what  ampere's circuital law states???
445109
Ampère's circuital law . This equation describes how an electric current generates a magnetic field. Ampère's circuital law states that \n\nThe line integral of the magnetic field around any closed loop is equal to  times the algebraic sum of the currents which pass through the loop.
97466
there is a metal strip inside that the current flows through. If too much current flows, the strip is actually made of two metals one that expands dirfferently than the other, then it will bend enough to cause the circuit to trip mechanically.\n\nThe GFI part measures the current out and the current back through the neutral with electronics. If they do not match by more than something like 4 miliamps then the current is flowing through ground. This is bad if you happen to be between ground and the circuit so to protect you it trips the breaker.
913471
In a purely resistive circuit, the power factor (PF) is 1 and you *can't* adjust it.\n\nAnd, since this is the 3'rd time you've asked the question and gotten the same answer from one other person and myself, do you even know what 'Power Factor' *is*?\n\n\nDoug
731049
resistance is said to be one ohm if 1 ampere of current flows through the circuit across potential difference of 1volt  becuse resistance =voltage /current\na 1 ohm resistor gives obstruction to the current such that only 1A\ncurrent pass through it when potential difference is1V
140428
x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
805935
Please don't take this as an attack or insult, it is meant to be helpful and comes from a professional Electrical Engineer:\n\nThe questions you are asking are all very basic algebra problems.  A solid grasp of college-level first and second semester algebra is an absolute requirement when it comes to understanding electrical systems.  These math fundamentals are used everywhere in describing, calculating the performance of, installing, and troubleshooting electrical circuits.  Without a solid foundation in mathematics, nothing within the electrical field is going to make any sense at all.\n\nThat said, here is how to do these problems:\n\n(1) Es=E1+E2+E3 solve for E2\n\nThis is calculation of the voltage across one element in a series circuit composed of 3 elements.  The total voltage across the series circuit (Es) is equal to the sum of the voltages across each of the elements of the circuit (E1, E2, E3).  So the voltage across any one element is equal to the total voltage minus the sum of the OTHER elements:\nE2 = Es - (E1+E3)\n\n\n(2) R1=I2R2 divided by I2 solve for R\n[double check that you have copied this right -- I think you mean "solve for R2" since it is already solved for R1]:\n(I2*R2)/I2, the I2's cancel out so you are left with R2.  R1 = R2.\n\n[you may have meant "R1=I2R2 divided by I1 solve for R2" which would be a formula for calculating an unknown resistance in a parallel circuit based on a the other known resistance and the respective currents]:\nR1 = (I2 * R2) / I1\nMultiply both sides by I1:\nR1 * I1 = I2 * R2\nDivide both sides by I2\n(R1 * I1) / I2 = (I2 * R2) / I2\n(R1 * I1) / I2 = R2\nSo, R2 = R1I1 divided by I2\n\n\n(3) R1 divided by Rx = R2 divided by R3 solve for Rx\nInvert both sides of the equation:\n1/(R1/Rx) = 1/(R2/R3)\nRx/R1 = R3/R2\nMultiply both sides by R1:\nR1*(Rx/R1) = R1*(R3/R2)\nRx = R1R3 divided by R2\n\n(4) y=mx+b solve for b\nSubtract mx from both sides of the equation:\ny - mx = mx + b -mx\ny - mx = b\nso, b = y minus mx\n\n(5) I2=I1R1divided by R2 solve for I1\nI2 = (I1*R1)/R2\nMultiply both sides by R2:\nR2 * I2 = (R2*(I1*R1))/R2\nR2I2 = I1R1 because the R2's cancel out\nNow divide both sides by R1:\n(R2I2)/R1 = (I1R1)/R1\n(R2I2)/R1 = I1\nSo, I1 = R2I2 divided by R1
674934
there r 2 parts to it:-\nA) the algebraic sum of the currents through a junction is zero..i.e (sumation)I=0...(where I=current)\nB)the algebraic sum of the emf (electro motive force) in a closed loop is equal to the algebraic sum of the products of current & resistance in each path of the closed loop...i.e (sumation)emf=(sumation)IR...(where I=current, R= resistance)
142477
Ampacity: Refers to the how much current a wire can safely carry. For example, a 12 gauge electrical copper wire can safely carry up to 20 amps.\n\nminimum ampacity is not really something that makes sense unless you are talking about the lowest safe maximum and calling that minimum. This might arise from various ways to calculate the safe maximum and then having to use the lowest one. \n\nThe definition above on the link in the sources is not always true for example. If a 12 gauge wire is used in a conduit with a bunch of others and the fill ratio (how much air versus how much wire) of that conduit is greater than the maximum allowed then the current limits on the wires are reduced based upon the amount the conduit is overfilled. Wires in high temperature environments can be a reason to derate them as well.
