Isn't a motor a fixed kW?  If so, Ohm's Law says that if the voltage is reduced during a soft start, then the current should increase.  However, I've heard that a soft start reduces the demand of the motor.  How does this work?
130341
Ever think about a motor? It's just a coil of wire. Therefore, when you apply a voltage, you get a huge current. What stops it is counter voltage induced in the coil that opposes the supplied voltage. This only occurs when the motor starts turning. By limiting the initial current, it gives the motor a chance to spin up slowly and develop this counter force. Then as you apply more current, the motor comes up to full speed under smaller current spikes. This extends the life of the coil. Now if they would just do this with light bulbs.
132812
If you change voltage say from 100 Volts to 200 Volts (perhaps by changing a switch position), the nearly instantaneous change will be a step change (like the steps on a stairs).  If you change the voltage from 100 to 200 Volts uniformly over a period of time (perhaps 10 seconds) for example by rotating a rheostat, the voltage will increase more like the ramp provided for wheelchairs than steps and will be a ramp change.  The terms are often used to describe the graphs or stripchart recorded traces of electrical signals relative to time.\n\nIn electric circuits, selecting various resistors with a switch may produce step changes and charging or discharging capacitors may produce ramp changes.
667686
motor converts electrical energy in to mechanical energy.\ntypes:\n  AC MOTOR--> it runs on ac current( eg :- ceilin fans, exhaust fans etc)\n  DC MOTOR-->  it runs on dc current(eg:- the ones inside toys)\n  SYNCHRONOUS MOTOR--> it runs at constant speed all the time(eg:-air compressors , gen set etc)\n  SERVO MOTOR---> it is precision device in the sense tht each   angle turned by it can be precisely controlled(printers , floppy drives , hdd etc)
578960
The voltage will be on the order of 10,000 to 20,000 volts on small engines.  The magneto is capable of producing such voltages on these engines as well.  If you happen to touch the plug while the engine is running, you will get a nasty, but non-fatal shock; let's just say it is a real attention getter, sort of like sticking your hand into a hornet's nest!!.\n\nBTW, standard ignitions on cars of yesteryear were about 18,000 volts.  Today's cars run 50,000 volts.
137022
Someone has used a gasifier and captive-piston engine to make a one-moving part alternating-current generator. The piston is magnetized and the coils are in the body of the cylinder. The trapped air in the cylinder at each end of the piston acts as an air shock to eliminate rebound and impact damage. No sensors needed. No make and break circuit. The voltage drops to zero at the end-points of travel of the piston. Search on this and you can probably find the same one I saw. I think this is where they got the idea for those new shake-up LED flashlights.
240135
Lower voltage means less electrical pressure, so likely less power consumption.\n\nProbably what you are seeing is line voltage fluctation due to induction devices coming on-line (motors, fluorescent lights, etc.)  Induction devices cause a power phase shift ("power factor") which definitely cause a negative power impact (higher kWh).\nWhen it gets bad enough, the power company will make you install capacitors to counteract the inductive load.  \n\nWhen the phase shift reaches 90°, all power goes into oscillating the current (and making heat) and none into the load.
9501
Some systems actually cool as they deliver power.  A rubber band is most common example I can think of.  If you expend energy to stretch a rubber band, it actually heats up, but it cools slightly as it contracts, which could potentially power some small device like a toy car.  Why this occurs is a little complicated, but the exact same principle of entropic cooling is used to achieve the lowest temperatures currently possible.\n\n***Response to first answerer***\nWhat basic law of motion are you talking about?  The second law of thermodynamics? Because that deals with entropic differences, and doesn't apply. The "laws of motion" aren't even relevant to this thermodynamics question.  \n\nI used the example of a rubber band because it is a clean example of entropic cooling, but in fact ANY  heat engine will result in a decrease in thermal energy of the system. It's true that there is some heat produced from friction in the engine, but on a whole more MORE heat flows into the engine than flows out.  The average temperature of the system decreases. That's why we can use heat engines to power things.
339622
The resistance of the filament of the lamp varies with temperature. When the filament is at room temperature the resistance is much lower than when it is white hot. \n\nCurrent = V/R. If the voltage is constant and the resistance increases, the current decreases.
174525
The back EMF that must be protected against is often called inductive kick. It's a voltage spike that happens when you interrupt the current flowing through an inductance such as a solenoid or relay coil. The diode is more to protect what's connected to the coil (and probably what interrupted the current to the coil) such as a switching transistor or a set of contact points. The diode is connected across the coil and its polarity is such as to short out the spike. This always works out not to short the current away from the coil when current is flowing.\nMost motors and pumps being AC, a diode can't be used for such protection although the inductive kick can still occur. Surge suppressors are used instead.
481130
Brushless generators have a fixed magnetic field which rotates through the AC output windings.  As the engine speed or load varies the output voltage will change.\nGenerators which are not brushless are able to control the magnetic field.  This supplies the ability to introduce a feedback loop to control the output voltage.\nRead my generator FAQ.  It may better answer your question.\nhttp://members.rennlist.org/warren/genfaq.html
302167
First there is a correction to your statement.\n\nResistance increases with the length of the wire or conductor; but it decreases with the area of cross section of the wire or conductor. \n\nIt will give a wrong meaning if you use the word breadth instead of area of cross section.\n\nConsider a wire of length L and area of cross section A connected in a circuit. Let n be number of charges flowing through it under a given condition.\n\nNow connect a similar wire side be side with the previous one. (This is called parallel connection)\n\nAs before in the first wire there will be n number of charges flowing through it.\n\nWhat about the second wire. Since the condition is the same as the first one, in the second wire also there will be n number of charges flowing through it. \n\nThus if the area of cross section is doubled current is increased or resistance is decreased.\n\nIn a high way there is a bridge which is very narrow; the flow of vehicles will be less.\nIf the bridge is broadened the flow will increase.\n\nHowever the above is an analogy to make the concept some what to understand. But the charge in a wire is different from the vehicle flow.  \n\n\n\n.
