# Linear Algebra/Topic: Analyzing Networks

|current=Topic: Analyzing Networks
|previous=Topic: Accuracy of Computations
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 drawn as
lines, shown straight and with sharp right angles for neatness. Each
light is a circle enclosing a loop.

```{=html}
<center>
```
![](Linalg_car_circuit.png "Linalg_car_circuit.png"){width=""
height="200"}

```{=html}
</center>
```
The designer of such a network needs to answer questions like: How much
electricity flows when both the hi-beam headlights and the brake lights
are on? Below, we will use linear systems to analyze simpler versions of
electrical networks.

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

The first fact about electricity is that a battery is like a pump: it
provides a force impelling the electricity to flow through the circuits
connecting the battery\'s ends, if there are any such circuits.
We say that the battery provides a
**potential** to flow. Of course, this network accomplishes its function
when, as the electricity flows through a circuit, it goes through a
light. For instance, when the driver steps on the brake then the switch
makes contact and a circuit is formed on the left side of the diagram,
and the electrical current flowing through that circuit will make the
brake lights go on, warning drivers behind.

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. The units of measurement are:
potential is described in **volts**, the rate of flow is in **amperes**,
and resistance to the flow is in **ohms**. These units are defined so
that $\mbox{volts}=\mbox{amperes}\cdot\mbox{ohms}$.

Components with this property, that the voltage-amperage response curve
is a line through the origin, are called **resistors**. (Light bulbs
such as the ones shown above are not this kind of component, because
their ohmage changes as they heat up.) For example, if a resistor
measures $2$ ohms then wiring it to a $12$ volt battery results in a
flow of $6$ amperes. Conversely, if we have electrical current of $2$
amperes through it 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 a electrical
circuits like the one above is that the battery provides a voltage rise
while the other components are voltage drops.

The two facts that we need about networks are Kirchhoff\'s Laws.

-   *Current Law.* For any point in a network, the flow in equals the
    flow out.
-   *Voltage Law.* Around any circuit the total drop equals the total
    rise.

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**.

```{=html}
<center>
```
![](Linalg_resisters_in_series.png "Linalg_resisters_in_series.png"){width=""
height="200"}

```{=html}
</center>
```
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**. This network is more like the car lighting diagram shown
earlier.

```{=html}
<center>
```
![](Linalg_resisters_in_parallel.png "Linalg_resisters_in_parallel.png"){width=""
height="200"}

```{=html}
</center>
```
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.)

```{=html}
<center>
```
![](Linalg_resisters_in_parallel_2.png "Linalg_resisters_in_parallel_2.png"){width=""
height="200"}

```{=html}
</center>
```
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
$8i_2-12i_1=0$.

$$\begin{array}{*{3}{rc}r}
i_0&- &i_1    &-  &i_2   &=  &0 \\
-i_0&+ &i_1    &+  &i_2   &=  &0  \\
&  &12i_1  &   &      &=  &20  \\
&  &       &   &8i_2  &=  &20  \\
&  &-12i_1 &+  &8i_2  &=  &0
\end{array}$$

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 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** way.

```{=html}
<center>
```
![](Linalg_wheatstone_bridge.png "Linalg_wheatstone_bridge.png"){width=""
height="200"}

```{=html}
</center>
```
This network is a **Wheatstone bridge** (see Problem 4 ). To analyze it,
we can place the arrows in this way.

```{=html}
<center>
```
![](Linalg_wheatstone_bridge_2.png "Linalg_wheatstone_bridge_2.png"){width=""
height="200"}

```{=html}
</center>
```
Kirchoff\'s Current Law, applied to the top node, the left node, the
right node, and the bottom node gives these.

$$\begin{array}{rl}
i_0     &=  i_1+i_2  \\
i_1     &=  i_3+i_5  \\
i_2+i_5 &=  i_4      \\
i_3+i_4 &=  i_0
\end{array}$$

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.

$$\begin{array}{rl}
5i_1+10i_3  &= 10   \\
2i_2+4i_4   &= 10   \\
5i_1+50i_5-2i_2  &= 0
\end{array}$$

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. For instance, networks of streets are given in the
exercises.

## Exercises

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

*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.*


