How to work induction furance?
354928
If you mean: how do induction furnaces work, then an answer is by heating the metallic contents by the circulating eddy currents caused by the ac in the surrounding coil. There is another type which uses induction heating or dielectric heating of the contents. In this case the contents are usually non metals.This method was used in an early type of heater for meat etc.
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
486492
THE EASIEST WAY TO DEMONSTRATE GENERATION OF ELECTRICITY IS TO GET SOME SMALL INSULATED COPPER WIRE AND WRAP IT AROUND A SMALL PILL BOTTLE,  AT LEAST 100 TURNS OR MORE (MORE IS BETTER) CLEAN THE INSULATION OFF THE ENDS OF THE WORE AND ATTATCH TO A VOLTMETER OR A MILLIAMETER.. INSERT A BAR MAGNET INTO THE PILL BOTTLE QUICKLY.  THE METER WILL GO ONE WAY WHEN YOU INSERT THE MAGNET AND WILL GO THE OTHER WAY WHEN YOU PULL THE MAGNET OUT.  THE FASTER YOU MOVE THE MAGNET THE MORE THE METER MOVEMENT.  HOPE THIS HELPS
856026
Use the gas law\nPV=nRT
207863
see:\nhttp://en.wikipedia.org/wiki/EngineerIng
604852
Here are a couple of web sites that approach this in different ways. \n\nhttp://www.circletrack.com/techarticles/84199/\n\nhttp://www.diracdelta.co.uk/science/source/e/n/engine%20excitation%20mechanisms/source.html\n\nHope these help.
310895
Magnetic flux \nweber: Wb = V s = W s/A = m^2 kg/s^2 A \n\nMagnetic flux density, magnetic induction \ntesla: T = Wb/m^2 = kg/s^2 A
616514
Generating static electricity.
53767
Inductance is the "definite answer" that I was given 30 years ago. I don't suppose that my confirmation is any better than yours, but it's a vote anyway.
170254
Try these sites. Some have better explanations than others.\n\nSwitching and inverters: http://www.kpsec.freeuk.com/trancirc.htm\nAmplifiers: http://www.tpub.com/neets/book7/25c.htm\nLogic gates: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/trangate.html
890360
I have already answered part 1. in another question you've asked. I think you probably didn't use the right units (like cm needs to be convered to meters) and got the wrong answer.\n\nNow for part 2:\n\nInduced EMF, ℰ = -L(di/dt)\n\nwhere i is the current and L is the inductance\n\nWe are given that the magnitude of the induced current,\n\ndi = 47 mA\n\nand the oscillation frequency is 100 Hz. This means that dt = 1/f = 0.01 sec\n\nWe have L = 8.44 × 10-^3 henries from before\n\nSo |ℰ| = L(di/dt) = -(8.44 × 10-^3)(0.047)/(0.01) = 0.0397 volts\n\n------------------------------------------------------\n\nNow part 3 has to do with mutual inductance:\n\nVs = VpNs / Np\n\nwhere \n\nVs is the voltage across the secondary inductor, \nVp is the voltage across the primary inductor (the one connected to a power source), \nNs is the number of turns in the secondary inductor, and \nNp is the number of turns in the primary inductor. \n\nTherefore Vp = VsNp/Ns and Vs = 0.0397 volts as calculated before, so\n\nVp = (0.0397)(197)/(2920) = 0.00268 volts
620886
(m)\n\ngoto the following website for your question with neat diagrams\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html
1086026
You can get an idea on counting the discs on the insulator string and multiplying by 12kV/disc.\nAgain it is only a general rule of thumb and is not accurate.\n\nSource 50 years as an Electrical Engineer
388634
Inductive reactance can be calculated using this formula: XL = 2πfL\n\nGood Luck
786730
Use the right hand rule, also known as the rule of thumb. If you have a wire (insulated!) with a current flowing through it, grab the wire in your right hand, with your thumb pointing in the direction that the current is flowing. Your fingers will curl around the wire in the direction of the magnetic field.\n\nIf the current alternates (periodically reverse direction) then the magnetic field also alternates. Envision it as a cylindrical zone that has the wire down its axis, and that the cylinder expands and contracts as the current of electrons increases or decreases. \n\nNow, suppose we wrap our wire around a coil form to make a coil. There is also a right hand rule, or rule of thumb, for coils. If you hold the coil (again, use insulated wire!) with your fingers wrapped around it pointing in the direction that the electron current flows, your thumb will point in the direction of the north pole of the coil.\n\nHow does that work? Remember, we are imagining the magnetic field as a cylindrical region around the wire. But the surface of the cylinder is rotating. (We'll stick with current in only one direction for the time being. It's easier to follow.) So think of the flow of the magnetic field as rotating around the wire. \n\nNow, if we wrap the wire into a coil, the rotation of the segements of the magnetic field add up in the same direction. They are working together. This is why the field is stronger inside the coil.\n\nWe should also think of the magnetic field as composed of individual lines of force that repel one another. That is why, when they exit one end of the coil and return to the other, they spread out in space. \n\nWe can see this if we hold a magnet under a sheet of paper or aluminum foil and sprinkle iron filings on top. They cluster most around the poles where the field lines are concentrated-- the field is strongest-- and more thinly around the outside of the magnet, where the field lines spread out, because the field is weaker there. This holds true for any magnet, not just electromagnets.\n\nI hope this is clear enough to help. It is a little tough trying to understand things we can't see.\n\n19 NOV 06, 1816 hrs, GMT.
815430
"Induction heating is the process of heating a metal object by electromagnetic induction, where eddy currents are generated within the metal and resistance leads to Joule heating of the metal. An induction heater (for any process) consists of an electromagnet, through which a high-frequency AC is passed. Heat may also be generated by magnetic hysteresis losses."\n\nThis and much more info is available on the free, online encyclopedia, Wikipedia:\nhttp://en.wikipedia.org/wiki/Induction_heating\n\nHope this helps!
891080
1. Xc = 1/2πfC\nSo 2πfC = 1/Xc\nSo C = 1/2πfXc\n\n2. ib = ibq + √2is\nSo √ = ib - ibq\nSo 2is = (ib - ibq)²\nie is = ½(ib - ibq)²\n\n3. Q = XL/R\nSo XL = QR\n\n4. Lt = L1 + L2 - 2Lm\nSo 2Lm = L1 + L2 - Lt\nie Lm = ½(L1 + L2 - Lt)\n\n5. R1=i2 R2 / i2\n??????? So R1 = R2\n\n6. Es = E1 + E2 + E3\nSo E2 = Es - E1 - E3\n\n7. R1 / Rx = R2 / R3\nSo Rx / R1 = R3 / R2\nSo Rx = R1 * R3/R2\n\n8. y = Mx + b\nSo Mx = y - b\nSo M = (y - b)/x\n\n9. i2=i1 R1 / R2\nSo i1 R1 = i2R2\nie i1 = i2 R2 / R1\n\nI would be extremely concerned about your progress if you find this simple algebra a serious chore!!!!!! (Especially if you are learning the electrical trade)