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Posted: Thu Oct 13, 2005 12:41 am
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Post subject: Part deux
Posted: Thu Oct 13, 2005 12:41 am
 TOC Moderator

Joined: Wed May 18, 2005 5:42 pm
Posts: 1297
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Location: Ames, IA
 We have discussed very briefly what electric current is and how it works. Now we will move to adapting the theory to more practical applications. We will discuss conductors, resistors and insulators, and how electrical properties are measured. We will limit the scope in this section to DC applications only. In dealing with electricity, there some terms that you must know and the relationship between them. If you understand these basics you will be able to analyze what is happening in a circuit when you are trying to troubleshoot it. E = Electromotive force, also called Voltage, measured in Volts. I = The flow rate of electricity, also called Current, measured in Amperes. R = Resistance, the opposition to current flow, measured in Ohms. P = Power, the quantity of electric energy, measured in Watts. To illustrate the relationships, it is often easier to visualize if you compare it to water. For example, if you had a dam and at the bottom you drilled a hole through it and put a pipe in it. If the pipe was capped you would have pressure on the pipe relative to the height of the column of water above it. This pressure in water is measured in PSI (pounds per square inch). Because the pipe is capped and there is no flow this is "potential energy". This corresponds to electromotive force (EMF) or Voltage, with the symbol V. Next, let's uncap the pipe. We now have water flowing through it. This flow is measured in Gallons Per Minute (GPM). This corresponds to current or the flow of electrons measured in Amperes, with the symbol I. Now, let's look at the size of the pipe. The larger the pipe the less resistance to the flow, so the more Gallons Per Hour (GPH). The longer the pipe, the more friction on the walls and the less water flows through it. So, both the size and the length of the pipe affect the flow rate (GPH). This corresponds to Resistance, measured in Ohms. The symbol for Ohms is the capital Omega symbol, W. The longer the wire or the smaller its size, the higher its resistance to the flow of current. If we want greater a volume of water delivered in a given time, we can do two things. One, we could raise the dam so the water column would be higher and the pressure would increase. Increased pressure would push more water through the pipe. The second option would to use a larger diameter pipe at the original level and by reducing the resistance we could increase the flow, or GPH. Electricity works in the same way. To increase the electrical current (measured in amperes), we can increase the electromotive force (measured in volts) or decrease the resistance (measured in ohms). Next we need to look at Power. If we hook the pipe to a hydraulic motor connected to a load, the amount of work that can be extracted from the motor (horsepower) is relative to the pressure PSI on the water and the rate of flow GPH. This corresponds to electric power, measured in Watts (W) or in the case of our homes, Kilowatt Hours, or KWH. Kilowatt hours equal Watts X Hours X 1000. Horsepower is often expressed in Kilowatts, particularly in Europe. One often sees engines rated as so many kilowatts [KW]. One horsepower is equal to 746 watts of electricity. A 3KW engine would be about the same as 4HP. A corresponding example would be a tractor you need to start. You have a 12v battery and you use a small wire and hook it to the starter. Because the small wire has a lot of resistance, most of the energy will be lost in the wire and the engine won’t crank because not enough energy is getting to the starter. The wire will possibly burn because most of the energy is lost in the wire. Now if we take a large cable and hook it up, it will have a very low resistance and most of the energy will get to the starter and crank the engine. This works just like the small pipe and the large pipe, and the same with the length of the pipe. See, it is not that mysterious! It is just that we can see the water and we can’t see the building blocks of atoms moving, so it is easier to comprehend examples using water. _________________Andrew TOC Moderator Mark Twain wrote:A man's character may be learned from the adjectives which he habitually uses in conversation.

