Q:How is the insulation of the transformer divided?

Transformer insulation can be divided into internal insulation and external insulation, the insulation is part of the tank insulation, external insulation is the upper part of the casing and the insulation between each other.

Q:I hate feeling the need to justify myself, but I've seen a lot of answers telling askers to do their own homework. I'm trying to do so, but can't find any equations in the book relating power to voltage as well as turn number. Is it PviN? I've just seen Pvi, but that wouldn't apply to multiple loops. This is for ideal transformers by the way, so power for the secondary and primary coils is the same right? But are their vi's equal, or their viN's?The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?

The turns ratio is 4:1 of the primary. The power supplied to the primary is 400W. So regardless of anything else, in a ideal transformer, 400W is available on the secondary. But voltage will be 4x higher, current 4x smaller, than the voltage and current in the primary. Disregarding all that, if there is 400W of power flowing in the primary, then 400W will be being dissipated in the 20 ohm resistor. Watts in watts out, but secondary load (resistance) controls the wattage. There is no way to tell the output voltage and current, without knowing the input voltage and current. But whatever they are, the product of voltage x current will be 400W, in this instance.

Q:How to distinguish between large, medium and small transformers

We usually put the capacity of 630kVA and below the transformer for the small transformer; 800-6300kVA transformer for the medium-sized transformer; 8000-63000kVA transformer for large transformers; 90000kVA transformer for large transformers. With the transformer capacity bigger and bigger, in the past that can only be called large-scale medium.

Q:i have an epcos 565-2 transformer and i want to know how much it will raise the electrical output of a standard 1.5 volt battery. i really do not know much about transformers, but i bought a motor, assuming that i would be able to do fun things with it, and i hooked it up to all the power i could(solar batteries, and numerous amounts of AA batteries) but it wouldn't spin anything that i attached to it more than probably 60 rpm. i bought a transformer to boost the power, and spin the motor much faster. the motor is graded for 3000 rpm maximum, but i just want it to be above 250. any help would be appreciated.

Transformers will only work with AC, not the DC from a battery. You will have to buy/build an inverter circuit to convert the 1.5 V to anything higher with a transformer (the transformer may become part of the inverter if it is the right kind). The other thing to think about is the power level. A 1.5 v AA battery is good for this much power: 1.5 volts * 1 Amp 1.5 Watts You might get 2 Watts out of it for a few minutes, but it can only sustain 1.5 to 2 Watts for about an hour. How much power is the motor rated for? If the motor is rated for 10 Watts, then 1, 2, 3, or even 4 AA batteries will not be enough to power the motor much above 250 RPM. There is still alot you need to know. Just for fun try a bunch of D cells in series (about 10 of them) The extra current provided by the D cells may spin the motor faster. .

Q:Branch circuit cables are rated for 75?C and feeder cables are rated for90°C. This will be a 3-phase, 575-volt system with four induction motors, specified asfollows:Motor 1: 60 hp 0.90 p.f. squirrel cage motorMotor 2: 60 hp 0.90 p.f. squirrel cage motorMotor 3: 40 hp 0.85 p.f. wound rotor motorMotor 4: 7-1/2 hp 0.80 p.f. wound rotor motorA) Assuming no line losses, find the capacity of the transformer required to supplypower to this system.B) Assume that it is desired to improve the overall power factor for this system to 0.95lagging. Determine, in kVAR, the required capacitance for this power factorcorrection.C) Assume these motors have their windings connected in a delta configuration. Whatwould be the line voltage if they were connected in wye?

