & & & & & &0 = &0 \end{bmatrix}

</math> Since the free variables are i_4 and i_7 we take them as parameters. Failed to parse (unknown function\begin): \begin{split} i_6 = i_7-30 \\ i_5 = i_6+20=(i_7-30)+20=i_7-10 \\ i_3 = -i_4+i_5+30=-i_4+(i_7-10)+30=-i_4+i_7+20 \\ i_2 = i_3-25=(-i_4+i_7+20)-25=-i_4+i_7-5 \\ i_1 = i_2+10=(-i_4+i_7-5)+10=-i_4+i_7+5 \end{split} {}

Obviously i_4 and i_7 have to be positive, and in fact the first equation shows that i_7 must be at least 30. If we start with i_7, then the i_2~equation shows that 0\leq i_4\leq i_7-5.

1. We cannot take i_7 to be zero or else i_6 will

be negative (this would mean cars going the wrong way on the one-way street Jay). We can, however, take i_7 to be as small as 30, and then there are many suitable i_4's. For instance, the solution (i₁,i₂,i₃,i₄,i₅,i₆,i₇) = (35,25,50,0,20,0,30) results from choosing i_4=0.