**Electrical Circuits**

This discussion will focus on using matrices to answer questions related to electrical circuits. We will provide the basic ideas from physics and see how matrices are useful in this subject. An electrical circuit consists of several components some of which are

- Batteries

- Resisters

- Wires

A *battery* provides current (in volts) to
the circuit, a* resister* converts electrical
energy into other useful energy such as light reducing the current, and a *wire*
allows the current to flow through it without increasing or reducing the
current. The example below shows an electrical circuit diagram.

In this circuit diagram there are three batteries and four resisters.
As you will notice the diagram contains plenty of additional notes. The
batteries' electrical potential difference is measured in *volts*.
The battery to the left is a 60 volt battery, the
middle battery is an 80 volt battery, and the
voltage of the battery to the right has yet to be determined. The
resisters are measured in *ohms*. We use
the Greek letter W as the unit
indicator. The currents I_{1} and I_{2}
are measured in amperes. In the diagram above the both currents have yet
to be determined.

You will also notice the letters a, b, c, d, e,
and f labeled on the circuit diagram. These
mark important points in the diagram. Points b
and e represent nodes. A*
node* is a point where three or more wires connect. Points
a, c, d, and f are not nodes, however as we
shall see next, it is convenient to label them.

A voltage *loop* is a closed connection
within a circuit. That is a piece of the circuit beginning at a point and
ending at the same point. We will be interested in loops that do not
contain any sub loops. There are two such loops in the above diagram:

a ---> b ---> e ---> f ---> a

and

b ---> c ---> d ---> e ---> b

A change in voltage occurs when a current passes through a battery. For example, on the 60 volt battery, a current flowing upwards (from "-" to "+") will gain 60 volts while a current flowing downwards (from "+" to "-") will lose 60 volts.

The voltage through a resistor is related by

V = IR

The sign is positive if the measurement is taken against the current flow and the sign is negative of the measurement is taken in the direction of the current flow.

Kirchhoff came up with two law for electrical circuits that will help us find the unknown quantities.

**Kirchhoff's Voltage Law:**
Around any voltage loop, the total electrical potential difference is zero.

**Kirchhoff's Current Law:**
At any current node, the flow of all currents into the node equals the flow of
all currents out of the node.

The first law is often called conservation of energy and the second law is often called conservation of charge.

We now have all the ingredients necessary to solve our problem. We will
use Kirchhoff's two laws and the help of matrices to find the unknown currents
and voltage. First we give a direction to the two currents. Later we
may change the directions. Let I_{1}
flow from e to f to a
to b and let I_{2}
flow from b to c to d
to e. We begin with the loops. Consider
the loop

a ---> b ---> e ---> f ---> a

From a to b, the
current I_{1} passes through a resistor of 3
amps hence the potential difference is

-3I_{1}

From b to e, the current passes though a battery with an 80 V voltage drop will have a potential difference of

80

From e to f the current passes through a resistor of 1 amp hence the potential difference is

-I_{1}

From f to a, the current passes though a battery with an 60 V voltage gain will have a potential difference of

60

Kirchhoff's voltage law tells us that

-3I_{1}
+ 80 - I_{1} + 60 = 0

or

-4I_{1}
= -140

Using Kirchhoff's voltage law on the loop

b ---> c ---> d ---> e ---> b

-2I_{2} -
E - 3I_{2} - 80 = 0

or

5I_{2} +
E = -80

Now we use Kirchhoff's current law. For the node b, the sum of the currents going in are

I_{1} +
13

and the current going out is

I_{2}

So that

I_{1} +
13 = I_{2}

or

I_{1} - I_{2}
= -13

The node e gives

I_{2}
= I_{1} + 13

Notice that this gives us no new information. We have the three equations

-4I_{1}
= -140

5I_{2}
+ E = -80

I_{1}
- I_{2} = -13

This can be written in matrix form as

or with the augmented matrix

Now we can rref the matrix to get

We have that

I_{1} = -35 I_{2}
= -22 E = 30

Typically
we want to present currents to be positive, so we change the orientation if I_{1}
to go from b to a to f
to e and say that I_{1}
= 35.

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