I need help with these electrical formulas?
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
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
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
65822
You'll have to describe the circuit a little better \n\nAre you adding either the 3.00 ohm or 2.50 ohm (but not both) in parallel with the 4.00 and 3.75 ohm, and need to know the total resistance with either the 3.00 or 2.50 ohm resistor connected?\n\nOr are BOTH the 3.00 AND the 2.5 ohm being placed in parallel with the 4.00 and 3.75 ohm?\n\nAnd without knowing the applied voltage, your first question (current through a 3.00, 4.00, 3.75, 2.50 ohm resistor) can't be answered.
1046550
Remember the equation v/i=r, Keep the voltage constant over all the areas. If the resistance goes up, then the current goes down.\nIf you have zero ohms, you have a short. IF you have infinite resistance, you have a break.  This seems general, but this is a great help for most of the electronics applications you will need. \nIf you want more you will need to give us more info.
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
246159
http://www.ee.duke.edu/~cec/final/node21.html#SECTION004101000000000000000\n\nhttp://www.electronics-lab.com/projects/test/014/\n\nI'm not sure you'll find all you need, but they would help.
837447
PROBLEM ONE\nThe electric resistance R of a wire varies directly as its length L and inversely as the square of its diameter.  So you mean R=cte*L/d2 where the constant depends on the metal used.\nfor that alloy, C=Rd2/L=36*0.6*0.6/20=0.648\nIn the second wire, you use the same constant\nR=0.648*60/(1.2*1.2)=27 ohms\n\nPROBLEM TWO\nhere, period=T=cte*sqrt(L)\ncte=T/sqrt(L)=2/sqrt 6=0.816\nL=(T/cte)2=(1/0.816)2=1.22*1.22=1.5
560737
En=-(13.6)/n^2eV\nE1=-(13.6)eV\nE3=-(13.6)/9eV\ndelta E=E3-E1=hv\nhv=12eV\ntherefore,v=(12*1.6*10^(-19))/(6.626*10^-34)\nv=2.8976*10^15hz
688944
View source of the that page.\nYou will see the formulas.
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.
318293
u can go to library.alldata.com and be able to find the wiring schematic. or sometimes u can call a local parts store and they can print one out for you.
924475
= 9u-10u + 6v = 6v - u
140136
Here's a study guide.  http://www.doctronics.co.uk/meter.htm
368244
V (Voltage) = I (Current) x R (Resistance)\n\nOr (although you probably don't need it):\n\nJ (current density) = sigma (conductivity) x E (electric field)\n\nI always use wikipedia to look up stuff like this, it's very helpful for all kinds of info.
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2
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.