Tinkering With Electronics

I remember when my parents bought their first TV set. It was in 1950. Vacuum tubes were pretty much the state of the art. Then the sixties rolled around. Solid state devices were the state of the art.

During the 1970's the first microprocessor appeared on the market. It was a four bit microprocessor. Today, we hear about 64 bit microprocessors. Back then, each memory chip had about 1 kilobyte of memory. Today, we have memory sticks that hold gigabytes of memory.

Prior to the microprocessor, programming was done with diodes. It is amazing how electronics has advanced. was applaud their desire to learn.

But despite all the advances, the safety issues remain the same. One act without thinking can cause serious injury or death. Every once in a while, a news report tells of a tragedy wherein somebody tinkering with electricity is accidentally electrocuted.

Years ago, I heard my supervisor warn a technician about high voltage within an instrument. Next thing I knew, the technician was thrown backwards. The chair fell to the ground and the technician was slammed into the wall. He was lucky. He did not need medical attention.

The golden rule is always be aware of what is within your reach when repairing any electrical equipment. Back in the days when plugs were not polarized, a person was working with two instruments that were isolated from ground. Both instruments had metal chassis. He did not realize that 110 AC volts existed between the two chassis. This was because one chassis was connected to common via its power plug while the other chassis was connected to 110 AC volts via its power plug.

If a bare wire that is connected to 110 volts AC comes too close to the common terminal, there may be arcing which will eventually heat up the wire enough to melt the insulation and possibly cause a fire.

Make sure your connections are solid and can't be undone via pulling on the wire. Inspect all your solder joints with a 3x magnifying glass to ensure ensure that none of the solder joints are cold.

Always be acutely aware of where high voltage exists within an instrument; Especially because if your hand gets too close to a high voltage contact, the high voltage may reach for you.

Don't assume that dangerous voltages do not exist in an instrument that is not powered up. A great example of this is the high voltage on the cathode ray tube of an old oscilloscope. Back in the early 1970's, thousands of volts were applied near the front of the tube. Some of that voltage still existed after the instrument power is removed. The old TV picture tube also had thousands of volts at the anode of the tube.

Isolation power plugs are appropriate in some situations. However using a isolation plug on a instrument wherein chassis is suppose to be grounded could cause problems.

Let's talk about troubleshooting an electronic circuit. The troubleshooter should be aware of the implicit assumptions made via his first attempts to resolve the problem. For example; if the troubleshooter views wave forms at various components in the circuit, the troubleshooter is implicitly assuming that the power supplies within that equipment are okay.

Always check for the obvious first. If you are troubleshooting an audio system, check the external cable and power cord connections first. If the instrument does not turn on, one might want to try switching power cords. Check the connectors too. For example, the old RJ-45 connectors might fail after being connected and disconnected numerous times. A USB socket may need a cleaning. A cable may have a broken connection at one the pins in it's connector. In the connectors, I use to work with, the pins were replaceable. I found cases wherein an edge connector on a printed circuit was at fault.

Sometimes I would start troubleshooting by measuring the power supply output voltages and the potential on the common and ground clads. In one case, I measured a half volt from one end of a ground clad to the other end. I found a very fine crack in the clad by using a magnifying glass.

One unusual problem I heard about was a broken lead on a transistor that was not noticed during initial inspection. When the person went to unsolder the transistor from the board, he found that one had disconnected from the bottom of the transistor.

It is always good to understand how an circuit works before trying to fix it. If a schematic is note available in the user's manual, one can always do an online search for it.

Let's consider a problem wherein the symptoms of the problem are changing. How does one troubleshoot this circuit? If one tracks down one symptom and the symptom changes while attempting to find its source, time has been wasted. The answer is to figure out why the symptoms are changing. It may be a bad connection that keeps making and breaking.

How about a problem wherein noise is on the power lines and the signal lines within a weighing scale? Then one has to figure out where the noise is originating. Does the noise consist of a bunch of transients? If so, do the transients coincide with the rising and falling edge of a pulsing signal? If the noise airborne? Is it coming from a drill or a motor nearby? Are there any external cables connected to the weighing scale? Do the cables come too close to a cable from some other equipment? Could the noise be coming from a walkie talkie? Or could some device be transmitting a very strong signal that the weighing scale cannot filter out?

Remember that any appliance containing a motor can be generating electrical noise. This includes a vacuum cleaner, an electric drill, etc.

Think of the simplest solutions first. If a monitor is not working, check if it is plugged in. Check if the fuses good. Do all the obvious checks first because those checks can save one a lot of time.

Good Luck!

