Linear Algebra/Topic: Analyzing Networks/Solutions

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Solutions[edit]

Many of the systems for these problems are mostly easily solved on a computer.

Problem 1

Calculate the amperages in each part of each network.

  1. This is a simple network.

    Linalg resisters in series 2.png

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

    Linalg circuit 1.png

  3. This is a reasonably complicated network.

    Linalg circuit 2.png

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.
  2. 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.

    Linalg circuit 1 2.png

    Using the variables from the diagram we get a linear system

    
\begin{array}{*{4}{rc}r}
i_0  &- &i_1  &- &i_2  &  &    &= &0  \\
&  &i_1  &+ &i_2  &- &i_3 &= &0  \\
&  &2i_1 &  &     &  &    &= &9  \\
&  &     &  &7i_2 &  &    &= &9
\end{array}

    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. Essentially, 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.

  3. Using these variables

    Linalg circuit 2 2.png

    one linear system that suffices to yield a unique solution is this.

    
\begin{array}{*{7}{rc}r}
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{array}

    (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, i_3=0.24658, i_4=1.10959, i_5=1.35616, i_6=4.35616.

Problem 2

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,

Linalg resisters in parallel.png

the electric current through the battery is 25/6 amperes. Thus, the parallel portion is equivalent 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?
  2. What is the equivalent resistance if the two are each 8 ohms?
  3. Find the formula for the equivalent resistance if the two resistors in parallel are r_1 ohms and r_2 ohms.
Answer
  1. Using the variables from the earlier analysis,
    
\begin{array}{*{3}{rc}r}
i_0&- &i_1    &-  &i_2   &=  &0 \\
-i_0&+ &i_1    &+  &i_2   &=  &0  \\
&  &5i_1   &   &      &=  &20  \\
&  &       &   &8i_2  &=  &20  \\
&  &-5i_1  &+  &8i_2  &=  &0
\end{array}
    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.
  2. 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.
  3. 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 resistance r satisfies 1/r=(1/r_1)+(1/r_2).)
Problem 3

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.
  2. If the hi-beam headlights and the brake lights are on.
Answer
  1. The circuit looks like this.

    Linalg car problem 1.png

  2. The circuit looks like this.

    Linalg car problem 3.png

Problem 4

Show that, in this Wheatstone Bridge,

Linalg wheatstone bridge 3.png

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

Not yet done.

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 networks of streets.

Problem 5

Consider this traffic circle.

Linalg rotary.png

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


\begin{array}{r|c|c|c}
&\textit{North}  &\textit{Pier}  &\textit{Main}  \\
\hline
\textit{into}      &100             &150            &25      \\
\textit{out of}    &75              &150            &50
\end{array}

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

  1. Adapt Kirchoff's Current Law to this circumstance. Is it a reasonable modelling assumption?
  2. 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.
  3. Solve that system.
  4. Interpret your solution.
  5. Restate the Voltage Law for this circumstance. How reasonable is it?
Answer
  1. 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.
  2. We can label the flow in this way.

    Linalg rotary 2.png

    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.

    
\begin{array}{*{3}{rc}r}
i_1  &-  &i_2  &   &     &=  &25  \\
&i_2  &-  &i_3  &=  &0   \\
-i_1  &   &     &+  &i_3  &=  &-25
\end{array}
  3. 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,
    
\{\begin{pmatrix} 25+i_3  \\ i_3  \\i_3 \end{pmatrix} \,\big|\, i_3\in\mathbb{R}\}
    of course, since the problem is stated in number of cars, we might restrict i_3 to be a natural number.
  4. 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.
  5. 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 endlessly.
Problem 6

This is a network of streets.

Linalg intersection.png

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


\begin{array}{r|c|c|c|c|c}
&\textit{east\ Winooski}
&\textit{west\ Winooski}
&\textit{Willow}
&\textit{Jay}
&\textit{Shelburne} \\
\hline
\text{into}      &80    &50    &65     &--    &40      \\
\text{out of}    &30    &5     &70     &55    &75
\end{array}

(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.)
  2. 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.

    Linalg intersection 2.png

    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.

    
\begin{array}{*{7}{rc}r}
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{array}

    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.

    
\begin{array}{*{7}{rc}r}
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{array}

    Since the free variables are i_4 and i_7 we take them as parameters.

    
\begin{array}{rl}
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{array}

    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.

  2. 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
    
(i_1,i_2,i_3,i_4,i_5,i_6,i_7)
=
(35,25,50,0,20,0,30)
    results from choosing i_4=0.