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Basic
Electricity-2
Before
reading this topic, you must first understand the basic
electricity-1 topic for best results.
Voltage
Drop Across a Resistance
When electrons are forced through a resistance,
some electrons pile up on one side waiting to get through. It
is a lot like water passing over a dam. The water piles higher
on one side than the other side. The higher the dam, the greater
the water level difference on the two sides. The higher the resistance,
the greater the electron difference on the two sides. Remember
that electrical potential (voltage) is the difference in charges
between two points.
When current passes through a resistance, a potential
difference develops across the resistance. This is called a voltage
drop, and is a direct result of electron flow being restricted
by the resistance to electron flow. The side where the electrons
are entering into is more negatively charged than the side where
the electrons are exiting. Therefore, the side where current enters
is negative, and the side where current exits is positive. The
resistance develops a voltage drop while current flows through
it.
Kirchoff’s Voltage Law
Louie Kirchoff
was another early scientist who devoted much time to the study
of electricity. One of his greatest contributions is his theory
about the voltage drops about a circuit. No, he's not from Australia,
and he is not referring to walkabouts.
To put it
simply, he says that, the sum of all the voltage drops about a
circuit is equal to the source voltage of that circuit. In other
words, is you have a 12-Volt battery, then any lighting circuit
you connect to that battery will drop 12 -Volts. As an example,
let us examine the circuit below:
The ammeter
is designed to have nearly 0-Ohms of resistance. The conductors
for this light bulb series circuit have a resistance of 0.01-ohms
per linear foot. Let us assume for the moment, that it takes 15
feet of wire to complete this circuit. Since every foot of wire
has 0.01-ohms, the total length of wire (15-feet) will have a
total resistance of 15 times 0.01-ohms = 0.15-ohms. Although the
wire resistance is really spread uniformly throughout the length
of the wire, it is represented as a single wire resistance component.
So now the
resistance of the bulb and the conductors equals 4.15-ohms. This
is called the equivalent resistance of the circuit (Req). Since
the battery voltage didn’t change, the current in this circuit
is now, I = E / R, therefore, 12-volts divided by 4.15-ohms (Req)
= 2.89-Amps. Notice how the battery changes its output current
to meet the demand. The circuit current is controlled by the battery
voltage and the circuit equivalent resistance (Req).
Using Ohm's
Law, the voltage drop across the bulb resistance (E = I * R) results
in 2.89-Amps multiplied by 4-Ohms = 11.56-Volts, and the voltage
drop across the wire resistance (E = I * R) results in 2.89-Amps
multiplied by 0.15-Ohms = 0.44-Volts.
Louie Kirchoff's
Voltage Law stated another way says that the algebraic sum of
the voltage about the circuit is equal to zero. E(source) - E(bulb)
- E(wire) = 0. In our example, let us start at the left side of
the battery, and travel clockwise. You will see +12-Volts, then
you see -11.56-Volts across the bulb, then you see -0.44-Volts
across the wire resistance. The algebraic sum of +12 -11.56 -0.44
= 0.
Now let us
start at the right side of the battery and go counterclockwise.
First you will see -12-Volts, then +0.44-Volts across the wire
resistance, then you will see +11.56-Volts across the bulb resistance.
-12 + 11.56 + 0.44 = 0. It doesn't matter how you travel the circuit
loop, the sum of voltage drops and voltage gains always equals
zero!
Power
Losses
Now that
we have considered the voltage drop of the wire, the bulb no longer
has 12-Volts across it. The new power value for the bulb is P
= E * I, therefore the power = 11.56-Volts times 2.89-Amps = 33.4
watts. Notice how the resistance of the conductors reduced the
power to the light bulb from 36 watts down to 33.4 watts. 2.6
Watts are now dissipated by the wires! This means that the light
bulb will get a little dimmer because the tungsten isn’t quite
as white hot as before. In fact the bulb is getting less than
36 watts because the conductor is also dissipating power in the
form of heat.