Collection of Electrical Formulas

Originally published on Yahoo Voices

Contents

DC Circuit containing one resistor (Figure One)

Rules of Series Circuits

DC Series Circuit containing two resistors R1 and R2 (Figure One)

Rules of Parallel Circuits

DC Parallel Circuit containing two resistors R1 and R2 (Figure One)

AC Series Circuit containing R, C and L (Figure Two)

AC Parallel Circuit containing R and L (Figure Two)

AC Parallel Circuit containing R and C (Figure Two)

AC Parallel Circuit containing L and C (Figure Two)

AC Parallel Circuit containing R, L and C (Figure Two)

AC Power

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DC Circuit containing one resistor (Figure One)

E: electromotive force - Unit of measurement: volts

I: current - Unit of measurement: amperes

R: resistance - Unit of measurement: ohms

P: power - Unit of measurement: watts

A^2 means A squared.

A*(1/2) means square root of A

E = I * R

R = E/I

I = E/R

P = E*I = (I*R) * I = (I^2) * R

E = P/I = (I^2) * R/I = I * R

I = E/P

Rules of Series Circuits

Total voltage = sum of voltage drops

Total current = current through each component

Total Resistance = sum of resistances

DC Series Circuit containing two resistors R1 and R2 (Figure One)

R1: Resistor

R2: Resistor

Vr1: voltage drop across R1

Vr2: voltage drop across R2

Rt: total resistance

E = I * R1 + I * R2

I = same through each component

I = E/(R1 + R2)

Vr1 = I * R1

R1 = Vr1/I

I = Vr1/R

Vr2 = I * R2

R1 = Vr2/I

I = Vr2/R

E = Vr1 + Vr2

Rt = R1 + R2

P = E * I

Rules of Parallel Circuits

Voltage = same across each branch

Current = sum of currents through each branch

Resistance = less than least resistance

DC Parallel Circuit containing two resistors R1 and R2 (Figure One)

I = I*R1 + I*R2

E = same through each branch

E = Vr1 = Vr2

1/Rt = 1/R1 + 1/R2

Rt = R1 * R2/(R1 + R2)

P = E * I

Rules of Series Circuits

Total voltage = sum of voltage drops

Total current = current through each component

Total Resistance = sum of resistances

AC Series Circuit containing R, C and L (Figure Two)

L: Inductor - Unit of Measurement: Henries

C: Capacitor - Unit of Measurement: Farads

XL: Inductive Reactance - Unit of Measurement: Ohms

XC: Capacitive Reactance - Unit of Measurement: Ohms

Z: Impedance - Unit of measurement: Ohms

o: angle - Unit of Measurement: Degrees

X: algebraic sum of XC and XL

j: (-1)^(1/2)

XL = 2 * pi * F * L

XC = 1/(2 * pi * F * C)

Z = R + j*XL - j*XC

Z = R^2 + (XL - XC)^2

Z^2 = R^2 + X^2

Z = Z * (cos o) + j*Z * (sin o)

I = E/Z = same through each component

E = I * (R^2 + (XL - XC)^2 )^(1/2)

angle o = arctan (X/R)

Rules of Parallel Circuits

Voltage = same across each branch

Current = sum of currents through each branch

Resistance = less than least resistance

AC Parallel Circuit containing R and L (Figure Two)

1/Z = 1/R + 1/( j*XL)

I = sum of currents through each branch

I = (I * cos o) + j * (I * sin o)

E = same through each branch

E = I*Z

angle o = arctan (XL/R)

AC Parallel Circuit containing R and C (Figure Two)

Ir: current through R

Ixc: current through XC

1/Z = 1/R + 1/(1/( j*XC))

I = sum of currents through each branch

Ir = E/R

Ixc = E/(- j*XC)

I = Ir - j*Ixc

I = I * cos o - j * I * sin o

E = same through each branch

E = I*Z

angle o = arctan (-XC/R)

AC Parallel Circuit containing L and C (Figure Two)

IXL: current through XL

IXC: current through XC

1/Z = 1/(j*XL) + 1/(1/(j*XC))

I = sum of currents through each branch

IXL = E/(j*XL)

IXC = E/(j*XC)

I = j*IXL - j*IXC

E = same through each branch

E = I*Z

If XL > XC then

angle o = 90 degrees

If Xc > XL then

angle o = -90 degrees

AC Parallel Circuit containing R, L and C (Figure Two)

1/Z = 1/R + 1/(j*XL) + 1/(1/(j*XC))

E = same through each branch

E = I*Z

I equals sum of curents through each branch

I = IR + jIX

IX = I * j*(IXL - IXC)

