Analyzing Networks

The diagram below shows some of a car’s electrical network. The battery is on the left, drawn as stacked line segments. The wires are lines, shown straight and with sharp right angles for neatness. Each light is a circle enclosing a loop.

Car lighting circuit with battery, door-actuated dome-light switch, brake-light switch, main light switch and dimmer; branches connect left and right headlights, rear lights and parking lights. Original English labels and circuit connections are retained.

The designer of such a network needs to answer questions such as: how much electricity flows when both the hi-beam headlights and the brake lights are on? We will use linear systems to analyze simple electrical networks.

For the analysis we need two facts about electricity and two facts about electrical networks.

The first fact is that a battery is like a pump, providing a force impelling the electricity to flow, if there is a path. We say that the battery provides a potential. For instance, when the driver steps on the brake then the switch makes contact and so makes a circuit on the left side of the diagram, which includes the brake lights. Once the circuit exists, the battery’s force creates a flow through that circuit, called a current, lighting the lights.

The second electrical fact is that in some kinds of network components the amount of flow is proportional to the force provided by the battery. That is, for each such component there is a number, its resistance, such that the potential is equal to the flow times the resistance. Potential is measured in volts, the rate of flow is in amperes, and resistance to the flow is in ohms; these units are defined so that $volts = amperes \cdot ohms$ .

Components with this property, that the voltage-amperage response curve is a line through the origin, are resistors. (Not every component has this property. For instance, as the light bulbs shown above heat up, their ohmage changes.) An example is that if a resistor measures $2$ ohms then wiring it to a $12$ volt battery results in a flow of $6$ amperes. Conversely, if electrical current of $2$ amperes flows through that resistor then there must be a $4$ volt potential difference between its ends. This is the voltage drop across the resistor. One way to think of the electrical circuits that we consider here is that the battery provides a voltage rise while the other components are voltage drops.

The facts that we need about networks are Kirchoff’s Current Law, that for any point in a network the flow in equals the flow out and Kirchoff’s Voltage Law, that around any circuit the total drop equals the total rise.

We start with the network below. It has a battery that provides the potential to flow and three resistors, shown as zig-zags. When components are wired one after another, as here, they are in series.

A 20-volt battery connected in series to resistors of 2, 5 and 3 ohms. The battery is labelled potential and each resistor is labelled resistance.

By Kirchoff’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 wire connecting the components is negligible), the current is $(20 / 10) = 2$ amperes. Now, by Kirchhoff’s Current Law, there are $2$ amperes through each resistor. Therefore the voltage drops are: $4$ volts across the $2$ ohm resistor, $10$ volts across the $5$ ohm resistor, and $6$ volts across the $3$ ohm resistor.

The prior network is simple enough that we didn’t use a linear system but the next one is more complicated. Here the resistors are in parallel.

A 20-volt battery feeds two parallel branches containing a 12-ohm resistor and an 8-ohm resistor.

We begin by labeling the branches as 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 $i_{0}$ . 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 get a negative number in the solution.

The same parallel circuit with directed branch currents i0 through the battery, i1 through 12 ohms and i2 through 8 ohms.

The Current Law, applied to the split point in the upper right, gives that $i_{0} = i_{1} + i_{2}$ . Applied to the split point 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 $12 i_{1} = 20$ . Similarly, the circuit from the battery to the right branch and back to the battery gives that $8 i_{2} = 20$ . And, in the circuit that simply loops around in the left and right branches of the parallel portion (we arbitrarily take the direction of clockwise), there is a voltage rise of $0$ and a voltage drop of $8 i_{2} - 12 i_{1}$ so $8 i_{2} - 12 i_{1} = 0$ .

$\begin{array}{rcrcrcr} i_{0} & - & i_{1} & - & i_{2} & = & 0 \\ - i_{0} & + & i_{1} & + & i_{2} & = & 0 \\ 12 i_{1} & = & 20 \\ 8 i_{2} & = & 20 \\ - 12 i_{1} & + & 8 i_{2} & = & 0 \end{array}$

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 can arise in practice.)

Kirchhoff’s laws can establish the electrical properties of very complex networks. The next diagram shows five resistors, whose values are in ohms, wired in series-parallel.

A 10-volt source feeds a Wheatstone bridge with arm resistances 5, 2, 10 and 4 ohms and a 50-ohm middle resistor.

This is a Wheatstone bridge (see Exercise 3). To analyze it, we can place the arrows in this way.

The preceding five-resistor bridge, labelled with directed currents i0 through i5 for the displayed Kirchhoff equations.

Kirchhoff’s Current Law, applied to the top node, the left node, the right node, and the bottom node gives these.

