Electrical Engineering

Two ways to connect an ammeter, You can use a direct reading type, by connecting the ammeter in series with the load. You can use a current transformer type or CT. The current transformer looks like a wire wrapped donut with two terminals on its side with a hole through the middle. you pass the wire carrying the load through the center. Connect the direct reading meter to the two terminals. A CT type meter allows you to measure higher currents.

The fault current of a power transformer will depend on the following;

Transformer Rating (in KVA/MVA)

per unit impedence of the transformer (%p.u.)

line/phase Voltage (VL/VP)

the following formula can be used to find the fault current on the secondary side of a transformer

Fault Current = Transformer Rating /(per unit impedance x phase voltage)

The Values of Transformer Rating, per unit impedance & phase/line voltage will usually be mentioned on the transformer rating plate / data sheet

As an example a 500kVA, 11kV/400V/3-Phase/50Hz transformer with 5% p.u impendence will have the following fault levels on the secondary side

Fault level = 500/(5%)=10000kVA

S=1.732 * VPP * IP

Fault current = 500/(5% x 400 x 1.732) = 14.4 kA

Remember to use 3phase voltage!

The power factor is a measure of the phase difference. If they are exactly in phase the PF = 1. If they are 180 degrees out of phase PF = 0.

Indeed yes. The magnetic core material does have some energy losses associated with its operation. These are known as hysteresis losses, and show up as the magnetic device (transformer, solenoid) becoming warm.

A transformer itself, merely reflects to its primary winding the conditions of the secondary winding. Their resistive losses, their phase change and so on.

Learning Electrical Engineering by yourself is much better than going to school, because you can choose the books and style of learning that you want. The most difficult part is finding out what works for you. I taught myself electrical engineering and now work in a firm as an electrical designer, and I am now on my way to getting my PE License. For me the most difficult part was teaching myself the mathematics and physics. It took me a long time to find the right way to learn these subjects on my own, but once I figured it out, I flew through the learning process, and was able to get my first job. I used these 2 books: Engineering Mathematics, and Advanced Engineering Mathematics by K.A. Stroud. These books teach you all the mathematics and physics that you could ever need, and they use a programmed approach to teaching that make it somehow painless to learn even the most complex subjects.

Now you have to learn how to apply your new found mathematical skills, which really you can do anything with, but since you want to go into Electrical Engineering I advise that you get the most basic easy to understand book about electricity ever written, Basic Electricity by Van Valkenburgh. After you have a good grasp of this book you should be ready to decide what specific area of electrical engineering you want to go into. Electrical Engineering is a very large field, but as long as you have a grasp on the mathematics and physics involved you should be able to go into any area you want. There are two major categories to choose from, Electronics or Electrical Engineering. Electronics is concerned with small circuits often found in modern technology. Electrical engineering is concerned with much bigger things like power supply, and distribution, and electrical infrastructure.

The next thing you need to do is start applying for internships or any kind of job at an engineering firm. If you can prove that you know the mathematics and have a grasp on electricity, you should be able to get your foot in the door some how. Once you are in you have to work under a licensed PE for 3 years, and then you will be eligible to take the Fundamentals of Engineering exam, which you should be able to pass as long as you have kept learning for those 3 years. Then a few years later, forgot exactly how many more, you will be eligible for the PE exam. Once you pass this exam you will be a Licensed professional engineer.

Seems like a long road, but even if you go to school, you still have to take the FE exam, and the PE exam. Becoming an engineer in general takes a long time, and takes a lot of passion. I got there this way, and would recommend it over school, because you start getting professional experience sooner, and you avoid a lot of useless classes and fees.

A good way to gauge your progress is to look up the FE exam requirements and see what you need to know.

a transformer require AC to function as desired it transform the AC to different levels. DC on a transformer can only see the actual primary or secondary resistance if the source is not limited in current it will burn the transformer by excessive heating since it will see only wire resistance.

Answer 2

the application of an AC voltage, V on one winding of the transformer produces alternating flux that links the entire core of the transformer. The changing flux induces an emf, E that opposes the main voltage. the current through the winding in this case is {(V-E)/R}; R= winding resistance

in case of application of DC, since there is no changing flux, there is no induced emf and hence the current will be V/R

since the resistance of the winding is very small, the current is very high and this can burn away the windings. hence DC is not used.

MVA is the apparent power.

MVA=( MW+ MVAr)1/2

I assume that by two-phase, you really are talking about a supply with two hot wires and a neutral, such as 120/240 volt. This is single phase power, it just has power taps on both ends of the transformer winding, the neutral is the center tap.

