Electrical Engineering

You can think of it as a motor that is very strong. If you were to try and hold the wheel of a 1000cc motorbike you would not be able to stop it from moving because it has tremendous torque in lower gears.

High torque means greater acceleration, which means you will need better tires to prevent wheel spins.

Alternating Current is measured in Amperes.

Both AC and DC are current methods in which electrons move along a wire

14

240 volts - the turns ratio of this transformer is 2:1, thus the primary voltage will be two times the secondary voltage.

No, we don't find the output force of a system is greater than its input. We've never seen this. There are losses associated with any system, whether they be mechanical, thermal or other. That prevents a situation where we can get more energy out of a system than we put in.

We don't know all there is to know about physics, but we know a lot. If we knew how to get more energy out of a system than we put into it, we'd have applied that to generate electrical power. Consider how we provide electricity ot the power grid now and think it through to confirm the idea that we don't know of a way to get more energy out of a system than we put in. Perpentual motion isn't something we've realized, either.

Your question in ambiguous. Are you asking what equipment you need to provewhether or not Ohm's Law applies to a particular load? If so, then you need a variable voltage supply, a voltmeter, and an ammeter. As you gradually increase the voltage across the load, you will need to record the voltage and current values. If the resulting graph is a straight line, then the load is 'ohmic' (or 'linear') and obeys Ohm's Law. If the resulting graph is a curved line, then the load is 'non-ohmic' (or 'non-linear') and does not obey Ohm's Law.

we proceed via the FT of the signum function sgn(t) which is defined as:

sgn(t) = 1 for t>0 n -1 for t<0

FT of sgn(t) = 2/jw where w is omega n j is iota(complex)

we actually write unit step function in terms of signum fucntion : n the formula to convert unit step in to signum function is

u(t) = 1/2 ( 1 + sgn(t) )

As we know the FT of sgn(t) we can easily compute FT of u(t).

Hope i answer the question

Current (amps)=Watts/Volts =2000/120 =16.75 =16.75 amps

A three-phase meter can be used for single phase while the reverse is not the case.

Because ammeters have low resistance. If you connected them in parallel, all the current would flow through the ammeter instead of the circuit you were trying to test, possibly frying the meter but certainly invalidating the measurement.

At STP(Standard Temperature and pressure), the temperature is zero degrees Celsius(273 Kelvin) and the pressure is 1 atmosphere. At RTP(Room temperature and pressure), the temperature is 25 degrees Celsius(298 Kelvin) and the pressure is 1 atmosphere.

For a very basic explanation, a manometer measures the pressure differential between a closed pressure applied at one end and an open pressure at the other.

For example. If you have a typical u-bend manometer and you apply an air pressure at one end, the liquid will raise correspondingly, as the pressure on the closed side increases more and more above the ambient air pressure.

The amount it rises depends on the density of the liquid used and the diameter of the tubing.

So basically a manometer uses the height of a liquid to translate the amount above ambient air pressure, which is considered zero, of your pressure source.

A transformer is an electrical machine whose purpose is to either 'step up' or 'step down' a voltage. For example, the voltage generated by an alternator in a power station is far too low for transmission, so a 'step up' transformer is used to increase that voltage to transmission levels. Transmission voltages are then gradually reduced, using 'step down' transformers at load centres (e.g. towns) where the electricity is to be used.

First of all, a megawatt is not a unit of energy. It is a unit of power, or the rate at which energy is used. 1 watt corresponds to 1 joule per second. For example, a sixty watt lightbulb uses 60 joules of energy per second. So a megawatt is one million watts, or one million joules of energy per second. Electrical energy is often measured in kilowatt-hours by the electric company (So that they know how much to charge you). This is equivalent to the amount of energy that a kilowatt power source could supply in one hour. The amount of energy in joules that a kilowatt hour is goes as follows: 1 kilowatt = 1000 watts, so 1000 joules per second. Multiply this by the number of seconds in an hour, and you have 3600000J.

Charge the capacitor with a voltage source through a resistor. Keep track of how long it takes the capacitor to charge to the voltage level. The value of the capacitor is c=time/5R where R is the value of the resistor and time is the charging time.

One horsepower (electric motor) is 746 watts, so ten horsepower is 7460 watts. Divide that by a power factor (assumed) of 0.8 and you get 9325 volt-amps. Divide that by a voltage (again, assumed) of 240 and you get 39 amperes.

For calculation purposes the electrical code book states that a 10 horsepower motor at 240 volts will draw 50 amps. The motor's nameplate amperage that is driving the rotary phase converter should all ways be used first.

The electrical code requires a breaker for a motor to be 250% of the motor's operating full load current. 250% of 39 amperes is 97 amperes. I would use a 100 ampere breaker for a 10 horsepower motor.

If the motor rating is used from the code book values, then 50 x 250% would equal 125 amps and that is the size of the breaker that must be used. This is the reason that the motor's exact name plate amperage rating should be used. In the long run feeder wire costs could make a saving in the overall projected costs.

Usually refers to a transformer. A step up transformer increases the voltage and a step down decreases the voltage by an amount proportional to the ratio of the windings between Primary and Secondary. A 1:2 ratio is a step up and doubles the voltage. A 2:1 would be a step down and halve the voltage.

This really depends on the scenario. What type electrical circuit are you talking about.

i love him because hes all wawys there for me and when i need him hes allways there and i allso love him because he keep it real and he don't lie to me and that's the things i love about him

one condition for the above question is either conductor or magnetic field must be rotating.considering rotating field and stationary conductor,the magnetic flux will be cut up by conductor resulting in the induced emf in the conductor.

Inductive reactance does NOT have it own sign or symbol. Rather, it uses Ohms as a quantifier. But Capacitive reactance ALSO uses Ohms as a quantifier. Fortunately, 1 Ohm of Inductive reactance is cancelled by 1 Ohm of Capacitive reactance at the same frequency of measurement.

Both the supply wires that enter a series motor is the field windings the field is split in two and is connected at the split to the brushes that is connected to the commentator of the armature, so the current pass thru the first half of the field then thru the armature and then thru the other half of the field. With a parallel DC motor the two fields is in series and the two wires is connected to the brushes so is the supply.

Ohm's Law: Voltage is resistance time current

So, 28 ohms and 3.8 amperes means 106.4 volts.