$\begin{aligned} i_{0} & = i_{1} + i_{2} \\ i_{1} & = i_{3} + i_{5} \\ i_{2} + i_{5} & = i_{4} \\ i_{3} + i_{4} & = i_{0} \end{aligned}$

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.

$\begin{aligned} 5 i_{1} + 10 i_{3} & = 10 \\ 2 i_{2} + 4 i_{4} & = 10 \\ 5 i_{1} + 50 i_{5} - 2 i_{2} & = 0 \end{aligned}$

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 $i_{5} = 0$ .

We can understand many kinds of networks in this way. For instance, the exercises analyze some networks of streets.

Exercises

Exercise 1 Worked answer

Calculate the amperages in each part of each network.
1. This is a simple network.
2. Compare this one with the parallel case discussed above.
3. This is a reasonably complicated network.
Back to Exercise 1

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 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 we can analyze it as in the prior item. We can also use linear systems.
  
  Using the variables from the diagram we get a linear system
  
  $\begin{array}{rcrcrcrcr} i_{0} & - & i_{1} & - & i_{2} & = & 0 \\ i_{1} & + & i_{2} & - & i_{3} & = & 0 \\ 2 i_{1} & = & 9 \\ 7 i_{2} & = & 9 \end{array}$
  
  which yields the unique solution $i_{0} = 81 / 14$ , $i_{1} = 9 / 2$ , $i_{2} = 9 / 7$ , 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’s Method, solving for some of the variables and then substituting.
3. Using these variables
  
  one linear system that suffices to yield a unique solution is this.
  
  $\begin{array}{rcrcrcrcrcrcrcr} i_{0} & - & i_{1} & - & i_{2} & = & 0 \\ i_{2} & - & i_{3} & - & i_{4} & = & 0 \\ i_{3} & + & i_{4} & - & i_{5} & = & 0 \\ i_{1} & + & i_{5} & - & i_{6} & = & 0 \\ 3 i_{1} & = & 9 \\ 3 i_{2} & + & 2 i_{4} & + & 2 i_{5} & = & 9 \\ 3 i_{2} & + & 9 i_{3} & + & 2 i_{5} & = & 9 \end{array}$
  
  (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 $i_{0} = 4.35616$ , $i_{1} = 3.00000$ , $i_{2} = 1.35616$ , $i_{3} = 0.24658$ , $i_{4} = 1.10959$ , $i_{5} = 1.35616$ , $i_{6} = 4.35616$ .
Exercise 2 Worked answer

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 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.
Back to Exercise 2

Answer.
1. Using the variables from the earlier analysis,
  
  $\begin{array}{rcrcrcr} i_{0} & - & i_{1} & - & i_{2} & = & 0 \\ - i_{0} & + & i_{1} & + & i_{2} & = & 0 \\ 5 i_{1} & = & 20 \\ 8 i_{2} & = & 20 \\ - 5 i_{1} & + & 8 i_{2} & = & 0 \end{array}$
  
  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 $i_{2} = i_{1} = 20 / 8 = 2.5$ 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 $i_{0} = 20 (r_{1} + r_{2}) / (r_{1} r_{2})$ , all in amperes. So the parallel portion is acting like a single resistor of size $20 / i_{0} = r_{1} r_{2} / (r_{1} + r_{2})$ ohms. (This equation is often stated as: the equivalent resistance $r$ satisfies $1 / r = (1 / r_{1}) + (1 / r_{2})$ .)
Exercise 3 Worked answer

A Wheatstone bridge is used to measure resistance.

Show that in this circuit if the current flowing through $r_{g}$ is zero then $r_{4} = r_{2} r_{3} / r_{1}$ . (To operate the device, put the unknown resistance at $r_{4}$ . At $r_{g}$ is a meter that shows the current. We vary the three resistances $r_{1}$ , $r_{2}$ , and $r_{3}$ —typically they each have a calibrated knob—until the current in the middle reads $0$ . Then the equation gives the value of $r_{4}$ .)

Back to Exercise 3

Answer. Kirchoff’s Current Law, applied to the node where $r_{1}$ , $r_{2}$ , and $r_{g}$ come together, and also applied to the node where $r_{3}$ , $r_{4}$ , and $r_{g}$ come together gives these.

$\begin{array}{rcrcrcr} i_{1} & - & i_{2} & - & i_{g} & = & 0 \\ i_{3} & - & i_{4} & + & i_{g} & = & 0 \end{array}$

Kirchoff’s Voltage law, applied to the loop in the top right, and to the loop in the bottom right, gives these.

