Q:Distribution transformer

The low voltage winding is placed closer to the core because the high voltage winding requires more space due to its MV insulation. If designers put the LV winding outside the MV winding, the tank would be larger because the _MV_ winding size would be about the same size. More insulating oil would be required, and the leakage reactance would be significantly higher. Remember, higher leakage reactance means that the voltage regulation (more aptly voltage drop) under load will be higher, which is not desirable. And, of course, a larger tank and more oil make for a more expensive transformer. Edit: MV winding is about the same size for either configuration, due mainly to insulation requirements.

Q:i rememeber when i was a lil kid i used to watch Beast Wars on friday nights. I think it was back in 2000 or so. It was pretty cool too; as opposed to transforming cars, etc, there were transforming animals. also there werent any humans.imo the new transformers animation sucks, though the movie was greatim sure Beast Wars and Transformers are related because they used names like Megatron and Octomus Prime (sp?)so yeah. what happened to Beast Wars?

yea beast wars was old school, and i think its better than cars. im not sure why they changed it

Q:What is the transformer absorption ratio (R60 / R15)

Generally, the absorption ratio should be greater than or equal to 1.3 as the absorption ratio is acceptable.

Q:i study transformer and found calculation for pf for open/short circuit test. but the value is bit confusing.pf at open test: 0.182pf at short tets: 0.135what the value means??why we need that??

The power factor and reactive volt-amperes are only important as intermediate steps to calculating the values of the transformer's equivalent circuit component values. The pf determined in the open circuit test is related to the magnetizing reactance and core loss resistance. The power factor determined in the short circuit test is related to the primary plus secondary leakage reactances and winding resistances. The equivalent circuit component values are needed for various transformer performance calculations and for calculating the influence of the transformer on the complete circuit. I don't believe that the open circuit and short circuit pf and VARs are used for anything once the equivalent circuit has been determined.

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?

That is probably the transformer to your door bell.

Q:Transformer secondary side is high pressure or primary side is high pressure

Transformer can be divided into two types of step-up transformer and step-down transformer, the primary side of the transformer is the voltage input, the secondary side is the output. Step-up transformer is the secondary side voltage greater than the primary side, step-down transformer is the secondary side voltage is less than the primary voltage.

Q:i need a transformers costume that can transform into a car, truck, etc i need to buy it for this kid for his birthday. medium size please. and he wants it as optimus prime.he wants it to look good and if he doesn't like it, well hes gonna be sad

wrap urself in tin-foil. lol

Q:I have a DC current transformer and it is a 2500:1. I do not know what this means but I have a rectifier hooked up to it and am measuring the output DC voltage. I have hooked up resistors in values of 20Ω, 2kΩ, 10kΩ, and 20kΩ. I do not know how to convert to amperes though the resulting DC volts. How can this be done?

When I say voltage I mean AC voltage. When I say current, I mean AC current. Let the input voltage you apply to the primary (the 2500 side) of the transformer Vin. Let the open circuit output voltage (no load across the secondary) of the secondary (the 1 side) of the transformer Voc +++ A certain magnetizing current, lets call it Im will flow in the primary at no load. The primary winding of your transformer has a resistance, lets call it Rp. The secondary winding of your transformer has a resistance, lets call it Rs. With no power applied, if you have the equipment you can measure Rp and/or Rs. If you have the equipment, if you do not apply a load to the secondary you can apply Vin to the primary and measure Im. Due to the magnetizing current, there is a voltage drop across ip of Im*Rp volts. Because of the voltage drop, if you apply an input voltage Vin to the primary the open circuit voltage can be calculated as Voc [Vin - (Im*Rp)]*(1/2500). +++ If you apply a load to the secondary winding, lets call the current drawn by the load Is. Because of the transformer action, the output current will be reflected into the primary. Lets call the reflected output current Ips. Then Ips (1/2500)*Is. Now there will be voltage drops in the primary due both to Im and Ips, so NEGLECTING THE EFFECT OF THE RESISTANCE OF THE SECONDARY the secondary voltage is given by Vo (1/2500)*(Vin - [Rp*(Ips+Im)]) But there will be a voltage drop in the secondary. So the actual Vo, NEGLECTING NOTHING is given by Vo (1/2500)*{Vin - [Rp*(Ips+Im)]} - Is*Rs}.

Q:i just obtained a transformer from a small ac-dc converter. usually, it converts the 240v ac (let's call it terminals a and c) to 4.5v dc (terminals d and c).what if i connect a 4.5 dc supply to the terminals d and c? meaning that where i would usually obtain the converted electricty, instead, i connect a 4.5 to it, leaving the terminals where i would usually connect to the wall empty. wouldn't the magnetic field step up to 240v?

TRANSFORMER BECOMES SMOKING IN A FEW SECONDS IF THE 4.5V DC HAS HIGH ENOUGH CURRENT.

Q:If I have a primary coil and secondary coil each with 400 turns that acts as a transformer, and the primary potential difference is 2.988 V, and the secondary potential difference is 0.0551 V.I need to find the efficiency. Ideally, it would be 100%, but this is the real world. So I did N(turns) and Volts and used the equation [N(p)V(s)/N(s)V(p) ] *100%p primarys secondaryNturns of coilV potential differenceI had an answer of 1.84% which seems awfully low. Help please. Thanks!

You are right. The efficiency is that low. Take a look at it another way: Pin Pout if 100% efficient. Since we know a 1:1 turns ratio will produce the same voltage from one side of the transformer to the other, the current produced on the secondary is extremely low to pull the voltage down to .0551V. It makes this transformer extremely inefficient! I can't even think of how you would design a transformer that was this inefficient. I think even a large OD iron nail with a lot of eddy currents at 7.5k Hz I believe would be more efficient. The next worst way I can think of how to produce an isolated output is with a TEC using the Seabeck effect that is about 6% efficient.