If XL > XC then

0 < angle o <= 90

If XL < XC then

-90 <= angle o < 0

If XL = XC then

Angle o = 0 degrees

AC Power

AP = Apparent Power

RP = Reactive Power

TP = True Power

PF = Power Factor

TP = V * I * cos o = I^2 * R

RP = V * I * sine o = I^2 * X

AP = V * I = V * I * cos o + j * V * I * sine o

(AP)^2 = (TP)^2 + (RP)^2

PF = cos o = TP/AP

                                                                Figure One

                                                                  Figure Two

                                    

The Electrician's Helper

I originally published this story on Yahoo Voices

This is a funny story about an electrician's helper who made a very bad mistake. Of course, the names in the story have been changed. You'll know why after reading this story.

The electrician's name was Joe and his helper's name was Jim. Jim had prior training in interior house wiring. They had a job installing a light fixture in the attic of a three floor house. The house had been built in the early 1900's. Probably somewhere around 1901. Joe did some investigation and determined that the best path from the service box in the cellar to the attic was an inside wall. The house was a brick house and hence the outer wall was probably clogged with mortar.

The houses back then were not as easy to wire as the newer homes today. The big difference was the fire blocks (in the walls) designed a fire from spreading too rapidly. The fire block was a solid 4 inch by 4 inch block of wood that extended horizontally across the wall.

The first task was to find that inner wall from the cellar. Sometimes one can determine where the wall is by observing the wooden joists above, stretching across cellar. But in other cases, one needs another method of determining where the wall is.

One method is to cut a hole in the wall on each floor. This had to be done because it was not possible to drop a snake from the attic to the cellar inside the wall. The reason was simple: fire blocks. This had suddenly become a daunting task. Jim and Joe needed the owner's permission to cut holes in the wall on every floor and chip away at the fire blocks to make enough space to allow the cable through. Furthermore, they needed permission to rip apart the floor in the attic. But how would they find out where to drill down from the attic to get into the wall? Since they had permission to chop holes in the wall, the task became simple. They went up to the third floor and cut a hole in the wall near the ceiling. Then Jim tapped on the wood above the hole while Joe was up in the attic determining the location of the tapping sound. After determining where the tapping sound was coming from, Joe drilled and gained access to the inside of the wall. Then Joe dropped a snake into the wall and hit the fire block. He tapped the snake on the fire block until Jim determined where the sound was coming from and dug a hole in the wall at that point. Jim chipped away at the fire block until there was enough room for a cable to drop past the fire block. This was repeated on the second floor and the first floor.

Finally, after they uncovered all fire blocks, Jim stuck a long stiff wire that was known as a snake through the hole and down the wall of the first floor. Joe was in the cellar. When the stiff wire hit something solid. Jim moved the wire up and down, tapping on the solid structure. Joe was in the cellar listening for the tapping sound. Once he found where the tapping sound was, he took an electric drill, climbed a ladder and drilled up.

Every snake had an open loop at it's end. Joe stuck a snake through the hole. Jim and Joe wiggled their snakes until the loops connected. Joe pulled Jim's snake down just far enough to work with. Now they were prepared to pull a BX cable up the wall. The end of the BX was stripped leaving the insulated wires visible. Then electrical tape was used to connect the BX cable to the snake. Next the cable was pulled up into the wall. Eventually the cable extended from the attic to the cellar. The cable was pulled out of the hole far enough to form and loop and tape it after disconnecting the snake. Now the cable could not slip back into the wall. The first part of the operation was complete. Or so they thought.

Now in order for the reader to understand what a big mistake Jim made, I'll start by stating that it took 24 hours (two days) to install a BX cable from the service box in the cellar to the hole and inside the wall up to the attic. Got that? Twenty four hours!

Finally Jim was up in the attic holding the cable. He taped the cable in a big loop to prevent it from slipping back into the wall. This is where everything went sour. Joe was way down in the cellar yelling up to Jim.

Jim heard Joe say "Let it go!" Jim couldn't believe his ears. Let it go? After all the trouble they went through? Jim yelled down trying to clarify what Joe intended. But to no avail. Now if Jim were thinking correctly, he would have went down the cellar and clarified what Joe wanted. But Jim was exhausted from two days hard work. He was tired, hot and sweaty and he wasn't thinking straight. Joe was unaware that Jim had already released the snake and it was stuck in the wall. Joe wanted Jim to let the snake go. Not understanding this, Jim let the BX cable go.

Joe swore so loud that Jim heard him from the attic. Joe was in the cellar. It was a three floor house. Everyone heard Joe's cursing!

