what is difference between voltage oand current?
96989
Voltage:-\n\nVoltage is a measure of the energy carried by the charge. \nStrictly: voltage is the "energy per unit charge". \nThe proper name for voltage is potential difference or p.d. for short, but this term is rarely used in electronics. \nVoltage is supplied by the battery (or power supply). \nVoltage is used up in components, but not in wires. \nWe say voltage across a component. \nVoltage is measured in volts, V. \nVoltage is measured with a voltmeter, connected in parallel. \nThe symbol V is used for voltage in equations. \n\nCurrent:-\nCurrent is the rate of flow of charge. \nCurrent is not used up, what flows into a component must flow out. \nWe say current through a component. \nCurrent is measured in amps (amperes), A. \nCurrent is measured with an ammeter, connected in series. \nTo connect in series you must break the circuit and put the ammeter acoss the gap. \nThe symbol I is used for current in equations.
211231
It depends on both the type of current and the circuit.  For example, AC and DC currents passing through a circuit with reactive elements (capacitors and inductors) will much diffreently from one another.... Ohm's law always applies however...
910320
There is something funny about your question as resistance and reactance is measured in Ohms.  However I believe that the answer that you are looking for is because the resistance part is dissipating 15 kW and the reactance part is drawing 10 kW so since the two are 90 degrees in phase apart the total power drawn would be sqrt( 15^2 + 10^2)  = 18 kW
168724
There is a relationship between power P, voltage V and current I. This relationship is best summarised by this formula: \n\nP=IV.\n\nCurrent is something else entirely. It is the rate at which electrical charge flows. The standard unit for charge is the Coulomb, and the standard unit for current is the Ampere.\n\nVoltage is the “push” that excites the electrons in a conductor and makes them flow (as current). Voltage is measured in volts and can be thought of as the potential for charge to do work. \n\npower, electric, energy dissipated in an electrical or electronic circuit or device per unit of time. The electrical energy supplied by a current to an appliance enables it to do work or provide some other form of energy such as light or heat. Electric power is usually measured in Watts
892848
One of the most frequently used is E=IR    ;E =voltage    I=current   R= resistance. U can change this every way depending on the unknown.\n    Good luck
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
254462
Georg Ohm formulated the relationship between voltage and current in a conductor (resistance). See the source for details.
291872
Volts is a unit of measure for Voltage\nAmps is short for Amperes and is a measure of Current\nWatts is a measure of Power\n\nNow Voltage, Current and Power are related quantities but not equivalent quantities. The relatioship is as described in the previous answer: Power=Voltage*Current\n\nSo Watts = Ampere * Volts\n\nAsking how many watts in a volt is like asking how many seconds in a meter
1170454
Inverter: DC to AC\n\nRectifier: AC to DC.\n\nthey do opposite functions.
820466
The voltage across each resistor will be 12 volts as each is effectively directly connected to the battery.\nIt is the current flow through the resistors which will be different.\nVolts/Ohms =Amps - that is\n12/7 amps or 12/5 amps now get your calculator out!\nRoyS
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.
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?
568661
Let's see if I can remember...\n\na) Find the current flowing through a DC circuit, from a 12V battery and a 3 ohm resistor.\n\nBy Ohm's Law, V = IR, where\n\nResistance = ohms = R\nCurrent = amperes = I\nVoltage = volts = V\n \nYou want to find the current, in amperes, so we switch the equation around a little: if V = IR, that also means that I = V/R (dividing both sides by the same thing gives the same answer).\n\nNow, plug in our values:\nI = (12 volts/ 3 ohms) = 4 amperes\n\nb) what is the current in an AC circuit, with 120 V and 60 W lightbulb.\nIn this case, Watts is a unit of power, and volts is a unit of resistance, and amperes is a unit of current. Electrical power is amperes times volts, or W = I * V\n\nIn this case, W = 60 and V = 120, so we rearrange it:\nW = I * V or \nI = W/V or \nI = (60 watts/120 volts) = 0.5 A\n\nAs for the atom model problem...hmmm! Si is the symbol for silicon. It has an atomic number of 14 (not 16) and an atomic mass of 28.085. According to the Bohr model of an atom, every atom has electrons (with no mass, and a negative charge), protons (with a mass, and a positive charge) and neutrons (with a mass, and no charge). The number of protons and electrons must be equal in order for the atom to have no charge, if one is greater than the other, it results in an ion (positive ion if protons are greater than electrons, negative ion if electrons are greater than protons). Electrons orbit around a nucleus of protons and neutrons.\n\nNow, an element's atomic number (14 in this case) is the same as the number of protons (and electrons, since the element is not an ion, in this case). So a silicon atom has 14 electrons orbiting around its nucleus, and its nucleus also has 14 protons and a number of neutrons.\n\nThe electrons orbit in shells around the nucleus. In the case of silicon, the inner shell has two electrons, the next shell has the maximum number possible or eight, and the third shell has the other electrons, or four. (To check, 2+8+4 = 14.) \n\nFinally, the number of neutrons in the nucleus can be calculated if we know how much a neutron weighs and how much a proton weighs. If N is the weight of a neutron, and P is the weight of a proton, and the number of neutrons is n1, and the number of protons is n2, and we say that the atomic weight is A, we have:\n\nA = (n1 * N) + (n2 * P)\n\nWe want to find n1, so we rearrange this like:\n(n1 * N) + (n2 * P) = A\n(n1 * N) = A - (n2 * P)\nn1 = (A - (n2 * P))/N\n\nNow all we have to do is plug in the relevant values, and we shall get the number of neutrons in an atom of silicon:\nA = 28.085\nn2 = 14\nP = \nN = \n\nSo, n1 = (A - (n2 * P))/N = \n\nSimple! Or at least, simpler! :)\n\nI know what you're going through...getting back into the swing of things after you've been away from school for a long time can be a pain. But keep on plugging away, it's worth it in the end! Good luck!
655128
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)
942921
Voltage is the amount of "potential difference" present - an analogy with water would be the pressure of the water\n\nCurrent - measured in Amps - (Coulombs per second) the flow of electrical charge - an analogy with water would be the thickness of the pipe.\n\nVolts x Current = Watts\n\nJust as\n\nWater pressure x pipe thickness = amount of water pumped\n\n\nSo a 12volt bulb taking 2 amps will be 24 Watts\nA mains bulb at 240Volts with 0.5 amp will be 120 Watts
855540
The circuit has a time constant which is calculated as\nT=L/R\n\nThe circuit will have the equation\nfor current of:\n\ni(t)=(Vs/R)*e^(-t/T)\nWhere\nVs is the supply voltage\ne is Euler's constant\n\nj
833466
Volts times amps = watts.