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
Each light is a circle enclosing a loop.
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
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
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,
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
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 ohms
then wiring it to a volt battery
results in a flow of amperes.
Conversely, if we have electrical current of
amperes through it then there must be
a volt potential difference
between its ends.
This is the voltage drop across the
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
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.
By Kirchhoff's Voltage Law, because the voltage rise is
volts, the total voltage drop must also be volts.
Since the resistance from start to finish is
ohms (the resistance of the wires is negligible),
we get that the current is amperes.
Now, by Kirchhoff's Current Law, there are amperes through
(And therefore the voltage drops are:
volts across the oh m resistor,
volts across the ohm resistor,
and volts across the 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.
We begin by labeling the branches, shown below.
Let the current through the left branch of the parallel portion
be and that through
the right branch be , and also
let the current through the battery be .
(We are following Kirchoff's Current Law; for instance, all points in the
right branch have the same current, which we call .
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 meets and , gives that .
Applied to the lower right it gives .
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 while the voltage drop is , so
the Voltage Law gives that .
Similarly, the circuit from the battery to the right branch and back to the
battery gives that .
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 and a voltage drop of
so the Voltage Law gives that .
is , , and , all in amperes.
(Incidentally, this illustrates that redundant equations do indeed arise in
Kirchhoff's laws can be used to
establish the electrical properties of networks of great complexity.
The next diagram shows
five resistors, wired in
This network is a Wheatstone bridge
(see Problem 4).
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
Kirchhoff's Voltage Law,
applied to the inside loop (the to to to loop),
the outside loop,
and the upper loop not involving the battery, gives these.
Those suffice to determine the solution
, , ,
, , and .
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.
Many of the systems for these problems are mostly easily solved on a computer.
Calculate the amperages in each part of each network.
This is a simple network.
Compare this one with the parallel case discussed above.
This is a reasonably complicated network.
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 ohms.
We can do a similar thing for parallel circuits.
In the second circuit analyzed,
the electric current through the battery is amperes.
Thus, the parallel portion is
to a single resistor of
What is the equivalent resistance if we change
the ohm resistor to ohms?
What is the equivalent resistance if the two are each
Find the formula for the equivalent resistance if
the two resistors in parallel are ohms and ohms.
For the car dashboard example that opens this Topic, solve
for these amperages
(assume that all resistances are ohms).
If the driver is stepping on the brakes, so the
brake lights are on, and no other circuit is closed.
If the hi-beam headlights and the brake lights are on.
Show that, in this Wheatstone Bridge,
equals if and only if the current
flowing through is zero.
(The way that this device is used in practice is that an unknown
resistance at is compared to the other three
, , and .
At is placed a meter that shows the current.
The three resistances , , and are varied— typically
they each have a calibrated knob— until the
current in the middle reads ,
and then the above equation gives the value of .)
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.
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?
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 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.
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.)
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?