Now you might be surprised to learn that Jim was not fired on the spot. Joe waited until the job was completed (another 8 hours.) During the final 8 hours of work on that job, Joe brought a couple of walkie talkies to ensure that that dumb mistake ( and it was dumb! ) never happened again. Of course, Jim was fired after the job was complete.

The Basics Of Practical Electricity

Originally published on Associated Content/Yahoo Voices

I attended a vocational high school and majored in electrical work. In my second year, we studied interior house wiring and wired up simple projects. For example, we would wire up a switch and light fixture in a wooden booth. Then the instructor would enter the booth and grade my work. In my third year the instructor took us out on jobs. My first assignment on a job was to install an outlet in a bedroom. The wall was the common wall between the bedroom and the kitchen. The instructor told me to drill a tiny hole in the bedroom wall and insert a still wire in that hole. Then wiggle the wire around to ensure that nothing was blocking the installation of the outlet. He did mention not to go through the kitchen wall with the drill. So I followed his instructions and stuck a still wire in the hole and wiggled it around. I heard a lot of noise. The noise caught the attention of the owner. She opened her wooden cabinet to find one end of a still wire flying around hitting all her pots and pans! During my fourth year I studied the National Electrical Code.

This article is a tutorial on the basics of electricity. The first section covers basic electrical theory without the complex math formulas. . The second section covers safety issues and the third section covers simple house wiring repairs.

This section will introduce you to four terms, force or pressure, current, resistance and power.

Electromotive force is defined as the energy per unit charge. EMF is the abbreviation for electromotive force. It's unit of measurement is the volt. For those familiar with physics, a volt is a newton-meter per coulomb. For purposes of the following analogy, I will use the term electrical pressure instead of electrical force. The classical analogy is the water pump. The water pump applies pressure to the water which results in the flow of water. Electrical pressure results in the flow of electricity. The flow of electricity is called current. Current is measured in amperes. Any material that opposes the flow of electricity is said to have resistance. In the sink drain, resistance is provided by food particles and grease stuck in the trap. In electricity, resistance is provided by the components in the circuit, The ohm is the unit of resistance.

In order to understand the flow of electricity, we have to understand the atom. The atom consists of protons and neutrons in it's nucleus and electrons circling the nucleus. Each proton has a positive charge and each electron has a negative charge. A little electrical pressure causes some electrons to break away from the atom. That electrons that break free of the atom are called free electrons. The medium of electric current is the electron. Materials with lots of free electrons are conductors. Materials with a lack of free electrons offer resistance to the flow of electricity. Materials with no free electrons are insulators.

Stated in another way: Any material that does not allow electrical current to flow through it is called an insulator. An insulator has an extremely high resistance. Air is an excellent insulator. Rubber is also an insulator. Any material that allows electrical current flow is called a conductor. Copper and aluminum are good conductors. A conductor does not have an extremely high resistance.

Ohm's law states that one volt of electricity will cause one ampere of current to flow through one ohm of resistance.

EMF = current X resistance

volts = amperes X ohms

The power is the heat dissipated. A watt is a unit of power.

Power = EMF X current

watts = volts X amperes

Equipment and appliance manufacturers provide the power and voltage ratings on a sticker or plate on the side or back of the equipment. 100 W means 100 watts. Likewise, 100 V means 100 volts. Some equipment lists the amount of current the equipment requires. 10 A means 10 amperes. The word ampere is often abbreviated: amp. Fuses are also rated in amperes. You may want to look at the electrical ratings on a new air conditioner to determine if it has to be connected to a separate fuse or circuit breaker. The rating also indicates alternating current or direct current. A battery delivers direct current. The current is always flowing in the same direction. The abbreviation for direct current is DC. Your electrical outlet delivers alternating current. The direction and magnitude of the current is constantly changing. The abbreviation for alternating current is AC. An outlet delivers 110 volts alternating current or 110 vac.

The prefix kilo means thousand. A kilowatt is a thousand watts. If you look at your electric bill, you will find the term kilowatt-hours. The abbreviation for kilowatt-hours is kwh. Kilowatt-hours equals the number of kilowatts multiplied by the number of hours those kilowatts were dissipated. For example, You have nine 100 watt bulbs in nine lamps and all lamps are turned on for two hours. The total wattage is 900 watts.

kilowatts = watts/1000

kilowatt-hours = kilowatts X hours

900 watts divided by 1000 watts per kilowatt equals 0.9 kilowatts.

0.9 kilowatts multiplied by 2 hours = 1.8 kilowatt-hours.

In general terms, the higher the power rating of an appliance, the higher the electric bill. My old 14000 BTU air conditioner was rated over 14 watts. The electric stove probably has the highest power rating. The refrigerator also has a high power rating.