Light bulb
manufacturers understand the power loss of wires, so they designed
the light bulb to compensate for these losses. In fact, the truck
light bulbs are designed to operate on 13.5-Volts when they are
rated as 12-Volt bulbs. We will explain why, later in this topic.
Battery
Resistance
Guess
what folks, that battery also has electrical resistance in it.
That current, which passes from the negative battery terminal,
through the conductors and connectors, through the light bulb,
and back to the battery positive terminal, must also flow through
the battery back to the negative terminal to repeat the loop.
Therefore, any battery resistance will also affect the current
in any circuit. When a battery is fully charged, its internal
resistance is quite low and on the order of 0.004 ohms. For our
light circuit, above, this battery resistance was negligible.
But for the starter circuit below, the battery resistance is very
significant.
When you
run the starter of your truck, then the small battery resistance
becomes quite significant. Using the circuit above, when the starter
draws 500 amps cranking power, the battery resistance drops E
= I * R, or 500-amps times 0.004-ohms = 2.0-volts. The battery
voltage drop due to internal resistance reduces the battery terminal
voltage by 2 volts. Battery resistance is why your lights dim
as you crank the starter, because the 12-Volt battery feeding
the lights, drops to 10-Volts at the battery terminals while cranking
the engine.
Those large
battery cables also have resistance of about 0.001-ohms, which
causes an additional 0.5-volts drop to the starter. So now, the
starter only gets 9.5-Volts to operate. The starter manufacturer
understands this and designed the starter to work well on 8-9
volts although the starter is rated as a 12-volt starter. The
starter motor has a resistance of about 0.016-Ohms. In this circuit,
Req = 0.016 (starter motor resistance) + 0.001-Ohms (starter cable
resistance) + 0.004-Ohms (battery resistance) = 0.021-Ohms. Therefore,
the cranking current = 12-Volts divided by 0.021-Ohms = 571.5-Amps.
As the battery
discharges, it’s internal resistance increases due to chemical
actions of the battery plates. This battery resistance increase
limits the available battery current and reduces the power delivered
to the starter. Now the starter is not getting the 8-9 volts that
it is designed for, so it starts dragging (arcing across the armature
windings) and repeated occurrences will shorten the life of the
starter motor. Weak batteries will destroy starter motors over
time because of the excessive current drawn and armature arcing
due to low cranking voltage.
Battery
Charging System
To charge a battery, the alternator or generator
must pump electrons back into the battery. When the electrons
are forced into the negative plate of the battery, a chemical
action takes place between both plates and the electrolyte, which
converts the electrolyte from water into acid. The acid is how
the battery energy is stored.
To make this transformation take place, requires
a voltage larger than 12-Volts at the battery terminals. Anything
greater than 12-Volts will have some charging effect, but 13.5-Volts
has the optimum charging effect.
Charging a battery is a compromise. If you charge
to fast, then hydrogen gas is vented by the battery. If you remember
the Hindenburg Blimp, then you realize that vented hydrogen gas
is dangerous. If you charge to slow, then it will take to long
to replenish current which was drained from the battery by electrical
devices, such as the starter.
Therefore, the charging circuit for most 12-Volt
trucks generates about 13.5 to 14 Volts. This is the voltage placed
across the battery terminals by the battery charger (alternator
or generator) while the truck is running down the highway.
If your battery has less than 13-Volts across its
terminals while the engine is running at high idle, then the alternator
is not charging properly.
If you think back to our light circuit earlier,
we now have 14-Volts feeding the light circuit, so, after the
wire losses, the light bulb still sees about 13-Volts. Remember,
the 12V light bulb manufacturer considered this when he designed
the voltage rating of the light bulb.
High
Resistance Problems
All electrical
connections add some resistance to the series circuit. Most electrical
connections add negligible resistance to the series circuit and
have no effect. The circuit shown below has a bad connector, and
the resistance of the bad connector does have adverse effect.
Houston ….. we have a problem!
