Electrical Engineering

It's for analyzing aperiodic waveforms.

An aperiodic waveform is one that occurs at...well, random intervals. The sine wave of a powerline is a periodic waveform: it runs all the time, so the period of the wave is either 0.0166 seconds (60 Hz power), 0.02 seconds (50 Hz power) or 0.0025 seconds (400 Hz aircraft power).

If the wave just comes up whenever it feels like it, that's an aperiodic waveform, and it's much easier to analyze them if the sweep only starts at the beginning of a wave.

In exactly the same way as you do so for a single-phase load. It's the product of the power of the load, expressed in kilowatts, and the during of operation, expressed in hours. (And the symbol of kilowatt hours is 'kW.h', not 'kwh')

since we need to get desired or rated voltage, so if it is connected in high voltage side of the transformer, voltage supply will be more, and hence the current will be more than the required amount. this is the reason why the instruments re connected at the low voltage of the transformer while performing no load test.

To answer this question a voltage needs to be stated and whether the load is three phase or single phase. Without the voltage the amperage can not be calculated. For single phase, Amps = kva x 1000/voltage, for three phase, Amps = kva x 1000/1.73 x voltage.

ac(Alternating Current ) has the high polarity.

by using an iron core

It is a transformer with No load attach to it.

Yes, but the sensitivity and frequency response will be very poor. The diaphragm on a microphone, corresponding to the voice coil and cone on a speaker, is extraordinarily light and compliant.

The terms potential and voltage are somewhat interchangeable when used to indicate electric pressure. The proper term is voltage, or volts, which is actually a combined term that means joules per coulomb or, in simpler terms, energy per charge.

Depends on what sort of motor it was. If it was the sort you find in an ordinary electric drill, then it would slow down. But rememer, the only way you can reasonably reduce the current in to motor is to reduce the voltage in the supply. A motor takes what current it can, dependent on Ohm's Law. To reduce the current, reduce the voltage.

But in a three-phase motor, the speed being dependent on the rotation of the phases, it would more or less stay the same. But this assumes it's not under load. A load would cause it to slow down.

It depends on how much power the lamps require, and on how they are connected.

Any hobby shop that deals in models should carry this type of motor.

This might be a homework question, in which case you find the power, which is 3 x 746 watts, and divide by the voltage to find the current taken if the efficiency is 100%, and that is 9.325 amps. So the efficiency is 9.325/12 or 77.7%.

In practice the question is complicated by the power factor of the motor, which means that the watts taken are less than the volts times amps. The power factor could well be 0.777, in which case the efficiency would be 100%.

Only if we are also in contact, directly or indirectly, with ground.

By ohm's law, current is voltage divided by resistance. Conductance is the inverse of resistance, so current is voltage times conductance. No conductance - no current.

It depends on the inverter. Keep in mind that the average starting current of a motor is four times its running current, so the inverter would need to be able to handle a 54 Amp transient. (Or higher, depending on design margins.)

There are two stages in the disease process. The first stage is during the active Leptospira infection and is called the bacteremic or septicemic phase. The bacteremic phase lasts from three to seven days and presents as typical flu-like symptoms.

Vfd is used for maintaing the pressure of whole chiller line.

It depends on the construction of the motor. If the field winding is connected in series with the armaturewinding, the motor is like a universal motor and will probably operate normally. If the field winding is connected in parallel with the armature, the motor will not run, but will draw current, make noise and get hot.

because when you plug in a power saw,it has to struggle from the inside the saw

your electrics trip out when there is to much power in use, otherwise if your using an adapter or extension they may be faulty or it could be a faulty saw, i would start by checking any leads have not been gouged or have cuts in them including the wire leading to your saw.

If you assume pressure stays the same (so volume is not limited in any way), you can use the ideal gas law, pV = nRT. I assume an initial condition of ~20C temperature (293K). Also, I assume the temperature you're giving is in degrees F. 35000F = 19427C. If the above assumptions are correct, air will expand 19427 / 293 = 66.3 times its' initial volume.

At 746 watts per horsepower (electric) a 35 kw generator would require about 470 horsepower. Round that up, and allowing for control loop margin and mechanical losses, I would guess that a 35 kw generator should need a 500 or 600 horsepower engine.

I am currently designing a planar transformer for a switch-mode power supply. From what I understand, planar transformer is better than a normal transformer due to its shape. It utilises flat E ferrite cores and winding tracks which saves space and is more compact for a power supply. It also has lower leakage inductance which makes it more efficient.