Wire is available in many sizes. The sizes are denoted by numbers. The smaller the number the thicker the wire and the more current it could handle. The two popular sizes for house wiring is AWG #12 and AWG # 14 copper wire. AWG stands for American Wire Gauge.

In order for current to flow in a circuit powered by a battery, there must be a conduction path from the plus terminal of the battery to the minus terminal of that battery. The air is not part of that path because the air has an extremely high resistance and will not allow current flow. A conduction path means a path through electrical conductors. Any electrical setup with a source of electricity and electrical components is called a circuit.

Lets examine a battery circuit. One wire is connected from the plus terminal of the battery to one side of a switch. Another wire is connected from the other side of the switch to the light bulb holder. The third wire is connected from the other terminal on the light bulb holder to the minus side of the battery. The light bulb offers resistance to the flow of current and dissipates power. The switch allows to user to make or break the electrical path to the light bulb. When the switch is open, no current flows through the circuit and the light bulb does not illuminate. When the switch is closed, current flows through the circuit and the light bulb illuminates.

You probably have often heard the term short. Ideally, a short means no resistance. Any two wires that have no resistance between them are shorted together. In our battery circuit, a short circuit exists if there is no resistance between the wire connnected to the plus terminal of the battery and the wire connected to the minus terminal of the battery. Another words, the resistance of a short circuit is 0 ohms. However, the wire does have a very small resistance.

Let's apply Ohm's Law to the definition of a short circuit. Assume that the voltage is ten volts and the resistance is zero ohms. According to Ohms Law,

10 volts = ? amps X total wire resistance.

In this case, when the voltage is divided by the tiny resistance of the wire, the result is an extremely large currrent.

One handy gadget to have around when you are working on electrical circuits is the multimeter. Most multimeters can read voltage, current or resistance. I recommend the digital multimeters. To read the voltage between a hot wire and ground, plug the multimeter leads into the plus and minus connectors on the multimeter. The red lead should be the plus lead and the black lead should be the minus lead. Set the multimeter range for the range that includes the voltage you expect to read. Connect the other end of the minus lead to ground and the other end of the plus lead to the wire whose voltage you want to measure. If you want to measure the resistance between two terminals, first ensure that power is turned off and no voltage exists between the two terminal. With the leads connected to the plus and minus connectors on the multimeter, set the scale to the appropriate ohms scale. Connect the other end of the leads to the two terminals and read the display.

Before talking about electrical house wiring, let's talk about safety issues. If at all possible, don't work on a live circuit. If you are replacing an electrical outlet or fixture, always pull the fuse for that circuit, You pull the fuse. Don't trust somebody else to pull the fuse. Take the fuse with you to ensure that no one puts the fuse back in while you are working. If you want to work on a light fixture, turn the light on, then pull the fuse. If the light extinguishes, you have the correct fuse. Always use a wooden ladder if you are working on a ceiling fixture. Never use a metal ladder. Remember that a very small current can kill you. Always be aware of where you are standing. Don't stand on a flooded cellar cement floor to work on electrical equipment. A simple experiment will demonstrate my point.

The explanation of this experiment is simple enough so you don't need to build the circuit. You can read the instructions and visaulize the circuit. If you want to build the circuit, you will need wire strippers, wire cutters, a screwdriver, long nose pliers, AWG 16 wire, a battery, a battery holder, a glass of water, salt, a bulb and a bulb holder. You will also need small alligator clips and a switch. You could disassemble a flash light to obtain bulb, bulb holder and battery. You could buy a battery holder for your size battery at any store that carries hobbyist supplies.

Connect a wire from the plus terminal of a battery to a glass of water. Make sure that the bare tip of the wire is in the water. Connect a second wire to a light bulb holder. Use the alligator clips to secure the connections to the bulb holder. Ensure that the other end of the wire is in the glass of water but is not touching the wire from the battery. Connect the third wire from the light bulb holder to the minus terminal of the battery. Note that the light bulb is not illuminated. Now pour some salt into the glass of water. Note that the light bulb illuminates. Pure water does not conduct but if you stand in water, your body provides the path to ground in two ways. Your body contributes salt to the water and your body conducts electricity. Cement also conducts electricity.

Equipment ground is another area where caution must be exercized. Consider an electric drill plugged into an outlet that does not have an equipment ground. In this case the electric drill has an internal short causing the metal case of the drill to be at 110 vac. The user of the drill climbs a ladder in the cellar and balances himself by holding on to a water pipe. As soon as the user starts drilling, the user receives an electrical shock. The reason is that the case of the drill is at 110 vac and the water pipe is ground!