I don't think it is possible to convert or rewire your motor. A large 3 phase motor must be run on a 3 phase electrical supply. It will not run and will overheat if you try to connect it to a single phase supply. The only way to make a 3 phase supply from a single phase supply is to use a converter like a motor-generator set. That would be way more expensive than getting a new motor.

115 Kohm = 115 *103 = 115, 000 ohm

all magnets have a north and south pole. opposite poles attract, like poles repel.

An electric motor converts electrical energy into mechanical energy. While an electrical generator does the opposite.

The fact a generator is A/C is a function of how the generator works.

Think of voltage as water pressure.

A generator works by sweeping a magnet around the inside of a coil of wiring. This means that the electrons are being moved first one direction, then the other.

the result is A/C power.

D/C sources generate a constant stream of electrons from one side to the other, like a battery.

It's possible to convert AC to DC by blocking flow of the electrons in one direction, but AC power has a lot of advantages for power grid use: it can be put through a transformer, making high-voltage transmission over lines more efficient, and it is better for running motors, which is the largest use of electrical power in terms of the amount of raw power consumption.

The short answer: because the physics of a generator makes it impossible.

Four 9v batteries connected in a parallel will still emit 9 volts because you are not increasing the voltage, you are increasing the life. To increase the voltage of four 9v batteries, you must connect them in a series; that series will emit 9v X 4(batteries), which equals 36 volts.

A shunt-trip coil is a built-in component of a shunt-trip circuit breaker. It is a magnetic coil that can be energized externally to make the breaker trip to shut off the flow of current.

For instance, fire sprinklers are sometimes required in the top of elevator shafts in case of a fire in the shaft. If the sprinklers were to spray water on the electric controls in the elevator cab, people could be hurt or killed, either from electrocution or from the elevator malfunctioning. In these cases, a shunt-trip breaker is installed in the circuit feeding the elevator controls, and the fire alarm system sends a trip signal if it detects waterflow from the sprinkler system. This trips the breaker and removes power from the elevator cab.

Once tripped, a shunt-trip breaker requires a person to reset it manually.

The normal phase of an element is:

What form it is (liquid ,solid ,gas)

what the classification is (nonmetal ,metal ,metalloid)

• The RELAY CIRCUIT used in electronics circuits for ,
• 1. It is used for general mechanical SWITCH.
• 2. It is used in High voltage switching mechanisms
  
• 3. It is used in High voltage and current Applications.

One ampere is defined as one coulomb per second. It can be calculated using Ohm's law as volts divided by ohms. Note, however, that in an AC circuit with reactive loads, the measurement is not necessarily straightforward, as the phase angle between voltage and current makes thing more complex.

Mega means million and a watt is a unit of power. A megawatt hour is the amount of power used if 1,000,000 watts are used for 1 hour, or 1 watt is used for 1,000,000 hours. If 100 light bulbs each using 1,000 watts of power are turned on for 10 hours, they will use 100 x 1,000 x 10 watt hours = 1,000,000 watt hours = 1 megawatt hour. Most electric companies charge for the number of kilowatt hours used. A megawatt hour is 1,000 kilowatt hours.

As close to zero ohms as makes no difference.
Zero, or close to it. It should be nearly the same as if you touch the probes together.

Can make electric motor more powerful by: 1) Increasing the number of turns of the rotating coil. 2) using stronger magnet. 3) multiplying the number of coils and commutator segments. 4) winding the coil on a soft iron core; the soft iron core becomes magnetised and concentrates its magnetic field in the coil, this increases the forces on the coil.

Yes, superposition theorem holds true in AC circuits as well. You must first convert an AC circuit to the phasor domain and the same rules apply.

Wire size is based on the amperage that the load requires. For an accurate answer the motor amperage is needed. That said you can get an approximate figure by this method. 1 HP = 746 watts. This formula for amperage when HP is known. HP x 746/1.73 x Voltage x %Eff x pf. Use a Efficiency % at 90 and a power factor of .85. A horse power of 5 x 746 = 3730/1.73 x 220 x 90% x .85 = 12.81 amps. Code states that motor supply conductors have to be 125% of the full load amps = 16 amps. Code also states that conductors can only be loaded to 80% on continuous load which a motor falls into. A #12 copper wire 90 degree insulation is rated at 20 amps x 80% = 16 amps. A #12 wire will do the job. It will need a 40 amp three pole breaker, 50 amp non time delay fuse or a 30 amp time delay fuse to feed the motor.

Any where between 220 and 240 volts are a nominal figure in the same voltage range. It is brought about by the power company, as they have a responsibility to keep voltages within a certain 10% range. The load will only notice a difference of 1% on the load current. eg. Wattage load of 2400. Amps = watts/volts. 2400/240V = 10 amps. 2400/220V = 10.9 amps.