how do you know how many volts and amps a capacitor discharges?
924272
the amount of voltage initially on a capacitor will be whatever voltage you charge it to. It will never be any higher unless there's other circuitry involved.  The current (Amps) will flow at a rate decided by the resistance it is emptied into.  To use the common water pipe analogy, the voltage is the amount of water pressure, and the amperage is the size of the pipe.\n\nA capacitor is like a water tower.  There's lots of voltage, because the water's trying to get to the ground.  If you put a small hole in the water tower, you have a high resistance path to ground.  Your voltage will slowly fall as more of the water drains out.  the current will start fairly slow, and gradually fall along with the voltage(there's less pressure pushing it).\nIf you were to blast a big hole in it with a stick of dynamite, you would have a huge amount of flow(current/Amps), but the voltage would fall quickly, and the current will follow.\n\nit all relates to ohms law.  Voltage = Current * Resistance\n\nor you could re-arrange it to calculate your current, based on the voltage you've charged the capacitor to, and the resistance you're draining it through.  Current = Voltage/Resistance\n\nI knew a physics teacher who would demonstrate this to his classes by shorting a very large capacitor, which was charged to 12V with a screwdriver.  It would make a huge shower of sparks, and often partially weld the screwdriver to the top because of the huge currents.  He'd then charge it up again and would ask if anyone wants to put their finger on top of it.  Nobody would of course, but he would then do it himself and get no shock at all.  The resistance of skin is high, so with only 12V there's almost no current flow, even though there's a huge current capability if you give it a low resistance.\n\nIf you want the capacitors you're playing with to last a long time, though, you're best charging and discharging through a resistor.  Very quick charges and discharges create hot spots inside the capacitor, and will eventually damage them.
362318
Capacitors store the applied voltage when the source voltage is removed.  Think of it like a bucket of water filled to the top with water running in it.  If you stop the running water the bucket will still contain the full amount, (the applied voltage).  Put a small hole in the bucket a the water will drain out slowly.  The hole represents a chosen resistance, the escaping water represents the current, and the dropping level the decreasing voltage or capacitors discharge.
92081
at uniform voltage  \n\nConverting Watts to Amps\n-----------------------------------------\nThe conversion of Watts to Amps is governed by the equation Amps = Watts/Volts\n\nFor example 12 watts/12 volts = 1 amp\n-----------------------------------------\n\nConverting Amps to Watts\n\nThe conversion of Amps to Watts is governed by the equation Watts = Amps x Volts\n\nFor example 1 amp * 115 volts = 115 watts\n--------------------------------------------\n\nfind out how many Amps your monitor needs (see power supply if it has a seperate one.) multiply by 115 or 120 to get the wattage....and that would be how many watt-hrs your monitor uses.
964185
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.
653976
You need to use Ohm's law.\n  E=I x R  \n  E=Voltage. I=Current/Amps  R=Resistance/Ohms\n  E= 9voltsDC\n  I= 3amperes.\n  R= ?\n  So:\n  R = 9volt/3amp\n  R=  3ohms
407443
100,000 volts for negative bolts\n\nhttp://en.wikipedia.org/wiki/Lightning\n\nAn average bolt of positive lightning ... up to 1 gigavolt (a thousand million volts)
238179
Resistance=Voltage/Amperes
154630
Voltage=Current*Resistance (V=I*R) \n\nV=voltage in volts, I=current in amps, R=resistance in ohms\n\nVariations: I=V/R when you know Voltage(V) and Resistance (R); R=V/I when you know Voltage(V) and Current(I)
329547
If there are 3 capacitors, Ca, Cb, Cc then use the following:\n\nSeries\nTotal Capacitance = 1 / (1/Ca + 1/Cb + 1/Cc)\n\nParalel\nTotal Capacitance = Ca + Cb + Cc
1020645
It's a defined standard.\nas are the following:\n10-50ma\n1-5vdc\n\nThe reason the low number isn't zero, is so if the instrument is working, you have a signal (4ma).
833466
Volts times amps = watts.
261339
For series resistances and inductances, and parallel capacitances, simply sum them.\nFor parallel resistances and inductances, and series capacitances, first sum the reciprocals of each one, then take the reciprocal of the sum.\nHow you do this depends on how your calculator expects you to enter numbers and operators. You should consult the calculator manual if in doubt. If you don't have the manual, set up a test case whose answer you know. For instance, assume 2 resistors of 5 and 10 ohms. In parallel they equal 1/(1/5 + 1/10), or 1/0.3, or 3.333... ohms. Try various ways to enter the numbers and operators to get the right answer.
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.
272621
v=voltage\no=ohm\nm=milliammeter\n\na tester which test those mentioned aboved. don't use it in measuring resistance(ohms) when the circuit is still connected to a source (voltage)
898015
Nominal= 36ohms\nTolerance=5% \nMultiply nominal by .05 , then add that figure to nominal for Max,.subtract it from nominal for Min.,,and study your ohm's LAW!! ;-)=
227248
between 120000 and 20000 volts= 70000 volts adv.\n15500 amps
224197
Amps and volts, while correlated, are different characteristics of electricity. You can determine one from the other, but you can't convert one into the other.\n\nHaving said that, here's OHM's law: \n\nV=I x R, where V is volts, I is Amps and R is resistance.