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!
402553
Your circuit diagram looks strange - some resistance are denoted in Volts. Hmm, anyway, otherwise it's rather simple. You got to use your formulaes for calculating resistance in a series and ohm's law V=IR. I hope I'm right - I last did physics in 1999.\n------------------------------------------------\n\nThe resistors are all in series, therefore:\nTotal Resistance = 33 + 23 + 2/I + 1/I + 43\n                 = 99 + 3/I Ohms\n\nUsing Ohms law, V = IR. Therefore:\n=> 100 = I . Total Resistance\n=> 100 = I (99 + 3/I)\n=> 100 = 99 I + 3\n=> I = 97/99 Amperes  --- ans\n\nSub. value for I into Total Resistance:\n=> Total Resistance = 99 + 3/(97/99)\n                    = 9900/97 Ohms (or 102.06 Ohms) --- ans \n\nUsing V=IR\n\nV(3) = potential difference from 100 V to when current I passes through the 33 Ohm resistor\n     = 100 - [(97/99 x 33)] \n     = 202/3 V or 67.33 V (The reading of V (3)'s galvanometer) --- ans.\n\nV(2) = potential difference from V(3) to when current I passes through the 23 Ohm resistor\n     = 202/3 - [(97/99 x 23)] \n     = 4435/99 V or 44.80 V (The reading of V (2)'s galvanometer) --- ans.\n\nV(1) = potential difference from V(2) to when current I passes through the 2V resistor\n     = 4435/99 - 2\n     = 4237/99 V or 42.80 V (The reading of V (1) galvanometer) --- ans.
821458
Use Ohm's Law.\n\nMake the network, put 10 volts across it, find how much current is being drawn across the 10 volts.\n\nE = iR so R = E/i, where E is 10 volts, and i is the current, amps, then R is the resistance in Ohms.
871831
What's to know?\n\nIt's the typical offset voltage, or DC component of the output.  The signal is AC.
248482
Ohm's Law is always obeyed.\nIf a conductor is non-linear it has a coefficient of resistance - thermal, pressure, tensile, even voltage-dependent.\nMeasure the current thru it and the voltage across it and you have the instantaneous resistance at the time of measurement.\nThe law stands, but the result may not be static.
971534
Well, u can assume that on every section of wire there is an amount of current flowing say i1,i2.... etc. and give any direction to them.\nKirchoff has two laws:\n1st law(current law) : At any junction the incoming current is equal to the outgoing current. So at any junction u can add all the incoming currents and equate it to the addition of all outgoing currents.\n2nd law (voltage law) : it states that the total voltage in a closed loop or mesh is zero. so if u consider first loop on left and the direction of current in both sections of wire clockwise. Then u can write -i1R1-i1R2+Ub1=0. if u r tracing the loop in the direction of current then voltage on resistance is negative(-iR) or u can take the vice versa it doesn't matter.For voltage source if the current is moving from positive to negative plate then voltage is termed as positive and the other way negative.\n\nSo form that many number of equation, that suffice u in finding the unknowns. And then by applying simple mathematics find the current in each segment and then the voltage across each resistor. U need to consider here the concept of voltage drop too.
917343
Current is the rate at which charge moves past a point on a circuit. \n\nTherefore the correct formula is:\n\n  (Current) = (Charge) / (time)\n\nNote that the definition of Ampere is "One Coulomb per second".
783633
YOU ARE ASKING A QUESTION THAT IS ANSWERED BY SOLVING THE EQUATION (E=IR) OR OHMS LAW. IF YOU GIVE ME TWO OF THE  UNKNOWNS THE EQUATION CAN BE SOLVED FOR A NUMERIC ANSWER.   LETS ASSUME THE VOLTAGE(E) IS 100VOLTS.  THEN THE CURRENT(IR1) WOULD BE (100VOLTS/110 OHMS) E1/R1 OR 0.909 AMPS.\nTHE SITE BELOW GIVE A LENGTHY DESCRIPTION OF THE SIMPLE CONCEPT AND JUST SEARCHING "OHMS LAW" WILL YIELD DOZENS MORE. YOU WILL FIND THAT THE VOLTAGE APPLIED TO EACH RESISTOR IS THE SAME AND THE AMPERES THROUGH EACH RESISTOR IS INVERSELY PROPORTIONAL TO THE RESISTANCE.
576166
the electrical circuit?