I need help with these electrical formulas?
817489
This is pretty basic algebra and if you are in an electrical certification program or any other program for that mater you may want to consider taking some math courses before you try to continue.  I'm not saying that you can't nuc your way through it, but it will be allot easier on you if you take a step back and strengthen your math skills before tackling an electrician course.
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.
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.
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
501994
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
100752
There are two relavent formulas,\nOhms Law: V = R*I\nwhere V is the voltage, R is the resistance, and I is the current.  Also,\nPower = V*I\n\nIf we know that power is 150 watts, then,\n150 = V * I,\nbut what is the current (I)?\n\nWe know the resistance is 95 ohms,\nV = 95*I --> I = V / 95\nSo plugging back into the power equation,\nP = V^2 / R \n150 = V^2 / 95 --> V = sqrt(150 * 95)\nso V = 119.4 volts --> 120 volts
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.
710087
Vt is the source with R1 and R2 in series\n\nIt=Vt/(R1+R2)\n(OHMS LAW)\n\nIt=12/(120)\n---------------\nIt=0.1 A\n-------------\n\n\nVr2=Ir2 *R2\n     = It *R2\n     =0.1 * 100=10v\n\n------------------\nVr2=10 v\n------------------\n(So,Vr1=Vt(total)-Vr2=12-10=2 V)
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.
80597
V = IR\nV---Voltage\nI--current\nR--- resistance\nIts also called ohm's law
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
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?
1130014
V=IR\n\nWire 1:\n50 = 2 * R1 => R1 is 25 ohms\nCurrent = 2 amps\npower = V*I = 2*50 = 100 watts\n\nWire 2: \n100= 2* R2 => R2 = 50 ohms\nCurrent=2amps\npower = V*I = 100*2 = 200 watts\n\nare you looking for ratios when you say "compare" ?\nif yes: ratio of resistances is  1:2\nratio of currents is 1:1\nand ratio of power is 1:2
331292
See the other postings of this question.  The equation is\n\nhttp://members.aol.com/logan20/elecspon.html
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
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
386317
V = voltage (volts)\nI = current (amperes)\nR = Resistance (ohms)