Let us assume
that the bad connector has a resistance of 2-Ohms. Now the total
circuit resistance is 0.15-ohms for the wire resistance, 4-ohms
for the bulb, and 2-ohms for the bad connector. Req = 6.15-ohms
total circuit resistance. Therefore the current flow is I = E
/ Req, therefore 12-volts divided by 6.15-ohms, results in 1.95-amps
current flow. Now the voltage across the bulb is E = I * R, or
1.95-amps times 4.0-ohms which equals 7.8-Volts. Not even close
to the 13-Volts the bulb was designed for. The power delivered
to the bulb is only (P = I * I * R) 1.95-Amps multiplied by 1.95-Amps
multiplied by 4.0-Ohms = 15.21-Watts. The bulb is now running
at less than half power, and looks very dim when you look at it.
This is the
classical high resistance circuit. The bulb works, but it doesn't
work right. The high resistance condition is proven when you measure
the voltage across the light bulb. You know that it should be
a little less than 12-Volts due to the wire voltage drop, but
it measures only 8.0-Volts when you place a voltmeter across it.
Imagine what
a 2-ohm bad battery cable connection resistance would do to the
starter circuit shown previously. The starter solenoid would click,
but the starter would not work. This is another classic problem
that will appear with vehicle starter circuit problems.
When troubleshooting
a problem like this, you first measure the battery voltage to
determine that it is in fact 12-Volts. Then you measure the electrical
device voltage to see if it is also approximately battery voltage.
In this case, it measures 4-Volts lower than the battery voltage.
Remember old Louie Kirchoff's Law? The voltage drops around a
series circuit must equal the source voltage. Therefore, 4-Volts
are being dropped (lost) somewhere in this circuit. If you place
a voltmeter across the bad connector, you will find the missing
4 volts there. You can find high resistance in corroded contacts
for light bulbs, in corroded connectors, in pinched wires which
have several strands broken, and in loose or corroded battery
cable terminals.
Stranded
wire is more flexible than solid wire, yet has nearly the same
resistance as solid wire of the same size. When a stranded wire
gets pinched, the pinching force will sometimes sever several
strands, thereby reducing the effective diameter of the wire.
This will increase the resistance of that segment of wire, and
may cause significant voltage drop which affects the electrical
device getting current from that wire.
Parallel
Circuits Theory
Isaic Newton
was a good mathematician, but he got caught up in watching apples
falling from trees and watching how the curve of our earth placed
a crescent shadow upon the moon. So he didn't have time for electricity.
Some other
mathematicians thought that electricity was a really neat thing,
and maybe they could find a way to make some money off of it.
So they set out dreaming up curious things to do to with that
basic series circuit.
Everyone
had already learned that resistances about a series circuit simply
added together. But then some wise guy said, lets put two bulbs
side by side and see what happens. Don't laugh, your trailer clearance
lights are all in parallel, so you had better pay attention.
They found
that when you placed two light bulbs in parallel, the current
drawn from the battery doubled. Now this seems reasonable, since
two light bulbs are now taking power from the battery. If the
current doubled, then the equivalent resistance must have halved,
and so it did half. But then one of the mathematicians said, wait
a minute, what would happen if we placed two light bulbs in series,
with that whole thing in parallel with another light bulb? Whoa
.... jump back, I'll bet that is really complicated. So they set
out writing mathematical formulas to prove what would happen.
When the
dust finally settled, they discovered that when two resistances
are placed in parallel, the equivalent circuit resistance is the
product of both resistances divided by the sum of both resistances.
Go figure .... who would have thought that! They learned how to
say this in two different ways:
1/Req = 1/R1
+ 1/R2 or just Req = (R1 * R2) / (R1 + R2) you do the equation
transformations yourself, because I just don't have the time.
Old Louie
Kirchoff wanted some of the action too, so he created Kirchoff's
Current Law. I think he had a brain fart or something because
he said that the algebraic sum of all currents entering into a
node is equal to zero. There he goes with that zero thing again!
There is definitely something wrong with this guy.