$\begin{array}{rcrcrcr} i_{3} r_{3} & - & i_{g} r_{g} & - & i_{1} r_{1} & = & 0 \\ i_{4} r_{4} & - & i_{2} r_{2} & + & i_{g} r_{g} & = & 0 \end{array}$

Assuming that $i_{g}$ is zero gives that $i_{1} = i_{2}$ , that $i_{3} = i_{4}$ , that $i_{1} r_{1} = i_{3} r_{3}$ , and that $i_{2} r_{2} = i_{4} r_{4}$ . Then rearranging the last equality

$r_{4} = \frac{i_{2} r_{2}}{i_{4}} \cdot \frac{i_{3} r_{3}}{i_{1} r_{1}}$

and cancelling the $i$ ’s gives the desired conclusion.
Exercise 4 Worked answer

Consider this traffic circle.

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

North Pier Main

into

out of

$100$

$75$

$150$

$150$

$25$

$50$

We can set up equations to model how the traffic flows.
1. Adapt Kirchhoff’s Current Law to this circumstance. Is it a reasonable modeling assumption?
2. Label the three between-road arcs in the circle with a variable: let $i_{1}$ be the number of cars going from North Avenue to Main, let $i_{2}$ be the number of cars between Main and Pier, and let $i_{3}$ be the number between Pier and North. Using the adapted 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?
Back to Exercise 4

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. 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.
  
  $\begin{array}{rcrcrcr} i_{1} & - & i_{2} & = & 25 \\ i_{2} & - & i_{3} & = & 0 \\ - i_{1} & + & i_{3} & = & - 25 \end{array}$
3. The row operations $ρ_{1} + ρ_{2}$ and $ρ_{2} + ρ_{3}$ lead to the conclusion that there are infinitely many solutions. With $i_{3}$ as the parameter, this is the solution set.
  
  ${(\begin{matrix} 25 + i_{3} \\ i_{3} \\ i_{3} \end{matrix}) ∣ i_{3} \in ℝ}$
  
  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 $i_{3}$ -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 we could find that a half dozen more cars entered than left, although the problem statement’s into/out table does satisfy this property. In any event, it is of no help in getting a unique solution since for that we would need to know the number of cars circling endlessly.
Exercise 5 Worked answer

This is a network of streets.

We can observe the hourly flow of cars into this network’s entrances, and out of its exits.

east Winooski west Winooski Willow Jay Shelburne

into

out of

$80$

$30$

$50$

$5$

$65$

$70$

–

$55$

$40$

$75$

(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 someone proposes construction for Winooski Avenue East between Willow and Jay, and traffic on that block will be reduced. What is the least amount of traffic flow that can we can allow on that block without disrupting the hourly flow into and out of the network?
Back to Exercise 5

Answer.
1. Here is a variable for each unknown block; each known block has the flow shown.
  
  We apply Kirchhoff’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.
  
  $\begin{array}{rcrcrcrcrcrcrcr} i_{1} & - & i_{2} & = & 10 \\ - i_{1} & + & i_{3} & = & 15 \\ i_{2} & + & i_{4} & - & i_{5} & = & 5 \\ - i_{3} & - & i_{4} & + & i_{6} & = & - 50 \\ i_{5} & - & i_{7} & = & - 10 \\ - i_{6} & + & i_{7} & = & 30 \end{array}$
  
  The row operation $ρ_{1} + ρ_{2}$ followed by $ρ_{2} + ρ_{3}$ then $ρ_{3} + ρ_{4}$ and $ρ_{4} + ρ_{5}$ and finally $ρ_{5} + ρ_{6}$ result in this system.
  
  $\begin{array}{rcrcrcrcrcrcrcr} i_{1} & - & i_{2} & = & 10 \\ - i_{2} & + & i_{3} & = & 25 \\ i_{3} & + & i_{4} & - & i_{5} & = & 30 \\ - i_{5} & + & i_{6} & = & - 20 \\ i_{6} & - & i_{7} & = & - 30 \\ 0 & = & 0 \end{array}$
  
  Since the free variables are $i_{4}$ and $i_{7}$ we take them as parameters.
  
  $\begin{aligned} i_{6} & = i_{7} - 30 \\ i_{5} & = i_{6} + 20 = (i_{7} - 30) + 20 = i_{7} - 10 \\ i_{3} & = - i_{4} + i_{5} + 30 = - i_{4} + (i_{7} - 10) + 30 = - i_{4} + i_{7} + 20 \\ i_{2} & = i_{3} - 25 = (- i_{4} + i_{7} + 20) - 25 = - i_{4} + i_{7} - 5 \\ i_{1} & = i_{2} + 10 = (- i_{4} + i_{7} - 5) + 10 = - i_{4} + i_{7} + 5 \end{aligned}$
  
  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 $i_{4} = 0$ .