Here is another case: A user has a piece of equipment with a metal case. The equipment ground is isolated via a broken prong on the power plug. It is a cold dry day. The user accidentally touches the piece of equipment discharging a static charge into the equipment. The equipment is damaged and no longer works correctly.

In your house, you will find two types of wire: solid and stranded. You can identify stranded wire because when you strip the insulation off the wire, you will find many strands of small wire. If it is solid wire, you will find one solid wire.

The outlet in your wall is mounted in a rectangular utility box. The lighting fixture may be mounted in a octagonal box. Inside the box, you should find information about the maximum wattage bulb to use in the lighting fixture. I was taught to make sure I made each wire [in a box] six inches long. I was also taught that when connecting an electrical cable to a box, always leave a service loop in the cable in case. If, during the next service, the wires are not long enough, the electrician could use the service loop to obtain more wire.

The equipment ground wire may be bare or it may be a green wire.

If you want to connect a wire to a screw, use wire stripers to strip some insulation off the wire. Some people use wire cutters or a knife to strip insulation off a wire, but this method often leaves a tiny groove in the wire which makes the wire prone to breaking. Loosen the screw and wrap the wire 270 degrees around the screw in the same direction that you tighten the screw [clockwise]. Make sure the the insulated portion of the wire ends at the screw. The bare portion of the wire should be under the screw. No other bare wire should be showing. Cut the end of the bare wire so that no bare wire will extend from the screw and tighten the screw. Repeat this process for each wire you connect.

In electrical wiring, two wires are connected together by making a pigtail splice. To make a pigtail splice with two solid wires, strip about one inch of insulation from each wire. Hold the wires so that they cross each other where the insulated portion ends and the bare portion begins. Grab the wire ends and twist them together,. The wire should be twisted together tight enough so that there is no stretch of straight bare wire. Twist a wire cap onto the pigtail splice in the same direction that you twisted the wire. This will insure that the wire does not untwist while you are putting on the wire cap. After the wire cap is on the splice, cover any remaining bare wire with electrical tape.

When I first attempted to make a pigtail splice with one solid wire and one stranded wire, I found that the stranded wire came off too easily. So the first thing I did, after stripping the wire and twisting the strands [of the stranded wire] together, was to tape the insulated portion of the wires together. I put a small bend at the end of the bare solid wire so that the wire cap would have something to catch on. Then I twist the stranded wire around the bare solid wire and screwed the wire cap on. The spring in the wire cap catches the bend in the solid wire. To ensure that the pigtail splice does not unravel, I use electrical tape. I start at the wire cap and wrap the tape tightly around the cap moving slowly towards the insulated wire. I keep wrapping it tightly until it covers a few inches of insulated wire. Never use masking tape on an electrical connection.

In your house wiring, the black wire is the hot wire. The hot wire carries 110 vac. The white wire is the circuit ground and the green wire is the equipment ground. I named the white wire the circuit ground to distinguish it's use from the use of equipment ground. The circuit ground is used for as a return for the equipment. The hot wire connected to the hot terminal of the equipment and the circuit ground is connected to the return terminal of the equipment. The equipment ground is connected to the metal case of the equipment. The long slot in your wall outlet is the circuit ground. The short slot is connected to the hot wire and the small hole is the equipment ground. Current flows through the hot wire to the equipment and back through the circuit ground. The equipment ground ensures that the metal case of the equipment will never be hot.

A lighting fixture has a black wire and a white wire extending from it. The black wire is the hot wire and the white wire is the circuit ground. The fixture's black wire should be connected to the black wire in the utility box via a pig tail splice. Likewise, the fixture's white wire should be connected to the white wire in the box via a pig tail splice.

Your standard outlet has one brass screw on one side and and a silver screw and a green screw on the other side. The silver screw is is circuit ground. The black wire is connected to the brass screw and the white wire is connected to the silver screw. The green wire is connected to the green screw labeled ground. GND is the abbreviation for ground.