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Posted: Thu Oct 13, 2005 12:43 am
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Joined: Wed May 18, 2005 5:42 pm
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 To understand electricity, one needs to be familiar with the common measurements and what they are. Volts: The unit of measure of electromotive force, E in Ohms Law 1 Microvolt = .000001 volt = 1 uv. = one 1/millionth of a volt 1 Millivolt = .001 volt = 1 mv. = one 1/thousandth of a volt 1 Volt =1 volt = 1 v. 1 Kilovolt = 1000 volts = 1 kv. Amperes: The unit of measure of electric current, I in Ohms Law 1 Ampere = 1 amp. = 1a. 1 Milliampere = .001 amp. = 1 ma. = one 1/thousandth of an ampere 1 Microampere= .000001 amp. = 1 ua. = one 1/millionth of an ampere Ohms: The unit measure of resistance, R in Ohms Law 1 Ohm = 1 ohm [omega symbol, not on keyboard] 1 Kilohm = 1 k ohms = 1000 ohms 1 Megohm = 1 meg. or 1 m. = 1 Million Ohms Watt: The unit of measure of electrical Power 1 Microwatt = .000001 watt = 1 uw. = one 1/millionth of a watt 1 Milliwatt = .001 watt = 1 mw = one 1/thousandth of a watt 1 Watt = 1 watt =1 w. 1 Kilowatt = 1000 watts =1 kw = 1 thousand watts 1 Megawatt = 1000000 watts = 1 megw = 1 million watts Horsepower: 746 Watts = the amount of energy in 1 horsepower There are only 2 formulas that you need to know and understand, in order to solve most common electrical problems on small power equipment. The first is Ohms Law: E = IR. This says that Voltage = Current X Resistance. Conversely, you can Find I with I = E/R and R = E/I. The second is the Power formula: P = EI. This says that Power in Watts = Voltage X Current (Note: current is often called amperage). Conversely, E = P/I or I = P/E and P = E2/R and P = I2R. Now, let's apply these formulas to a couple of tractor examples to see how they work. Example 1: Your tractor has two 12 volt headlights rated at 36 watts each, connected in parallel. Your tractor has a 6 amp charging system. You would like to add two more lights, can you do this with the charging system you have? There are two ways to solve this. 1st Method, using the Power formula: P, or Power = 12volt system, so E [12 Volts] X I [Current 6 amps.] = 72 Watts. This is the capability of the charging system. Two lamps in parallel with 36 watts each = 72 watts. No extra power is available for more lights. Connecting more light would result in the battery not staying charged and likely "smoke" the rectifier or stator in the charging system. 2nd Method, using Ohm's Law: Your lights are rated at 36 Watts. Current available, I = P [36W] /E [12V], so Current, or I = 3 amps each. Two lamps, at 3 amps each, equals 6 amps, the total capacity of the charging system. Example 2: Your mower has an electric clutch with 4 ohms resistance and the two lights above. You are looking at a replacement engine. What size alternator does it need? To solve this we need to use Ohms Law to find how much current the clutch draws. You have a 12 volt system, so: I, Current = E [12 Volts] / R [4 ohms] = 3 Amps You would need a charging system that would provide current to charge the battery after starting, plus 3 Amps for the clutch, plus 6 Amps for the lights or more than a 9 Amp charging system. In this case an engine with a 6 amp alternator would not be a deal even if it was cheap, as it would need to be converted to a different charging system. Next I will go to some basic circuits, and explain how resistance, voltage and current interact in a circuit. Once you understand these principles, how you would use a voltmeter to locate troubleshoot wiring problems in you tractor, or actually any electric circuit. First we need to know some common symbols and what they are. Then later we can use them to draw a diagram and analyze how it works. These symbols are used to create schematic diagrams, these show electrically how a circuit works. Some industrial and other diagrams use some variation of these symbols. In the next article [4] we will look at series and parallel circuits and how they work, using some of these symbols. _________________Andrew TOC Moderator Mark Twain wrote:A man's character may be learned from the adjectives which he habitually uses in conversation.

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Posted: Thu Oct 13, 2005 12:47 am
 TOC Moderator

