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.

Kirchoffís 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 didnít 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 isnít 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, itís 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 Ö.. 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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