A step-up transformer has a primary coil with 100 loops and a secondary coil with 1,500 loops. If the primary coil is supplied with a household current of 120 V and 15 A, (a) what voltage is produced in the secondary circuit? (b) What current flows in the secondary circuit?
213657
This transformer will provide 1800 volts in the secondary.\nConsidering 15 amp flowing in the primary, it indicates 1 amp of current in the secondary.\n\nReal world losses not considered..........
49569
Its using induction charge principle. Essentially the primary coil side of the transformer is built into the base and the secondary coil is built into the toothbrush. \n\nThe primary winding of the inductive charger transmits charging power to the secondary winding, inside the target device, without making direct electrical or mechanical contact with it. In the target device, power is received with a coil and then rectified by a diode. This DC voltage is used in charging the battery of the device.
419127
It is a step down transformer.\nNs/Np = 6/220\nwhere Ns is the number of turns in the secondary and Np is the number of turns in the primary.
488095
Ohm's law:   V=RI\nBy changing I you will have V adjust to keep R constant.\n\nIf I goes up, so will V, because it's a direct relation.\n\nTry it with 100 ohm.  You can have\n100=R=V / I=500 volt / 5 amps\n100=R=V / I=700 volt / 7 amps
317350
b) and c)\n\nthe transformer works in such a way that electric current flows through the secondary circuit only when the electric current in the primary circuit is changing.\n\nwhen you close the switch the electric current in the primary circuit suddenly rises from 0amp to some value\nwhich causes electric curent to flow through the secondary circuit\n\n.. b) is true\n\nwhen you open the switch the electric curent in the primary circuit changes from the maximum value to 0amps thus electric curent flows through the secondary.\n\n.. c) is true\n\nbut \n..a) is false\n\nsicne a battery is a dc source wich means that as long as the switch is closed it supplies a steady amount of electric current through the primary. thus ther will be no current through the secondary.
500516
it depends on the load and the nature of the large appliance\nthe flush example was great but it does not describe dthe momenteraly effect ( it was nice though).\n\nwhen you deal with inductive loads there is something that is called start up current.  it can be 2 to 6 times the nominal current under normal operation. that means that if a device needs 2 amperes of current when it is originally powered the consumption may reach till 10A. then the current returns to normal copnditions. \nall due to inductive and self inductive effects where the device opposes to the change and so on till an equilibrium is reached.\n\nthe demand of the current during start up is great and that is way momentarely  the lghts dim.
406375
a) 2900 * (54 + 0.29 v)\nb) 9 * 2900/v\nc) 2900 * (54 + 0.29 v + 9/v)\nv = 5.57
117333
a) E=v/l ---> E = (12/3)*1000=4000 (Volt/meter)\n\nb) v ~ 1/r ---> v2=12*3/5=7.2 Volt\n   E=(7.2/5)*1000=1440 (Volt/meter)\n\nc) w=int(f.dx)=int(Eqdx)=int(vqdx/x)=int(36qdx/x^2)\n\n    = 36q*(-1/5+1/3)=72q/15\n\nq=4*pi*e*l*v/2=2*3.14*8.854*10^(-12)*0.003*12 = 200*10^(-12)\n\n\n---> w=960*10^(-12) J
602003
You're confusing resistance with pressure. If indeed there is a difference in voltage between A and B then there will be flow.\nThink of a tank filled with water held a few feet above another tank that has a pipe joining the two. Water will flow to the lower container thru the pipe. In this case the pipe has no resistance, which means water flows at maximum possible speed.\nIf both tanks are at the same level, then, sure, there will be no water flow. That's not how the problem was stated, however.
374608
1.a) receives the alternating current. \n b) the magnetic properties are altered by the electromagnetic field. . \nc) by the alterations in the core induces other alternating current in this coil. \n2.a) because the alternating current induces the varying magnetic field.\nb) that magnetic flux is passed by the core to the secondary coil.\nc) yes an alternating one. \n\nBecause radioactivity destroy organic tissues.
128782
in a series circuit, current is constant across every resistor.  Intensities (brightness, I assume) of A and B to C should be the same as the power across all the resistors is the same.  \n\nHowever, by the sound of your question, it sort of would make more sense if A and B are in series, but then C is connected in parallel when switch S is closed.  Sorry, these questions are hard without diagrams.  I'll watch this question and edit my answer if you clarify the question.\n\n*edit* for Sky's question...to close a circuit is to "turn it on," allowing electricity to flow.  In an open circuit, the circuit is "off," electricity cannot flow.