Louie says
that if you have three wires tied together, and you have 3-Amps
going into the connection from one wire, and you have 2-Amps leaving
the connection by another wire, then you had damn well better
have 1-Amp leaving the connection on the third wire. It's that
simple, no magic or formulas involved. Gimme a break ....
The rest
is history, so lets take a look at typical parallel circuit with
unequal resistances and see what happens. We'll ignore wire resistance
and battery resistance, and just look at this from a theoretical
perspective.
Notice that
the two light bulbs are of different resistance values. If they
were the same resistance values, then the Req solution would be
easy, it would simply be 1/2 the value of one bulb. But since
they are not equal, tis not that simple.
Remember
what the mathematicians discovered? Req = (R1 * R2) / (R1 + R2).
Therefore, Req = (4-Ohms times 6-Ohms) / (4-Ohms plus 6-Ohms)
which equals 24/10 = 2.4-Ohms. So now, we can determine the current
leaving the battery. I = E / Req = 12-Volts divided by 2.4-Ohms
= 5-Amps. If we just needed the current draw of this circuit,
then this is the simplest solution.
But what
if we want to know how much current goes through each light bulb?
Remember old Louie Kirchoff's Voltage Law? If we ignore wire resistance
and battery resistance, and we know that the battery is 12-Volts,
then we also know that the voltage drop across the parallel bulbs
has to be 12-Volts, because the bulbs are the only voltage drop
in the circuit loop. And Louie Kirchoff says that the sum of the
voltage drops about a circuit must equal the source voltage.
Notice the
voltmeter in this circuit. What voltage would this voltmeter read?
12-Volts you say ..... WRONG answer, the voltmeter is installed
backwards so it wont read anything. Gotcha! When using a voltmeter
and ammeter, always place the negative lead on the side closest
to the negative battery terminal because that is where the electrons
are coming from and they will pile up when they hit the resistance,
causing the voltage drop that we intend to measure. Yes, we did
cover that back in Basic Electricity-1. For additional meter use
information go to using multimeters.
Let's see
now, I = E / R, so bulb #2 current = 12-Volts divided by 4-Ohms
= 3-Amps. That was pretty easy. Bulb #1 current = 12-Volts divided
by 6-Ohms = 2-Amps. It looks like old Louie Kirchoff was right.
The current leaving the negative terminal of the battery is 5-Amps,
and it splits with 3-Amps going through the top bulb#2 and 2-Amps
going through the bottom bulb#1. Then when the split current joins
before hitting the ammeter, it is 5-Amps again. Louie Kirchoff
is happy and I am happy.
Wait a minute,
there is another problem with this circuit. One of the bulbs just
burned out, which one would it be?
Charlie Ohm
and his laws about power still apply, yes, even in parallel circuits.
Lets check this out, what is the power into the top bulb, bulb
#2? P = E * I, therefore, the power is 12-Volts times 3-Amps =
36-Watts. Since the bulb is rated at 36 watts, that bulb should
be OK.
Now check
out the bottom bulb, bulb #1. Let's see, P = E * I, therefore,
the power is 12-Volts times 2-Amps = 24-Watts. This bulb was only
rated at 6 watts, so 24 watts burned it out! How could this happen;
you may ask? It was a 6V bulb and was never intended to have 12V
placed across it. If it had 6-Volts across the bulb, then power
= P = (E * E) / R = 6-Volts times 6-Volts divided by 6-Ohms =
6-Watts. So, doubling the voltage placed four times the power
into the bulb and burned it out.
This concludes
our basic electricity-2 topic. Understanding this topic and the
basic electricity-1 topic, will make you a very effective electrical
troubleshooter. If you don't understand any section of these two
topics, reread that section until you have it figured out. Congratulations
for completing both of these topics. You have learned many things
which will make you much more effective in fixing electrical truck
problems than most other truck mechanics.
We recommend
that you study the magnetism topic next.
It will help you to understand the relays, motors, generators,
and alternator topics. If you have any problems or comments with
this topic, please feel free to contact webRider.
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