I would look in the NEC for the full load amperes of these motor sizes. These are: 7.5 HP9 amps 40 HP.41 amps 60 HP.62 amps The 7.5 HP motor KVA will be KVA 575 * 9 * 1.732 8.9631 The 7.5 HP motor KW will be KW 575 * 9 * 1.732 * 0.8 7.17048 The KVAR of this motor is Sqrt(8.9631^2 - 7.1705^2) 5.3778 The 40 HP motor KVA will be KVA 575 * 41 * 1.732 40.8319 The 40 HP motor KW will be KW 575 *41 * 1.732 * 0.8 5 34.707115 The KVAR for this motor is Sqrt(40.8319^2 - 34.7071^2) 21.5095 The 60 HP motor KVA will be KVA 575 * 62 * 1.732 61.7458 The 60 HP motor KW will be KW 575 *62 * 1.732 * 0.8 5 55.5712 The KVAR for this motor is Sqrt(61.7458^2 - 55.5712^2) 26.9141 The total KVA requirement for all motors running at once is 8.9631 + 40.8319 + 61.7458 + 61.7458, which is 173.2866 KVA (note this is the requirement from a transformer, not the size of the transformer The total KW of all the motors is 7.1701 + 34.7071 + 55.5712 + 55.5712 153.0196 KW The total KVAR of all the motors is 5.3778 + 21,5095 + 26,9141 + 26.9141 80.7145 KVAR The power factor for all the motor is KW / KVA 153.0196 / 173.2866 0.883 I took a short cut at this point and used an application i wrote to calculate the required capacitor KVAR. The results follows: At 0.95 power factor, the motors' KVA will be 161.065 The motors' KW will be the same. The motors' KVAR will be 50.2926 The required capacitor's KVAR will be 53.988 The reactance of the capacitor will be 10.65 ohms The capacitance will be 3.49059248 E -4 Farads You can calculate these values as I did above, if you want to. EDIT Forgot the last answer. The line voltage is the same for both delta and wye. The leg voltage for the wye motor will be 575 / 1.732 331.98 volts TexMav

Q:I own a 1973 house and want to do a little remodeling. It has a 25foot hallway with 3 (old fashioned) ceiling lights that I want to replace with recessed lights. I went into the attic to look over things - looks like each e-box is made of bakelite which makes sense due to its age, attached to a rafter. However, there appears to be a transformer with what looks like low voltage wires -redwhite- on top of the box.Can a qualified electrician please tell me what this is and do I need to worry about it when I retrofit to recessed cans?

I'm not an electrician, but I have a few ideas. First, Do these lights use standard incandescent bulbs, or something else? If they are plain ole incandescent bulbs, I'm left wondering what is going on. If I understand you correctly, each of the three lights has an old bakelite electrical box with its own transformer, with regular wiring coming into it, and low voltage coming out of it feeding one light fixture below. If thats the case, then you are going to remove the transformers along with the fixture. However, if its one transformer that feeds the three fixtures, then you have a low voltage light system and will want to remove the fixtures and transformer and run new electrical wire (Romex) to each of the three new fixtures. The previous answer was correct in that most houses only have one low voltage transformer that powers the doorbell buttons and chime(s). Some homes back then had intercom systems and/or security systems that would require a transformer. Another point: check the smoke detectors. If they are over ten years old, replace them. They only work for about 10 years (even the new ones).

Q:How to connect in parallel the following transformers:a) Dy1 Dy11b) Dd0 Dd6

does this make sense to any one? delta /star 1 and delta start 11? is thi 1 transformer with a turns ratio of 1:11? question doesnt make sense?

Q:Production of high and low voltage switchgear transformer industry belongs to what industry category

Electrical, power distribution equipment? On the "switch talent network" or "electrical excellence network" to check it.

Q:do you know when the knew transformers film is coming out ( the year as well)

Transformers: Revenge of the Fallen (2009) Release Date: 24 June 2009 (UK) :] Hope this helped. I soo wana' see this movie.

Q:A single phase transformer operates from a 230V supply. The primary and secondary resistances are 0.03 Ω and 1.12Ω respectively while the corresponding leakage reactances are 0.1Ω and 6.4Ω; the magnetising branch can be neglected. The secondary winding has four times as many turns as the primary. Calculate :i) The equivalent impedance referred to the primary circuitii) The secondary terminal voltage when the load has a resistance of 200Ω and an inductive reactance of 100Ω.iii) The secondary terminal voltage when the load has a resistance of 200Ω and a capacitive reactance of 100Ω.Can someone show me how to tackle this question please?

The is something wrong with your leakage reactances or the turn ratio, because N^2Lls/Llp. I'll neglect this issue. i) The total impedance of the transformer referred to the primary is Zt .03+j.1+(1.12+j6.)/16. ii) VloadVin*ZLoad/16/(Zt+ZLoad/16) ii) Same as above, except ZLoad now becomes the parallel combination of 200 and -j100.