Electrical Engineering

The stator and rotor of a machine are not made from sold pieces of metal. They are constructed by clamping stacks of thin sheets of silicon steel, called 'laminations', together. Laminations are essential to minimise circulating currents due to voltages induced into the core/rotor when the machine is operating. These laminations are created by a stamping machine, which cuts the required shape from large sheets of thin metal.

CT ratio is the ratio of primary (input) current to secondary (output) current. A CT with a listed ratio of 4000:1 would provide 1A of output current, when the primary current was 4000A.

Voltage is the electrical potential difference between two points.

The armature has the stationary (not physically moving) magnetic field, which attracts the magnetic field in the rotor. Since DC does not alternate, a split ring is used to alternate the current (and resulting magnetic field), so that the rotor will spin.

due to the extension of loads they are using the relays for protection.........

you can get an idea by looking at the insulators

they must function in rain or snow

dew can collect on the insulator

they are made of porcelain with a glazed surface

3" diameter gives about 3" length of dew coated glaze and is safe at about 5000V

so if the insulator has 3 skirts its about 15,000V

10 skirts about 50,000V

the drops to the houses are low voltage and may be as little as 1/2 of glazed surface 600V or less

The power consumption is the same in both cases. For a high-power motor it is more efficient to feed a 3-phase motor because the amount of power wasted in the wiring that feeds the motor is less, assuming the same size of wire.

An emergency generator is attach to a location power supply and electricity. When the power supply or electricity shuts down unintentionally the emergency generator will send power to the things that it is attached to that is designated, until the main power is back on.

The causes of over loading in generator comes from the action of applying more load that the generator is rated for. An example using a 5000 watt generator and a connected load of 6000 watts.

110 vs 230 Vac - The lower voltage is safer, but requires more current and larger wires. (BTW, the transmission loss is not worse because of 110 Vac because everything is distributed at much higher voltages, it only becomes 110 at the transformer near your home).

60 vs. 50 Hz - The higher frequency allows for the use of less iron in transformers and motors, but at the cost of more losses in the distribution system.

The symptoms of an interturn fault are loss of power output (blown fuse) and distortion as in audio power transformer. In power transformer there will be excessive heat as power is being comsumed. Using LCR meter to determine these 2 values of each winding 1) DCR 2)inductance. Compare the value obtained with a good one. There should be some differences in DCR and particular noted the low or no inductance in case of interturn shorting as if output winding is being shorted. The symptoms of an interturn fault are loss of power output (blown fuse) and distortion as in audio power transformer. In power transformer there will be excessive heat as power is being consumed. Using LCR meter to determine these 2 values of each winding 1) the DCR 2) the inductance. Compare the value obtained with a good one. There should be some differences in DCR and particular noted the low or no inductance in case of interturn shorting as if output winding is being shorted.

The question has to be more specific. Full load amps, watts or voltage. Please restate your question.

There are two types of power - real power and reactive power. Real power is power that is used, such as the power used to light a light bulb. Reacitve power is power that is held and released by a reactive element (capacitor or inductor), thus is not actually used to do any work. The reactive elements cause a phase shift between voltage and current, which manifests itself as a change in power factor.

Power companies must supply both reactive and active power. Total power is equivalent to sqrt( Reactive^2 + active^2). Not only do they need to supply the reactive power, their equipment must be sized to handle a larger total power. Reactive power is generated by installing extra equipment - capacitor banks or inductor banks - or by running generation in such a way that more reactive power is created (this will lower the power plants' real power output).

Depending on who the customer is, they may not be billed for the reactive power, thus the oversizing of equipment, and the supplying of reactive power does not generate any revenue (but costs them). If power factor is bad enough, the power company will lose money, and thus require power factor correction, or will bill on both real and reactive power usage.

It's simple economics (money)!

That might be to make it run the right way.

It depends how they are connected. If they are connected between line conductors then they are measuring line voltages. If they are connected across phases then they are measuring phase voltages.

Stabilise the arc and optimize the operation of the furnace.

Because as the signal voltage increases on the base, the base current increases this causes the collector current to increase, this decreases the voltage drop Vc to ground. As the base voltage decreases the base current decreases causing the Collector current to decrease, increasing the voltage drop Vc to ground.

i think dis ans will clerify d doubts of students...

... in CE config..

as Vo=Vcc-Ic Rc nw as Vcc is const. v cn say Vo is directly proportional 2 (-Ic Rc) so nw when d base voltage is increased Ib will inc. thus, Ic will inc. cuz Ic=beta Ib thus Ic Rc drop increases... as Vo is prop 2 -Ic Rc thus, Vo will decrease...thus d curve for d +ve half cycle in which Ib is increasing drawn in 180 deg phase showing Vo is dec. n vice versa 4 -ve half cycle...

nw in CC...

as d o/p is taken frm emitter... Vo= Ie Re so der's no -ve sign..n for d +ve cycle Ie will incr(dat's leakage current) as der's no-ve sign it'll b in phase wid i/p..hence no phase shift as leakage current doesnt incr. so much thus,d o/p voltage Ie Re doesnt inc much thus, d o/p doesnt amplify much thus dis config knwn as emiter follower...

nw in CB,...

d I/P is appiled at emitter n o/p is taken frm collector... so, 4 d +ve half cycle d Eb junc will b more FB thus causing incr. in Ib n thus, in Ic... so inc. in Ic causes inc. in o/p voltage..n during -ve half cycle d junc will b less FB so o/p voltage will b decreasing..hence der's no phase shift...

for any other help..

cont.-

shrey.dhingra51@gmail.com

A diode allows electrons to pass only in one direction. During the negative half cycle of an alternating current the diode will not allow electron current to pass. It is like a one way valve. This property of diode enables it to convert AC into pulsated DC

In Pakistan, ranking of universities is determined by the HEC. The ranking list is available at the following website:

http://www.paked.net/higher_education/hec_university_rankings.htm

There are 360 degrees in a full cycle so 120 is one-third so each of the three phases is equally spaced delivering power evenly to the load.

In England they use Type G plugs and sockets for their standard household electricity supplies. For more information see the Related Link shown below and also the answer to the Related Question.

The current density for a copper busbar will vary as the size of the bar. And no, that's not a trick answer. More information about the application is needed to determine what size the busbar is, since only the ampacity was cited here. There are different specifications (sizes) for busbars, as you know. The deal is that there is usually a "temperature rise" factor that delimits what you're supposed to use. Higher temps mandate bigger bars, which you probably already figured out. The basic busbar size is set on how much heat the busbar will generate when conducting it maximun rated current. But factors regarding the thermal environment in which the unit is operating must be considered. Minimum size for 4000A is probably 1/4 x 10, but some applications will require up to 3/4 x 10 with some allowing something in between. With a specified size bar, you can easily calculate the current density by dividing the current (4000A) by the cross-sectional area. Pick one and go for it.