I need help with these electrical formulas?
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
688944
View source of the that page.\nYou will see the formulas.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
827050
Yupp... but unfortunately is a hardcopy on my desk. Perhaps you may straight to tell us what\n\n1. You are designing from HT to Transformer 11kV / 415V\n2. You are designinig from Transformer to LV ??\n3. What is the kVA of the transfomer?
652086
1. volts\n2. (b) negatively charged.\n3.  short.  Could also be "conductor," but the lack of control seems to indicate an unwanted short.\n4.  watt-hours.\n5.  \n6.  (a) The formula is P=V*V/R.  Where V is the voltage and R is the resistance.  So P=45*45/15=135 watts.  Actually the object is consumming the power, not producing it.\n    (b)  135*10 = 1350 watt-hours, or 1.35 kw-hr.\n    (c) 0.08*1.35 = $1.08\n7.  (a)  V=IR, so if the voltage is increased while the resistance is kept constant, then the current will also increase. (I=V/R)\n    (b)  If the resistance is increased and the voltage is kept constant, then the current has to decrease. (I=V/R)
186229
Here are some sites that may help.  Both have circuit examples utilizing LVDT's.  Hope this helps.
534594
1.  The time constant for the circuit is Re * Ce\n\nCeffective = Ce = c1c2/(c1 + c2) since they are in series.\n\nReffective = Re = R1 + R2 since they are in series.\n\nSo Ce = 3 * 6/(3+6) = 18/9 = 2µF\n\nRe = 1 ohm + 5 ohms = 6 ohms\n\nSo the timeconstant T (tau) = RC = (2E-6F)(6 ohms) = 12E-6 sec or 12 microseconds.\n\nWith two capacitors in series, the charge on both are the same.  The voltages will be different, but the charge is the same.\n\nq =C1V1 = C2V2 \n\nAlso in your circuit using Kirchoff's law: (Vs = powersupply or battery, r = int resistance, R = 5ohms)\n\nVs = ir + V1 + iR + V2  However, we can write V2 in terms of V1:\n\nV1 = q/C1   V2 = q/C2   V2 = C1V1/C2  (substituted for the q in V2)\n\nSo the above equation becomes:\n\nVs = ir + V1 + iR + C1V1/C2\n\nVs-ir-iR = V1 + C1V1/C2\n\nVs-ir-iR = V1(1 + C1/C2)\n\nV1 = Vs-ir-iR/(1 + C1/C2)\n\ni = Vs/Re[EXP(-t/ReCe)]\n\nNow solve for V1 using i in the V1 equation.
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
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2
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.
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.
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.
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!! ;-)=
437730
See http://members.aol.com/logan20/elecspon.html for the equation to use.
32772
V=IxR where V = volts, I = Amps and R = the resistance in the circuit.  This is called Ohm's Law.\n\nManipulate the equation to get I = V/R
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.
61446
http://babbage.cs.qc.edu/courses/cs341/IEEE-754references.html\n\nhelpful info\ncheck 3 links..make sure u also learn the calculations dont rely on calculators