171951
a ditisbution transformer is much larger . but they both do the same job .they either step up or step down current.
165082
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
467314
Well, you see, that isn't done.  When you need single phase power, you use only one phase of the supply circuit.  Each of the phases of the 480 volt 3 phase power will be run through a separate transformer to step the supply down into 3 phases at 120 volt.  Typically, a house is served by two of those three phases.  You use two phases with two hot wires and a neutral to get 240 volt power for your oven, range, dryer and water heater.  The other circuits in the house are served by one phase or the other at 120 volts with one hot wire and a neutral wire.
473056
First thing to do: with no power applied, check continuity through the magnetic chuck with an ohm meter.  If there is no continuity between the power leads, there is a broken electrical connection that needs fixing.  There should be no continuity to ground.\n\nSecond thing to do:  check the nameplate electrical rating to make sure you're applying the correct voltage (110VAC, single phase).  If the nameplate requires a different voltage, apply that.  If no nameplate, you might want to check out the Rockwell website (if it is a Rockwell component; I suspect it isn't, probably aftermarket so check out their website).\n\nOne of the things to consider is that while the magnet might be able to operate on AC, the actual electrical current going through the chuck coils may be DC, so you need to rectify the current somehow.  I'd look for some diodes or rectifiers inside the chuck and check those for proper operation - if they failed there would be no magnetic chucking even if you had it plugged into the wall.  Failed diodes will fail the continuity test in step one.\n\nYou probably have to remove the chuck to take it apart; you'd only do this if it failed the continuity check because you'll need to regrind the surface of the chuck after reinstalling to make the chuck surface parallel.\n\nThe other possibility is that external DC current needs to be supplied through some power supply that you don't have; this is where the nameplate comes in handy telling you if you need DC or AC current.\n\nI would think you'd need an electrical current limiting device on the input side to the chuck (like a transformer) rather than running AC from the wall straight into the chuck - too much current would blow the circuit breaker or cause a fire.  Be careful!
20448
If you happen to turn on at peak voltage the voltage across the transformer or generator will be 1.4 times 115. This is your PEAK VOLTAGE. Consider the uncharged capcitor to be a direct short. The formula for the IMPEDANCE of your coil is: impedance = 2 times 3.1416 times 440 times 8.6 dived by 1000. Divide your PEAK VOLTAGE by the IMPEDANCE of your coil. This will give your MAXIMUM PEAK CURENT. The actual peak curent will be lower because of DC resistance in the coil, diodes, and transformer. You can measure this with an ohmeter.The DIODE RESISTANCE will be 2  times the forward resistance of one diode because in a full bridge the curent has to flow through 2 diodes. You probably can't get an accurate diode resistance because it is nonlinear and goes down with increased curent. Forget the diode resistance it is low anyway.  Add the TRANSFORMER RESISTANCE or GENERATOR RESISTANCE to the IMPEDANCE to this to get the TOTAL IMPEDANCE. Divide the PEAK VOLTAGE by the TOTAL IMPEDANCE to get the REAL WORLD PEAK CURENT. Parts are not perfect. Give a tolerance of two times in your specifications to avoid stressing them and use a heatsink on the diodes. I was going      to calculate this for you but the batteries in my calculator were dead.
536079
The voltage can reach levels of over 500,000 volts  when it comes out of an eletric plant.  \n\nThe high voltage electric current travels through transmission lines. A second set of wires is used to carry the current back to the generator. These wires are there because electric current travels in a closed circuit. High-voltage substations are places where very high voltages are reduced to lower levels. High-voltage power lines enter and leave the transformer through large insulators. These insulators are used to keep the high voltages away from the metal tank containing the transformer coils. Oil flows around the coils to keep them cool. \n\nThe electricity now travels from the substations to transformers in your neighborhood. These transformers, called "Step-Down Transformers," lower the voltage still further to make the voltage safe for ordinary use in your home. Inside these transformers are coils of wire which help reduce the voltage to 120/240 volts which then can be used safely in homes. \n\nThere is no energy wasted along the process. Remember that energy cannot be destroyed but can be transformed. There is no transformation of energy on the process thus it is not wasted.☺