184145
If you know how many watts the motor can provide then dividing that by 746 will give you horsepower.\n\nThe Amperes and knowing the supply (ie. 220V) can give you horsepower since amps times volts gives watts.\n\nThings get complicated if these readings are not a measure taken just before the point of stalling the motor with a load on the motor to induce it's full power output.
114210
A resistor reduced the voltage in an electrical curcuit at the same time as is applied some load it increases the amps.\nThe formualr volt = amp x ohms can be used to calculate the required resistance.\nA resistor had a measurement unit of ohms.\nVolts = I x R\nResistance = V/I\nI(Amps) = V/R\n\nSo if you had a simple circuit with a battery and a lamp, if you wanted to make the lamp dull you would put a resistor in the curcuit you then could change the value of the resistor to make the lamp brighter or duller.
493693
Generator sizing becomes more difficult when electric motors are involved, due to their large starting demand.  Typically, there are two motor styles: CODE G requires 3 times running watts to start, and CODE L requires 5 to 6 times running watts to start.\n\nIt is not practical to think that all loads will start at the same time.  Nor is it practical to size a generator using all starting watts for all selected motors, as some motor loads will already be running (or OFF), when others are ready to start.  As you can see, it can be difficult to exactly size a generator for a group of specific loads, as they are constantly changing in their "ON" or "OFF" cycle.  Always try to limit your load selections to the bare essential items.\n\nTO FIND:\n1 PHASE KW = V * A * PF / 1000\n3 PHASE KW = V * A * 1.73 * 0.8 / 1000\n1 PHASE KVA = V * A * 1000\n3 PHASE KVA = V * A * 1.73 / 1000\nMOTOR - REQD 1 PHASE KW = ( MOTOR HP) * (EFF)\nMOTOR - REQD 3 PHASE KW = (HP) * (0.746) * (EFF)\nMOTOR - REQD 1 PHASE KVA = (MOTOR HP) * (EFF) * (0.8)\nMOTOR - REQD 3 PHASE KVA = (HP) * (0.746) * (EFF) * (0.8)\n1 PHASE PF = 3 PHASE PF = KW / KVA\n1 PHASE AMPS = (KW * 1000) / (VOLTS * PF) =  KVA * 1000 / VOLTS\n3 PHASE AMPS = (KW * 1000) / (1.73 * VOLTS * PF) = (KVA * 1000) / (1.73 * VOLTS)\n \nREQUIRED PRIME MOVER (ENGINE) HORSE POWER\n KW / (GENERATOR EFF. X .746)\n \nFREQUENCY (HERTZ)\n NUMBER OF POLES X R.P.M / 120\n \n ~ ~\nEXAMPLES IN GENERATOR SIZING FOR LARGE ELECTRIC MOTORS \n\nAll AC electric motors require large amounts of electric power to start up, due to it's starting winding. After approximately 3/4 - 1-1/2 seconds, the starting winding drops out, and the running winding continues, at a much lower power demand.  A generator must be sized to handle the load of the starting winding.\n\nSizing a generator for three phase motor starting LRA : \nThe formula is: Motor HP x KVA per HP x (1000 / (Volts x 1.73)) = LRA. \nGiven the size of your machines, estimate KVA/HP at approximately 6.\nTherefore, for 11 kW ... LRA = 134.6 = 135 amps. Generator must produce this current at 380 V, so (1.73 * V * A * 0.8 / 1000) = KW = 71.\nSimilarly, for 15 kW .... LRA = 184 A and KW = 97.\n\nBOTH SINGLE PHASE AND THREE PHASE MOTOR SIZING EXAMPLES ARE BASED ON INFORMATION IN NATIONAL ELECTRICAL CODE HANDBOOK, ARTICLE 430.7 AND IS INTENDED FOR MAXIMUM VOLTAGE DIP OF 25% UPON INITIAL MOTOR START-UP.  IF 35% VOLTAGE DIP IS ALLOWABLE, REDUCE LRA, THEREFORE KW SIZE, BY 25% (LRA or KW X .75). IT IS NOT RECOMMENDED TO CHOOSE ANY GENERATOR KW SIZING THAT YIELDS MORE THAN 35% VOLTAGE DIP.\n\nStarting more than one motor, at one time: make a list of all motors, totaling all running watts.  Calculate the starting demand when the largest motor is started while all others are running.  Add the largest motor starting watts, to the running watts of all smaller motors already running.
627265
8.5x1000=8500 watts.  415 volts.  W=V*A A=W/V\n\nso 8500watts/415 volts =20.48... Amps\n\nthis is the basic relationship between watts volts and amps.  I dont know if the 3 phase motor part would change that answer.
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
338807
Partly, it's the path the electricity takes.\n\nBut remember, it's not the volts that kill ya, it's the amperes.\n\nI once shorted out the entire electrical system of an old car engine with my bare hands (don't ask why; I was an idiot)  This was thousands of volts - but there's so few amps going through the distributor that all I got was a nasty jerk and a really funky hairdo.  \n\nBut just because I survived that stupidity - don't try it.  You could get a big boo-boo.