What is the basic design features of a dual voltage transformers?
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hermetically sealed, high voltage....
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I am not sure but I would guess that the polyester will hold up a little better than the paper through all of the movement of the windings. Adding to that, the mineral oil used for cooling the transformer may deteriorate the paper as well.\n\nAll of the above is speculation based on what I have observed
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Here is a good article. To understand them, you need to get a multimeter, and electronics projects book, a breadboard (to mount circuits on), and a selection of electronic parts from Radio Shack. Use batteries initially, and then when you are more confident a power supply and rectifier to do experiments, and measure voltage, current, and resistance values around the circuit.\nhttp://en.wikipedia.org/wiki/Electronics\nHyperphysics. (Google)\nhttp://www.mathworld.wolfram.com\nhttp://www.refdesk.com\n"You can do it electronics," Needham Heights, MA\nEnjoy!
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n1/n2 = V1/V2\nFor step down transformer n1>n2\n\ntherefore, 11000/400 = 120/n2\n\nn2 = 400 * 120 / 11000\n   = 48/11\n   = 4.36\n   ~ 4 turns
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Depending on the utility company, some use Dyn11 and others Dyn1 or some other vector group that is universal to their\nsystem.\n\nThe reason being is that  they can be interconnected with the low voltage side of another transformer that is connected to the same high voltage supply circuit.\nDyn 11 means that the vector group is at the 11 o'clock position. If for instance two transformers on the same high voltage line side had different vector groups say Dyn11 and Dyn 1 they would always be 30° out of phase, this would mean that should they be interconnected they would immediately trip the incoming CB/fuses.\n\nSource: Electrical Engineer (ret.) with 45yrs experience
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A transformer style welding machine converts the high voltage and low current electricity from the utility into a high current and low voltage, typically between 17 to 45 volts and 190 to 590 amps. This type of machine typically allows the welder to select the output current by either moving the core of the transformer in and out of the magnetic field or by allowing the welder to select from a set of taps on the transformer. These machines are typically the least expensive to purchase for hobbyist use.
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Most AC electrical sources can be converted to other phases, voltages and currents but not to increased power (P = IE) either using semiconductors, transformers or motor generator sets (and transformers).  Usually 240 Volt AC is obtained using two phases of a three-phase distribution system where a single phase is 110 Volts nominal (perhaps 120 Volts actual or stepped up) to double voltage to 240 Volts.  However, in the lab, many different phases, voltages and currents may be required for special applications and suitable converters must be found (or built?).
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Your question is missing.  However, I can say that there are no power losses in an ideal transformer.  Therefore, if the ratio of the currents between primary and secondary windings is 100:1, then the ratio of the voltages must be 1:100.  That is to say that the voltage across the secondary is 100 times greater than the voltage across the primary.
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You need two of those as fans. Connect them in series.\n\nOr...\n\nIf you don't have two of it, get an appliance which has the same rating and connect them in series.\n\nOr...\n\nPlug it to a voltage half its rating.
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How about this one?\n\n"This calculation module allows you to quickly calculate fault currents for a complete radial system. It can simultaneously handle a main distribution transformer, service lateral, second transformer, up to 40 main-feeders, with 8 main-feeders having 40 sub-feeders for a total of 360 nodal points. \n\nAccommodating both single-phase and three-phase circuits, the module will calculate from either infinite or known fault current values at either the transformer or the main-bus. \n\nThe module accommodates multiple conduit types, bus types, multiple conductors and motor contribution. The program is based on the popular LICE (point-to-point) calculation method. \n\nOur unique Help file (of which we are quite proud) includes complete practical guidance on Fault Calculations, and even includes a theoretical derivation of the LICE method. "\n\nhttp://www.edreference.com/fault.asp
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What voltage dc, or do you need to be able to vary it?  What current capacity do you need?  What ripple voltage is acceptable?
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You'll need to provide a little more information in order to have your question answered properly.  AC or DC? Input voltage?  Fixed or variable output voltage? If variable, over what range? If fixed, what is the desired output voltage?  Output current?  How critical is the load regulation?\n\nAs you can see, the answer to your simple question can be a bit complicated :)
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independent: voltage, length of wire, number of coils\ndependent: magnetic effect (paper clips)
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I am not sure I understand your question. Are you asking, what are the effects of core losses in relationship with frequency and supply voltage?\n\nThe total power loss in a transformer is a combination of three types of losses. One is due to the DC resistance in the primary and secondary windings. This is called Copper loss. The two other losses are due to Eddy Currents and Hysteresis in the core of the transformer. Copper loss, Eddy Current and Hysteresis loss result in unwanted conversion of electrical energy into heat energy.
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It is DC voltage.\n\nhttp://www.ecs.umass.edu/ece/allerton/papers1999/36/
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Hi! A transformer works on the principle of mutual induction which is valid only when the voltage (or EMF, to be more appropriate) on one of the coils of the transformer is changing with time, as in the case of A.C. voltage. As the magnitude of the voltage changes on one of the coils, the corresponding flux on the other coil also changes, thus giving rise to a voltage on the other side.\n\nAn AA battery provides a constant voltage, as a result of which there is no change in flux and no voltage on the other coil (120 V side, in this case).\n\nHowever, there will be a flux change for a very short duration of time as the current in the 24 V side rises from 0 to its full value after the battery has been connected. During this time, a voltage will be induced in the 120 V side and you'll get an A.C voltage on this side. But this would be for a very short time duration.\n\nSo, the answer: After some time, there is no voltage (A.C. or D.C.) on the 120 V side.
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105W\n\nA transformer changes the voltage, but inversely alters the current to the same degree. Therefore, using P=VI, the power is always the same.