what is the correct formula for electrical current??
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".
1083571
1) Magnetic field due to a current passing in a circular loop:\n        B=(PIN)/2r where\n  P is the permeability of the medium\n  I is the electric current\n  N is the number of turns\n  r is the radius of the loop\n\n2) Magnetic field due to a current in a soleniod:\n         B=(PIN)/L\n         B=PIn     where\n   P is the permeability of the medium\n   I is the electric current\n   N is the number of turns\n   L is the length of the coil\n   n is the number of turns per unit length\n\nAbout V & B\ni don't know any formulas that contain V & B \n\ni hope i could helped u
905553
Current = Charge per unit time = 14/7 Ampere = 2 A
386317
V = voltage (volts)\nI = current (amperes)\nR = Resistance (ohms)
713269
Faraday's Law!\n\nMass Plated = t * I * MW / (n * F)\n\nt = time\nI = current (Amps)\nMW = mol weight of metal\nn = valence charge of metal\nF = Faraday's Constant (96,500 Coulombs/mol e-)\n\nSo....\n\nt = 2 days * 24 hours/day * 3600s/hour = 172800s\nI = 7.60 A\nMW = 51.94 g/mol\nn = 3 mol e- / mol M\n\nPlug it all in and you should get 235.62g of Cr.
387680
Ok...you guys got this all wrong.  Voltage doesn't travel.\nThis is simple:\n\nSeries:\nAll series stuff see the same current\n\nParallel:\nAll parallel stuff see the same voltage
890360
I have already answered part 1. in another question you've asked. I think you probably didn't use the right units (like cm needs to be convered to meters) and got the wrong answer.\n\nNow for part 2:\n\nInduced EMF, ℰ = -L(di/dt)\n\nwhere i is the current and L is the inductance\n\nWe are given that the magnitude of the induced current,\n\ndi = 47 mA\n\nand the oscillation frequency is 100 Hz. This means that dt = 1/f = 0.01 sec\n\nWe have L = 8.44 × 10-^3 henries from before\n\nSo |ℰ| = L(di/dt) = -(8.44 × 10-^3)(0.047)/(0.01) = 0.0397 volts\n\n------------------------------------------------------\n\nNow part 3 has to do with mutual inductance:\n\nVs = VpNs / Np\n\nwhere \n\nVs is the voltage across the secondary inductor, \nVp is the voltage across the primary inductor (the one connected to a power source), \nNs is the number of turns in the secondary inductor, and \nNp is the number of turns in the primary inductor. \n\nTherefore Vp = VsNp/Ns and Vs = 0.0397 volts as calculated before, so\n\nVp = (0.0397)(197)/(2920) = 0.00268 volts
914458
i=C(dV/dt)\n\nwhere\n i= the instanious current through the cap\nC= the cap siz in farads\ndV= the change in voltage\ndt= the change in time\n\nI use this equation in my op-amp class where were are doing PWM power supplies and motor drivers right now.
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
291589
Use the node voltage method. Label each node with a voltage, then apply Kirchoff's current law at as many nodes as you need to generate enough equations to solve for any unknowns you have introduced.\n\nLet the node at the bottom of the circuit be ground, and then try to figure out the voltages at the other nodes relative to ground. Clearly the voltage at the node on the top will be 14V (I'll just call this V1) but no other voltages are apparent. So call the voltage at the node between R1, R3, and R4 Vx and the voltage at the node between R2, R3, and R5 Vy.\n\nNow that we have labeled all the node voltages, you can figure out every single current in the circuit in terms of V1, Vx, and Vy (using Ohms Law, the voltage drop across a resistor is equal to the resistance times the current in the direction of the voltage drop). For example, we know the current going down through R1 is (V1-Vx)/R1, the current going right through R3 is (Vx-Vy)/R3, etc. Now we just have to solve for the unknowns Vx and Vy.\n\nKirchoff's current law (KCL) says that the sum of the currents leaving a node is equal to zero. If we apply this law to the node with voltage Vx and the node with voltage Vy, we'll get a pair of equations that we can use to solve for Vx and Vy.\n\nApplying KCL at the node with voltage Vx should give you the following equation:\n\n(Vx-V1)/R1 + (Vx-Vy)/R3 + Vx/R4 = 0\n\nThe first term is the current going up through R1 (we already figured out the current going down through R1, but we want currents that are leaving node Vx so we had to flip the sign), the second term is the current going right through R3, and the final term is the current going down through R4.\n\nI'll leave it to you to figure out the equation for node Vy. Once you have that you'll have two equations and two unknowns and you can solve for everything. Once you have Vx and Vy, you can easily figure out a numerical value for any current in the circuit.\n\nBTW, the node-voltage method is pretty powerful and can solve most linear circuits. Check out the link below, it was written by an old EE prof of mine who is very thorough.
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
200995
by electrolisis\nH2O------>H2+O2\ni.e by passing electric current
238179
Resistance=Voltage/Amperes
969011
You might care to explain what some of the letters are, not all of them have much to do with electronics.\n\nR I guess here is resistance, but what of l m x and s?  Is l for inductance (I think not here), what of s? seconds?\n\nSome of these (all ?) are simpley equations (eg: c=q/E solve for E) would apply whatever C, ! and E stood for.\n\nIn this case if you gave C an arbitar value or 2, Q an arbitrary value of 6 ad E an arbitary value of 3, you would see that C (2) = q (6) / E (2).   So E would be Q/C\n\nThis is usually the I=v/r fourmal where I=current, v=voltage and r=resistance. R=v/I is also the same but the other way around. V=IR is valid also.\n\ny=mx+b?  what electrical formla is that? Again it is easy to solve but I can see nothing to with electronics here. Again try assigning sinple numbers to the letters so that the equation works - then you can re-arrange the sum so it still works:\n\nGive m, x and b a different value: 3, 4+ 5\n\nNow 3*4 +5  = 17\nso Y=17.\n\nNow you can play about with re-arranging your sum:   \n\n5 = 17   -    3 * 4 \n\n5 represents b, 17 represents Y, 3 represents m and 4 represents x\n\nSo swapping numbers for letters:    b=y-mx
543060
I'd say that you already know.  If my memory serves (at 6 AM) it is an equation where you can determine the unknown if you have two values.  The equation can be expresssed multiple ways.  Ohms are given in impedance or interference/resistance, and I think the other values are volts(or watts) and amperage (amps).  Whether it's home, commercial or electronics, the law applies to the flow  of electron in any circuit.
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!! ;-)=
768635
Well we are given voltage, and power.  We are asked to find the current. Well the formula we have to use here is P=VI.\nRearranging this formula we get; I=P/V\n\nHowever; there is only one thing wrong with this problem.\nWe do not know what the voltage is correctly.\nIf this diode is a ideal diode. An ideal diode acts like a short circuit with no voltage drop across it. Then we would use .8V as the Voltage Source across the diode.\nIdeal Diode:\nI = .05 W/.8V\nI = 62.5mA\n\n\nIf it is not an ideal diode and there is a .75V drop across the diode then we would use .05V as the Voltage Source across the diode.\nI = .05W / (.8V - .75Vdiode drop)\nI = .05W / (.05V)\nI = 1A\n\nHere this might give your more information!\nhttp://en.wikipedia.org/wiki/Diode_modelling