How do you find voltage changes and current changes in the number of turns on the primary/sec of transformers?
472581
The equation for this is relatively simple.\n\nVp/Vs = Np/Ns\n\nWhere: \nVp=Voltage primary winding.\nVs=Voltage secondary winding.\nNp=Number of coil turns on the primary winding.\nNs=Number of coil turns on the secondary winding.\n\nThus taking that into consideration you should be able to get the voltage for both sides.  To get the current on both sides.  Just realize that you need the resistance.\nOr if you have the current on one side, use the conservation of Power equation:\n\nVpIp=VsIs\n\nWhere: \nVp=Voltage primary winding.\nVs=Voltage secondary winding.\nIp=Current in the primary winding.\nIs=Current in the secondary winding.
129261
Kilowatts  Equations\nDirect Current  =        I x E/1000\nAC / 1phase 115v or 120v =I x E x PF/1000\nAC / 1phase 208,230, or 240v= I x E x PF/1000\nAC 3 phase All Voltages     =    I x E x 1.73/1000\n\nE = Voltage \nI = Amps \nPF = Power Factor
448145
Transformers do not store energy, that would be a battery or a capacitor.  Transformers work by wrapping wire around an iron core that is basically a circle.  For each wrap of the wire on the input side, you divide the total number of wraps into the voltage. If the input is 120 volts, and you have ten wraps, you get 12 volts per wrap.  Then, on the output side, you get an equivalent voltage per wrap.  So, that means on the example I gave, if you have five wraps on the output side you get 60 volts.  (5 x 12 volts = 60) This example would be a step down transformer, because you have a lower voltage than you started with.  If you have 20 wraps on the output side, you would get 240 volts. (20 x 12 volts = 240)  This is a step up transformer.  Note: Using Ohm's Law, you do not actually increase the total amount of power available, only change it's configuration.  Thank you, Nikola Tesla, inventor extraordinaire.
644810
Your circuit has 6 resitance R1, R2, R3, R4, R5, R6 and correspond currents are: I1, I2, I3, I4, I5, I6 and voltage are: U1, U2, U3, U4, U5, U6.\nYou can make many kind of circuit(direct, parallel or both) depend on total voltage as well as resistance.
575172
V=IR, voltage = current times resistance. If you know the current flowing in the circuit, just plug that and the known resistance into the formula to get the voltage. If you don't know the current, you need to measure it. One way to measure it is to measure the voltage across the resistor and plug that and the resistance into the equation and calculate it. But if you are measuring the voltage across the resistor, then you don't need to calculate it, do you?
1105216
This question has a simple answer observe\n V=IR now R=V/I\n  so i.e.120/0.6 =200 ohms \n  if you heating effects please see the joule,faradays heating effects of electricity
869383
(m)\n\n transformer is an electrical device used to convert AC power at a certain voltage level to AC power at a different voltage, but at the same frequency. \n\nThe construction of a transformer includes a ferromagnetic core around which multiple coils, or windings, of wire are wrapped. The input line connects to the 'primary' coil, while the output lines connect to 'secondary' coils. The alternating current in the primary coil induces an alternating magnetic flux that 'flows' around the ferromagnetic core, changing direction during each electrical cycle. The alternating flux in the core in turn induces an alternating current in each of the secondary coils. The voltage at each of the secondary coils is directly related to the primary voltage by the turns ratio, or the number of turns in the primary coil divided by the number turns in the secondary coil. For instance, if the primary coil consists of 100 turns and carries 480 volts and a secondary coil consists of 25 turns, the secondary voltage is then: \n\nsecondary voltage = (480 volts) * (25/100) = 120 volts \n\nA transformer may have multiple secondary coils to feed a number of electrical loads; however, power must be conserved, so the sum of the output power must equal the sum of the input power minus losses. Energy losses in transformers are due to a number of factors: these are copper losses in the coils themselves due to material resistance, core losses due to hysteresis (the reluctance of the material's magnetic domains to reverse during each electrical cycle), and eddy currents.
386317
V = voltage (volts)\nI = current (amperes)\nR = Resistance (ohms)
52773
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.
1172658
Actually it's a question in physics. I will try to explain it anyway:\nassume we have two windings w1 and w2 with  n1 rounds and n2 rounds respectively. If an A.C. voltage of V1 is connected to w1 then a voltage V2 will be induced in w2, satisfying the following equation:\nV1/V2=n1/n2 and for currents we have: I1/I2=n2/n1\nand it doesnt convert ac to dc to do so we should use a rectifier such as bridge rectifier etc
107334
This question needs a little work. By convention, current flows from + to - although in real life, current is electrons which go from - to +. But only physics types do it that way. Electrical engineers say + -> -.\n\nNow, that being said, if you have a non-zero resistance across a voltage source, you can find the current flow by measuing the voltage across the load (resistance). The lower  the resistance, the lower your voltmeter has to read. The sign of the voltage would indicate the current flow. \n\nSo say you have a 10V battery and two 5 ohm resistors in series across it. The total current will be 1A (ohm's law). Now if I measure the voltage across one resistor, I should read 5V (kirchoff's law by the way) and the sign (+ or - 5V) will tell me the current direction. Using conventional + to - flow, if the voltage is + the current is flowing from the red lead to the black lead.
423967
first connect all grounds then take the  neutral on the transformer to your  receptacle neutral then  take one live wire from the receptacle to the switch and from the switch to the transformer being that it is two pole you will take a wire from each pole and take each "hot leg" wire to a separate "live" connection point on the transformer, provided that it is a two phase connection
71357
Volts-drop test.\n\nPut the armature on a V-block powered by a regulated voltage source.  Check the voltage across 2 adjacent com segments and repeat as you rotate thru 360 degrees.  They should all be the same.
517654
Need to know the impedance of the solar panel to determine current.
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Assuming that we have 2 series resistors R1 and R2. And the voltage applied on them is V.\n\nTo get voltage applied on R1 (V1)= V * R1/(R1+R2)\nand so V2= V * R2/(R1+R2)
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http://www.mikeholt.com/mojonewsarchive/EC-HTML/HTML/WyeTransformerCalculations~20030916.htm
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The resistance of the coil which is connected to the D.C supply will add to other resistances in the circuit.\n\nNo voltage will be in the secondary.\n\nWhen the circuit is closed and opened there will be potential at the secondary.\n\nWhen the current is steady it will not work as a transformer