Joined: Wed May 18, 2005 5:42 pm
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Location: Ames, IA
 n this section we will look at some simple DC series and parallel circuits. The purpose will be to observe the relationships occurring between Voltage, Current, Resistance and power. We will explore how changing the Voltage affects the Current and the fact that Power is a square law function relative to Voltage and Current. In all of these examples we will assume that there is NO resistance in the wires, ammeter or the battery [Not possible in the real world] . Figure 1, below, demonstrates a simple complete circuit with one resistor. It could be viewed as a Series circuit because it is in series with the ammeter, and could also be considered a Parallel circuit because the load resistor is connected directly across the battery. Example1 1 P = EI = 10W E = IR if V = ??, I = 1 Amp, R = 10 W A:10V Example2 2 P = EI = 40W I = E/R if V = 20V, I = ?? Amps, R = 10 W A:2 Amps Example3 3 P = EI = 576W R = E/I if V = 12V, I = 48 Amps, R = ?? W A:25 W In this circuit the current flows from the Battery through the Ammeter, R1 and back to the Battery. Note the current is equal in all parts of this circuit. In the first example, if the voltage was unknown, we could find it using Ohms Law E = IR. [E] ?? = 1a [I] X 10 ohms [R] = 10 volts The Power [P] is found using the Power formula P=EI P = [E]10V X [I] 1a. = 10 Watts In the 2nd example we will change the Battery Voltage to 20v and leave the resistor 10 ohms. Now to find the current this circuit will draw we will use Ohms Law again. I = E/R [I] ?? = [E] 20v / [R] 10 ohms = 2 amps Now if we calculate the Power P=EI we see that: [P] ?? = [E] 20v X [I] 2 a. = 40 watts Note that when the Source Voltage is doubled and the Resistance remains constant, The Power increases by 4. This is a "Square Law" function. If the Voltage had been increased by 3 times the Power would have increased by 9 times. In the 3rd example we have a 12 Volt Battery and the ammeter is indicating 48 amps of current flowing. What is the value of R1? [R] ?? = [E] 12 v /48a [I] = .25 ohm The Power would then be: [P] ?? = [E] 12v X [I] 48a = 576 watts The purpose of this exercise was to illustrate the effects of changing Voltage, Current and Resistance in a circuit and their relationship to each other and to observe how these changes affect the total Power. Figure 2, below, illustrates two resistors connected in series. Note that in this circuit the Current will be the same in all parts of the circuit. In this circuit the current path is from the battery through the ammeter through R1 then R2 then back to the battery. The value of the two resistors ADD together when they are in SERIES: R1 + R2 = 5 + 15 = 20 ohms With 20 volts applied, E = IR = 20 v = 1a [I] X 20 ohms [R]. Note: The voltage drop across R1 would be 5 V, as shown by: [E] 5v = [I] 1a X [R] 5 ohms The same would be true for R2. NOTE: The voltage drop in a SERIES circuit is always proportional to the resistance. If one was to assume R2 was a light and that R1 was a piece of wire that had 5 ohms resistance, it is obvious that ¼ of the power would be lost getting to the light. This is how one adapts or utilizes these concepts in practical applications. If you understand what happens in a circuit like this, then when you take a voltmeter and start measuring you can understand what it is telling you and locate the problems. Figure 3, below, illustrates a circuit with 2 resistors in parallel, and what happens. In this circuit we will not go through all of the E = IR calculations. Note that Both Resistors have 20 volts across them. If we take the Current flowing through R1, which is 1amp, and the Current flowing through R2 and add them together we get 1a + 4a = 5a, or the total circuit current is 5a. If we look at E = IR, then: 20v @ 5a = 4 ohms. Using P = EI the total power provided by the battery is: 20v X 5a =100 watts Of these 5 watts are dissipated as heat in R2 and 20 are dissipated in R1. Note the parallel resistance formula in line 4, calculates to 4 ohms which agrees with the power calculations previously done. If more than 2 resistors are connected in parallel the formula would be the one shown in the bottom line. Note: IN THIS CIRCUIT WITH 2 RESISTORS IN PARALLEL THE CURRENT THROUGH THE RESISTORS IS INVERSELY PROPORTIONAL TO THE RESISTANCE OF THE RESISTORS. Note also that when we insert Ohms Law [E = IR, and I = E/R] into the Power Formula: P = EI, then P = [IR]I, therefore P = [I]2 R. If we insert P = E[E/R], we get the formula P = [E] 2/R. You can see these calculations used in line 2 and 3 above. This circuit should demonstrate how current flow is inverse to resistance values in parallel circuits and that the current flow is not the same in all parts of the circuit. Further Note: If a resistor were to replace the ammeter, you would have a series parallel circuit and the 2 parallel resistors R1 and R2 would look like a 4 ohm resistor to the series resistor and the voltage drop across the two would be proportional to added resistor and the 4 ohms for this pair. _________________Andrew TOC Moderator Mark Twain wrote:A man's character may be learned from the adjectives which he habitually uses in conversation.

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Posted: Thu Oct 13, 2005 12:49 am
 TOC Moderator

Joined: Wed May 18, 2005 5:42 pm
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Location: Ames, IA

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Posted: Thu Oct 13, 2005 12:51 am
 TOC Moderator

Joined: Wed May 18, 2005 5:42 pm
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Location: Ames, IA
 This is alot of material. My old boss wrote all this stuff down, with a bit of my help, and he taught it all to me. his site is www.edensltd.com if you want to see this stuff in its original form. Hopefully this helps out! Andrew _________________Andrew TOC Moderator Mark Twain wrote:A man's character may be learned from the adjectives which he habitually uses in conversation.

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Posted: Thu Oct 13, 2005 12:51 am
 TOC Member

Joined: Wed Oct 12, 2005 12:40 am
Posts: 37
Location: moorseville
 wow this stuff looks all too familiar. i dont know if i said it or not but i am currently attending a school called Nascar Technical Institute. i just got done with my corse on electronic fundamentals and have now started on electronic technology. this stuff is very accurate to what we are using great job on it man. -Intimidator68

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Posted: Thu Oct 13, 2005 1:38 am
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Joined: Sun Jun 20, 2004 1:39 pm
Posts: 67
Location: San Antonio, TX
 Good stuff! Intimidator, where's the school at? Sounds like a cool deal. C.J.

