Electrical Engineering

Yes, you should be able to employ a buck boost Xformer to boost your 208V to the motors specified 240 V. Make certain when your buy or order the Xformer that it has taps to supply 240V specifically. Most will in that voltage range. The next important thing is that the new Xformer must be capable of supplying the full horsepower required by the motor. Transformer output capacity is measured in VA or volt-amps. In all praticality, this means watts. To get the number of volt-amps or watts your motor needs, multiply the number of horsepower of the motor times 746. ( 1 Horsepower = 746 Watts) example: a 3/4 hp. motor would need 3/4 or .75 X 746 W or 559.5Watts or volt-amps.Therefore a standard 500VA or .5KVA transformer wouldn't have quite enough output and you'd have to go to the next standard size 750VA. But the bottom line here is cost. Buck Boost Xformers are expensive. Figure out what the correct size of transformer you need to run your motor. Then price it out. Then price out a new 208volt motor of the same hp. rating as the one you have. If the motor is cheaper, well then there you go, just buy a new motor. Unless the BB Xformer is a lot less than a new motor, I'd advise buying a new motor and then you have a NEW motor instead of an old motor with a transformer and the hassle of wiring it up. ( For the record, the difference between the pure value of volt-amps and watts or power is that in alternating current, the CEMF, (counter electromotive force) creates a small power or wattage loss because of increased resisatnce. This is insignificant to smaller motors, i.e., less than 10 h.p.)

It will work, but the transformer will not be able to supply its full rated load, because of the harmonic distortion of the non-sine wave output of the inverter. The transformer will also probably run a little hotter too. If you are talking a small load fed from a transformer much larger than required, probably nothing to worry about. But if the power level is large, or you are loading the transformer to more than 30 or 40% of its rated VA, you could run in to trouble. Monitor voltages and temperatures carefully (and have a fire extenguisher handy...)

The actual act, meaning how bad the person was hurt. Then there is intent, or weather the perpetrator was trying to hurt the victim. If it is done with intent to kill or with a dangerous weapon it is aggravated battery, and the sentence can be greater.

A lower-case Greek letter 'mu', followed, without a space, by an upper-case A. Incidentally, when written in full, it's 'microamperes', not 'micro amps'!

u can take 4 pole. . . Bcoz v cnt take 3 pole,

either 2, 4, 6 8, 10

There are 2 ways you can go You can hook them up in series or in parallel

If you hook them up in series That is you connect the positive side of the battery to the negative side of the next battery and so on you en up with the voltage being increased by the voltage of each battery So if you hook up a 12 volt battery to the end of a 6 volt battery you end up with 18 volts.

Or you can hook them up in parallel That is you hook the abaterys together on one side going from positive to positive and on the other side negative to negative When you do this you end up with the same voltage but increase the amperage.

because of high resistance in the load.

If you get 220 volts instead of 208 volts it means that the transformer supplying the circuit is wired in Delta with one of the legs center-tapped to ground.

The cell has no more electricity and the bulb has blown.

each appliance should have a rating label showing the amps or watts used, Add up these figures to see if they exceed the capacity of the circuit. Amps =watts/voltage

No. Going out of the manufactures recommended rating is not recommended. The dimmer will build up heat,that the heat sinks can not dissipate. The additional heat will eventually destroy the device.

At resonance, the L and C impedance cancels out, so the current can be calculated based on the resistance and applied voltage.

Imagine increasing frequency of the supply from 0 Hz to very high. At low frequency, the impedance of the inductor is ~0 (defined as Zl = w*L*j), and the impedance of the capacitor is very large (defined as Zc = 1 / (w*C*j)). As you increase the frequency, the impedance of the capacitor will decrease, as the impedance of the inductor increases. At some point (the resonant frequency), these two will be equal, with opposite signs. After crossing the resonant frequency, the inductor impedance will continue growing larger than the capacitor impedance until the total impedance approaches infinite.

Nothing. The reason a circuit works is that there is an imbalance between the positive and negative (cathode and anode) ends of the battery. two positive ends wouldn't move anything.

I don't know im doing this same worksheet lol probably mr kullens:)

33 kV line, it should be 3.7 mtrs

If the photo cell was on during the short circuit then the chances of damage to the photo cell are at a minimum. If the short circuit occurred when the photo cell was operated then yes damage to the internal contacts of the photo cell could have happened. What happens is, the short circuit is downstream from the photo cell and when the contacts of the relay close in the photo cell the dead short is applied to the contact faces of the relay. This is somewhat like striking an arc with a welded with the same results, the contact faces weld close and the circuits breaker trips to open the short circuit. With the pole faces of the photo cell's relay closed it gives the illusion that the photo cell is not operating. The photo cell's circuitry will be working but the relay will not drop the circuit open because of the welded contact in the relay.

You keep the tip nicely sharp, and prevent them from rusting.

The torque on the moving coil will reverse at twice the frequency of the supply, causing the pointer to vibrate. High-quality a.c. measuring instruments that incorporate a moving coil are, of course, common but incorporate a rectification circuit to supply the coil.

If you are describing a circuit breaker then its electromagnetic trip component will operate in the event of a line*-to-ground (*not 'phase'!) short circuit. The thermal overload component will only operate in the event of a sustained overload.

Yes a shunt trip breaker can be activated manually.

absolutely NO! the batchalor electrical limited is a bogus.

The names mr. cha, david vibol and mike lins are all bogus names. beware of them.

they will offer work then they will send contract and will promise you that they will pay your visa fees and free everything. but later they will say you have to send money for the transfer of your visa in the UK embassy in manila. do not engage with them.

The excitation system puts the load on the generator, so if it fails, the generator is liable to overspeed.

Think of it as the gears on a bicycle, where your legs are the power source, the gearing is the excitation system. If you're riding along, and suddenly your gears fall off, your legs will start to move very fast. You might hyper extend your knee, your foot might slip off the medal, you could hurt yourself in a bunch of different ways because you expect the resistance to be there and it is suddenly gone. The same applies to a generator - it's designed to run at specific speeds. Going significantly faster can cause all kinds of problems.

be connected in parallel

There is no 'total voltage' in a three-phase system. There are three line voltages and three phase voltages.