Q:What is his calculation? Trouble you prawn.

250KVA transformer, the maximum capacity to load how much KW power? This is a bit of a problem, it is difficult to correct answer. Is it possible to ask how much the maximum load of the 250KVA transformer is KW? Or, how much amperes the maximum output current? If you ask, then answer you, the maximum output current is about 250 × 1.445 ≈ 361A or so

Q:BQ: Which director do you like more: Michael Bay or Steven Spielberg?BQ2: Are you excited about Transformers 3 or worried that it can be not good?thanks :)))) lt;3

Indiana Jones. I think it is more adventurous and entertaining. I don't really know Michael Bay's work, but Steven Spielberg is a little to creepy in his films for me. Personally, I've never seen any of the Transformers.

Q:coz in a step up transformer the voltage is increased and the and the current is decreased and vice-versa . and the ohm's law states that voltage is directly propotional to the current .now in a transformer we r making the current inversly propotional to the voltage

A transformer does not make the current inversely proportional to the voltage; the currents in the two coils are inversely proportional to the turns ratio Q. If the secondary (output) coil drives a resistance, the current depends on Ohm's law; Is Vs/R. The resistance equals the ratio Vs/Is. But the ratio V/I in the primary (Vp/Ip) is multiplied by Q^2; that is, V is proportional to Q and I is proportional to 1/Q. So the power source driving the primary sees a load resistance Q^2R. The power, or product of V*I, is the same in the primary and secondary. But no matter what you do with the input voltage, the currents in both windings change proportionally to voltage. You can think of a transformer as similar to a lever, where V and I correspond to F and d (distance), F/d correwponds to R, and Q corresponds to the applied-work and output-work lever lengths. Work Fd is constant on both ends of the lever, and F/d has a Q^2 relationship between the two lengths of the lever.

Q:i have learnt from engineers that welding transformer is different from power transformer. welding transformer's current will be high.but its load will be low.my actual doubt is current is always proportional to load.then how we can justify this logic that welding transformer's current rating is high and its load is less.

The power transformer is used for transformation of electrical power from one circuit to another but welding transformer is used for obtaining the very high current at secondary windind for melting the metals of different types for welding INZMAM SIDDIQUI

Q:Do you think if I asked this. which I am maybe people will pick Transformers because more Americans watch Trans instead of Gundam but if people watched both, would they still pick Transformers?

What, in a battle or as in an anime? If it's anime, I pick Gundam. If it's a battle. Well Transformers got someone called Unicorn who's the size of a Planet. So you can imagine how powerful he is.

Q:I am building a simple circuit and was wondering what is the difference between a coil and a transformer? Also, in my circuit I have a generator. My generator is made up of a coil and magnets. I would like to know do the amount of turns my coil is determine the power of the generator? When I select the gage wire I am going to use we that determine the amount of volts and current I can gain from my generator?

A transformer is used to transfer power and either step up or step down the voltages. Hence for this purpose, the mutual inductance must be very high. The leakage inductance needs to be minimum. Thats why iron core is used. A Coil is a conductor generally wrapped in a helicoidal form by certain number of turns.

Q:How do I show the effects of putting a resistor in series with the primary of a transformer.Currently the transformer drops 120 to 6Putting the resistor on the 120 side what would happen? I am assuming it would drop the 120 down to some value which would then affect the secondary but I would like some numbers

An ideal transformer changes impedances by a ratio which is the square of its voltage ratio (which is also the winding ratio). So in your case you drop from 120V to 6V. The voltage ratio is 20:1. If we square that, we get a ratio of 400:1 for impedance transformation. This means a 400 Ohm resistor in the primary circuit will look like a 1 Ohm resistor in the secondary circuit. Please note that putting resistors in front of or behind transformers is generally not a good thing. It rarely ever makes things better and it almost always screws things up. If you need a different voltage, you have to change the ratio of the windings.

Q:A step up transformer has 5,000 turns in the secondary coil and 200 turns in the primary coil. The primary is supplied with alternating current with an effective voltage of 900V. (A) What is the voltage in the secondary coil? (B) If the current in the secondary coil is 20A, what current flows in the primary coil? (C) What power is developed in the primary coil? (D) What power is developed in the secondary coil?

A step up transformer has 5,000 turns in the secondary coil and 200 turns in the primary coil. N1 200 turns N2 5000 turns The primary is supplied with alternating current with an effective voltage of 900V. V1 900V (A) V2 ? V2 N2 / N1 x V1 V2 5000 / 200 x 900 25 x 900 22,500V (B) I2 20A ; I1 ? V1 x I1 V2 x I2 I1 (V2 x I2) / V1 I1 (22500V x 20A) / 900V I1 500 A (C) P1 ? P1 V1 x I1 P1 900V x 500A 450,000W (D) P2 ? P2 V2 x I2 P2 22,500V x 20A 450,000W Yep, the power in the primary is equal to power in the secondary.

Q:Find the transformer ratio of a transformer having 350 turns on the secondary winding and 7000 turns on the primary. What RMS voltage must be applied to the primary of the transformer to develop an open-circuit secondary voltage of 63 V? if a load of 100 ohms is connected to the secondary winding what is the primary current, ignoring resistive and reactive losses in the transformer?

Primary/secondary ratio 7000/350 20/1 For 63 V on secondary: Primary volts 20 * 63 1260 V.

Q:Transformer total circuit breaker rated current 380v, rated current 1000a, the rated capacity and output power how to calculate, I did not learn more points

KVA is apparent power, kW is active power. inspecting power: S = 1.732UI = 1.732 x 0.38 x 1000 ≈ 658 (kVA) Active power: P = 1.732UI cosφ = 1.732 × 0.38 × 100 × 0.8 ≈ 526 (kW) AC power, power divided into three kinds of power, active power P, reactive power Q and apparent power S, at any time these three power always exist at the same time. ????? The cosine of the phase difference (Φ) between the voltage and the current is called the power factor, denoted by the symbol cosΦ, where the power factor is the ratio of the active power to the apparent power, that is, cosΦ = P / S ????? Three power and power factors cosΦ is a right-angle power triangular relationship: two right-angled edges are active power P, reactive power Q, and oblique edge is apparent power S. ????? S? = P? + Q? S = √ (P? + Q?) ????? Apparent power S = 1.732UI ????? Active power P = 1.732UIcosΦ ????? Reactive power Q = 1.732UIsinΦ ?? ????? DC = P = UI = U? / R = I?R