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Posted: Thu Oct 13, 2005 1:41 am
 TOC Member

Joined: Wed Oct 12, 2005 12:40 am
Posts: 37
Location: moorseville
 The school is in Moorseville N.C. It is about 20 miles North of Charlotte and about 20-30 miles southeast of Hickory. It is a great school and I have learned a whole lot more in the short time ive been there then i ever did in my home town -Intimidator68

Posted: Thu Oct 13, 2005 4:29 am
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Joined: Sun Feb 15, 2004 3:25 am
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Location: Southern California
 Hi andrewk, hello everyone, Quote: ........ THERE IS ONE VERY CONFUSING ITEM: ELECTRON FLOW IS FROM NEGATIVE TO POSITIVE. IN GENERAL PRACTICE CURRENT FLOW IS SAID TO BE FROM POSITIVE TO NEGATIVE ......... Current flow in our Oldsmobiles, is from -ground to +battery? If you reverse the coil wires, you make the spark jump from ground to the center electrode on the plug, and the engine doesn't run very well. I believe the change from positive to negative ground, in automobiles, was to make the flow go to ground for a more efficient electrical system. Does that mean lightning goes from the earth (ground) to the clouds? Looks like good reason for the confusion. Regards, Norm _________________Harry S. Truman wrote:When you have an efficient government, you have a dictatorship.

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Posted: Thu Oct 13, 2005 7:29 am
 TOC Member

Joined: Sun Jun 20, 2004 1:39 pm
Posts: 67
Location: San Antonio, TX
 Norm and all, For the whole negative/positive thing, this is from one of my old texts: "Although the negative charged electrons move through the wire toward the positive (+) terminal of the source of electricity, the current is indicated as going from positive to negative. This is an unfortunate and confusing convention. Ben Franklin originally named charges positive (+) and negative (-) when he was studying static electricity. Later, when scientists were experimenting with electrical currents, they said that electricity travels from (+) to (-), and that became the convention. This was before electrons were discovered. In reality, the negative charged electrons move toward the positive, which is the opposite direction that people show current moving. It is confusing, but once a convention is made, it is difficult to correct it." For a good explaination of how that relates to spark plugs, look on page 3 here: http://www.mightyautoparts.com/pdf/articles/tt105.pdf I just try not to fill my head with too much and I'm ok. C.J.

Posted: Thu Oct 13, 2005 1:21 pm

Joined: Fri Jun 06, 2003 3:58 pm
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Location: Michigan, USA
 88 Coupe wrote:Does that mean lightning goes from the earth (ground) to the clouds? yes And that IS a LOT of info, looks like last years physics class... Definately a sticky. TOC Admin Brando _________________1999 Oldsmobile Intrigue GLS1987 Oldsmobile Delta 88 Royale

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Posted: Thu Oct 13, 2005 1:54 pm
 TOC Moderator

Joined: Wed May 18, 2005 5:42 pm
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Location: Ames, IA
 Norm, on older cars and most farm tractors, current flow goes neg to pos. They use a positive ground system. When they changed from generators to alternators, the electrical system changed to negitive ground, IIRC. And yes, lightning goes from the earth to the clouds... speed of light is pretty damn fast to the human eye... _________________Andrew TOC Moderator Mark Twain wrote:A man's character may be learned from the adjectives which he habitually uses in conversation.

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Posted: Thu Oct 13, 2005 3:35 pm
 TOC Village Idiot

Joined: Mon Jun 09, 2003 5:06 pm
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 Regarding the lightning theory, from the ground to the clouds... I have been told this is very true, but if you have ever looked at slow motion lightning stikes on TV? It APPEARS to be the other way around. The strike seems to build in the clouds, and then at the threshold, jumps downward to the ground. Does the visual appearance of lighting not have anything to do with how the current actually flows? Just an observation... _________________"I know that you believe that you understood what you think I said, but I am not sure you realize that what you heard is not what I meant." http://www.cardomain.com/id/88delta88

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Posted: Thu Oct 13, 2005 6:08 pm
 TOC Member

Joined: Sun Feb 15, 2004 3:25 am
Posts: 3196
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Location: Southern California
 Hi andrewk, All, Quote: ........ on older cars and most farm tractors, current flow goes neg to pos ......... When they changed from generators to alternators, the electrical system changed to negative ground ........ Change to neg ground was made with the change from 6 to 12Volts. Don't know about farm tractors, but Ford was the last to change. '55 was 6v pos, '56 was 12v neg. Alternators did not appear until the early 60s. Regards, Norm _________________Harry S. Truman wrote:When you have an efficient government, you have a dictatorship.

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