a chart that converts amps to voltage and the other way around?
815065
I tried to import a spreadheet for you but I coouldn't make it come through, if you email me I'll send it to you, like1m@yahoo.com  \n\nYou can fill in the two variables, just write it in excel and you will be able to calculate your values for voltage and current dependent on the resistance for any given values.  All answers here are correct as it is the basic premise of \nOhms law, but you asked for a chart.  You do need to know the resistance though.  The interrelation is such that for at a given voltage with infinate resistance there will be no current (flow), for that same voltage with minimum resistance there will be maximum current (flow) this would be referred to as a dead short and is comparable basically to welding. Either extreme is not normally practical unless you are turning the power "off" in one case or welding in the other (dead short).  Dependent on the requirement usually the interrelation is desired to be somewhere in the middle.  There is also the element of reactance which is slightly more dynamic that static resistance. but for now this is what your auto calc chart should look like....  \n\nInput ohms & volts to autocalc Amps\n\nInput OhmsInput VoltsI=V/R\n      200                     24                 0.12\n\nInput ohms & Amps to autocalc Volts\n\nInput OhmsInput AmpsV=I*R\n    200                     0.12                   24
621328
ohm, V/m, Amp  <--- with the exception of flux, those units were set as those by the SI
994340
V(voltage)=I(current)*R(resistance)\n\nVoltage measured in volts\nCurrent measured in amps\nResistance measured in ohms.
252526
none. amps has no weight, its a flow of current. It's like calculating tonnage to speed. How many tons does 100 mph represent?
525922
Before you can answer this you need at lease to variables. You have the voltage of 1.5 vdc. You will need Amp the clock pulls. Wattage = Volts X Amps.
709498
The amp-hour rating is the current flowing out of the battery in Amps multiplied by the time in hours it is flowing until the battery is dead.\n\nYou could measure this by putting a resistor across the battery and an ammeter in series with it - measure the current - take readings periodically - until the battery dies. Average the readings then multiply the average by the number of hours it took to die.
434630
dont forget the formula I= R/U\nsort them and u will get the answer\n\n Current: 625mA (milli amperes)\nResistance: 19.2 ohms
1064803
http://www.epanorama.net/circuits/parallel_output.html
960408
http://www.eas.asu.edu/~holbert/ece201/kvl.html
938006
divide the voltage by R\nI=60sin(120(pi)t)\n\nThe amplitude is 60 amps\n\nThe period is 1/120 seconds\nj
823030
You could use a resistor in series for the voltmeter and a shunt resistor in parallel with the ammeter.  You can also use a signal amplifier to transform the input signal to one that is within the range of the meter.  If it is a panel type meter, do not forget to change the scale behind the meter.\n\nJeffM
142836
Sorry dude I can't able to arrive at a answer.\n\nThis conversion tool looks good check it.
844099
It depends on the applied voltage.  Divide the voltage by the resistance which gives you the amperage.  Now multiply the amperage  with the voltage, and that is your wattage.\nW = V²/R\nThis is straight forward for DC; with AC things become a little more difficult, as because of the fluctuation of the voltage, the average is only .707 of the nominal voltage.
783144
dBpowerAMP Music Converter (dMC)\nhttp://www.dbpoweramp.com/dmc.htm\n\nMichael\nhttp://www.webmasterautomation.com/forums/
386317
V = voltage (volts)\nI = current (amperes)\nR = Resistance (ohms)
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
789710
15 Amps? Looks like 10 gauge, see the link, take a grain of salt. This site says 15 amps, regardless of voltage. Tough to believe, since we care about heating, which is power, which is P=IV, so stating the current doesn't mention the power. But power is also RI^2, so I guess they know R, so they know power with current squared.