Phasor Analysis of a Balanced Three Phase Wye Circuit

INTRODUCTION
This article treats phasors in a three phase wye circuit. It shows how to  solve for current in the neutral line of a three phase wye circuit using polar coordinates. It also demonstrates the power savings in a three phase circuit.
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BRIEF REVIEW OF PHASORS
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CONVERTING PHASOR TO A COMPLEX NUMBER
magnitude/angle = (magnitude * cos angle) + j*(magnitude*sin angle)
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MULTIPLICATION OF PHASORS
mag1/angle a * mag2/angle b = mag1*mag2/angle a + angle b
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DIVISION OF PHASORS
(mag1/angle a) / (mag2/angle b) = mag1*mag2/angle a - angle b
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ADDITION OF PHASORS
mag1/angle a + mag2/angle b =
 ((mag1 * cos angle a) + (mag2*cos angle b)) +
  j*((mag1*sin angle a) + (mag2*sin angle b))
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SUBTRACTION OF PHASORS
mag1/angle a - mag2/angle b =
 ((mag1 * cos angle a) - (mag2*cos angle b)) +
  j*((mag1*sin angle a) - (mag2*sin angle b))
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WHAT IS THREE PHASE?
A three phase source is a source providing three voltages that are 120 degrees out of phase with each other. (See figure one) This article is focused on the three phase wye system with a neutral line.
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THREE PHASE SYSTEMS WITH BALANCED LOADS
The purpose of this example is to show that the current in the neutral line of a balanced three phase load is zero amperes and to show that the total power dissipated in a balanced three phase load is also zero watts.
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RESISTIVE LOADS BETWEEN EACH PHASE AND NEUTRAL
The voltage in each phase of the three phase wye source is 100 volts.. The resistive load 25/0  exists between each phase and neutral of a three phase circuit. (See Figure Two) Calculate the current in the neutral line and the total power dissipated.
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V1 = 100/0
V2 = 100/120
V3 = 100/240
R1 = R2 = R3 = 25/0  = (25 * cos 0) + j(0 * sin 0) = 25 * cos 0
---
CALCULATE CURRENT IN PHASE ONE
I1 = (100/0) / (25/0)
I1 = 4/0  = (4 * cos 0) + j(0 * sin 0)
The current on the real axis is
4*cos 0 = 4
I1 = 4 + j0
---
CALCULATE CURRENT IN PHASE TWO
I2 = (100/120) / (25/0)
I2 = 4/120 = (4*cos 120) + j(4*sin/120)
4 * cos 120 = 4*(-0.5) = -2
4*sin 120 = 4*(0.866) = 3.46
I2 = -2 + j3.46
---
CALCULATE CURRENT IN PHASE THREE
I3 = (100/240) / (25/0)
I3 = 4/240 = (4 * cos 240) + j(4* sin/240)
4*cos 240 = 4*(-0.5)
4*sin 240 = 4*(-.866)
I3 = -2 -j3.46
---
CALCULATE CURRENT IN NEUTRAL LINE
In = I1 + I2 + I3  (Algebraic Sum of currents on each axis.)
In = (4 + J0) + (-2 + j3.46) + (-2 -j3.46)
The algebraic sum of the currents is
In = 0 amperes
---
POWER DISTRIBUTION
Calculate the total power Pt
P = V*I
---
CALCULATE POWER IN PHASE ONE
P1/0 = V1*I1
V1 = 100/0
I1 = 4/0
P1/0 = 100/0 * 4/0 = 400/0
---
CALACULATE  IN PHASE TWO
P2 = V2* I2
V2 = 100/120
I2 = 4/120
P2 = 100/120 * 4/120 = 400/240
---
CALCULATE POWER IN PHASE THREE
P3 = V3*I3
V3 = 100/240
I3 = 4/240
P3 = 100/240 * 4/240 = 400/480
---
CALCULATE TOTAL POWER
Pt = P1 + P2  + P3
Pt = 400/0 + 400/240  + 400/480
---
CONVERT PHASE ONE POWER TO COMPLEX NUMBER
P1/0 = 400/0 = (400 * cos 0) + (j*400 * sin 0)
P1/0 = 400 + j0
---
CONVERT PHASE TWO POWER TO COMPLEX NUMBER
P2 = 400/240 = (400 * cos 240) + (j*400 * sin 240)
cos 240 = -0.5
sin 240 = -0.866
P2 = 400*(-0.5) + j400*(-0.866)
---
CONVERT PHASE THREE POWER TO COMPLEX NUMBER
P3 = 400/480 = (400 * cos 480) + (j*400 * sin 480)
cos 480 = -0.5
sin 480 = 0.866
P3 = 400*(-0.5) + j400*(0.866)
---
ADD PHASE ONE POWER, PHASE TWO POWER AND PHASE THREE POWER.
Pt = P1 + P2 + P3
P1/0 = 400 + j0
P2 = 400*(-0.5) + j400*(-0.866)
P3 = 400*(-0.5) + j400*(0.866)
Pt = (400 - 200 - 200) +j(0 + 400(-0.866) + 400(0.866)
Pt = 0

---


                                                  Figure One: Three Phase Source

                                           
                                                 
                                                   Figure Two: Three Phase Resister Load

Figure Three: Phasor Diagram

Solving For Total Current In Complex Circuits

Problem: Find the total current in the circuit shown above.  Assume that R1 = R3 = R4 = R5 = R6 = 2 ohms and V1 = 8 volts.

Normally, one would resolve this problem using a circuit emulator like LTSPICE.   However, in this post, the equation will be derived via defining three loops around the circuit.  The DC power supply V1 will be a part of every loop.

So lets define the three loops.

The current through the first loop is arbitrarily labeled i1.

The first loop includes the power supply labeled V1 and the two resistors labeled R6 and R4.

The current through the second loop is labeled i2.

The second loop includes the power supply labeled V1 and the two resistors labeled R3 and R5.

The current through the third loop is labeled i3.

The third loop includes the power supply labeled V1 and the three resistors labeled R3, R1 and R4.

For this problem, we'll assume conventional current flow from the plus terminal of V1 through the resistors to the negative terminal of V1.

First lets develop the equation for the first loop;

The total current through R6 is i1.  The total current through R4 is the sum of i1 and i3. Hence the loop equation is
V1 = i1*R6 + (i1 + i3)R4
V1 = i1*R6 + i1*R4 + i3*R4
V1 = i1(R6 + R4) + i3*R4

The second loop includes the power supply labeled V1 and the two resistors labeled R3 and R5.
V1 = (i2 + i3)R3 + i2*R5
V1 = i2*R3 + i3*R3 + i2*R5
V1 = i2(R3 + R5) + i3*R3

The third loop includes the power supply labeled V1 and the three resistors labeled R3, R1 and R4.
V1 = (i2 + i3)R3 + i3*R1 + (i1 + i3)R4
V1 = i2*R3 + i3*R3 + i3*R1 + i1*R4 + i3* R4
V1 = i1*r4 + i2*R3 + i3(R1 + R3 + R4)

Since all resistors are equal, we could use the label R to represent any resistor
For the first loop
V1 = i1(R6 + R4) + i3*R4
V1 = i1* 2R + i3*R
For the second loop
V1 = i2(R3 + R5) + i3*R3
V1 = i2*2R + i3*R
And for the third loop
V1 = i1*r4 + i2*R3 + i3(R1 + R3 + R4)
V1 = i1*R + i2*R + i3*3R

So our three loop equations are
For the first loop
V1 = i1*2R + i3*R
For the second loop
V1 = i2*2R + i3*R
For the third loop
V1 = i1*R + i2*R + i3*3R

The loop current i2 does not flow in the first loop. Therefore we can write the first loop equation as follows
V1 = i1*2R + i2*0 + i3*R
Likewise we can write the second loop equation as follows
V1 = i1*0 +  i2*2R + i3*R

Now our three loop equations are
loop one
V1 = i1*2R + i2*0 + i3*R
loop two
V1 = i1*0 +  i2*2R + i3*R
loop three
V1 = i1*R + i2*R + i3*3R

Now we can use determinants to solve for the loop currents and then add the three loop currents to find the total current.

Using determinants to determine i1

First, we'll solve for the denominator

= 2R*2R*3R + 0*R*R + 0*R*R - R*2R*R - 0*0*3R - 2R*R*R
= 4*4*4 + 0 + 0 - 2*4*2 - 0 - 4*2*2
= 64 - 16 - 16
= 64 - 32
= 32

Next, we'll solve for the numerator

= V1*2R*2R + 0*R*V1 + R*V1*R - R*2R*V1 - 0*R*3R - V1*R*R
=  8*4*4 + 0 + 2*8*2 - 2*4*8 - 0 - 8*2*2
= 128 + 32 - 64 - 32
= 160 - 32
= 128

i1 = 128/32
i1 = 4 amps

= 2R*V1*3R + V1*R*R + R*0*V1 - R*V1*R - V1*0*3R - 2R*V1*R

= 4*8*6 + 8*2*2 + 0 - 2*8*2 - 0 - 4*8*2

= 4*8*6 - 4*8*2

= 192 - 64

= 128

128/32 = 4 amps

i2 = 4 amps

Solving for the numerator

= 2R*2R*V1 + 0*V1*R + V1*0*R - V1*2R*R - 0*0*R - 2R*V1*R
= 4*4*8 + 0 + 0 - 8*4*2 - 0 - 4*8*2
= 128 - 64 - 64
= 0
 i3 = 0 amps

The total current
it = i1 + i2 + i3 = 4+4+0 = 8 amps

Here is a link to an excellent article on simulateous equations and third order determinants.

http://math-equation.net/syst2.html