graph of hall voltage vs current?
140428
x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
472581
The equation for this is relatively simple.\n\nVp/Vs = Np/Ns\n\nWhere: \nVp=Voltage primary winding.\nVs=Voltage secondary winding.\nNp=Number of coil turns on the primary winding.\nNs=Number of coil turns on the secondary winding.\n\nThus taking that into consideration you should be able to get the voltage for both sides.  To get the current on both sides.  Just realize that you need the resistance.\nOr if you have the current on one side, use the conservation of Power equation:\n\nVpIp=VsIs\n\nWhere: \nVp=Voltage primary winding.\nVs=Voltage secondary winding.\nIp=Current in the primary winding.\nIs=Current in the secondary winding.
112062
Heat conductivity? electrical?
168710
http://www.graphcalc.com/
655128
Assuming that we have 2 series resistors R1 and R2. And the voltage applied on them is V.\n\nTo get voltage applied on R1 (V1)= V * R1/(R1+R2)\nand so V2= V * R2/(R1+R2)
960408
http://www.eas.asu.edu/~holbert/ece201/kvl.html
833466
Volts times amps = watts.
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.
191157
V = IR\nI = V/R\nR = V/I\n\nV = voltage    / V (or E)\nI = current    / A\nR = resistance / ohm\n\nOhm's Law = The voltage across a circuit is directly proportional to the current and the resistance.
621328
ohm, V/m, Amp  <--- with the exception of flux, those units were set as those by the SI
224197
Amps and volts, while correlated, are different characteristics of electricity. You can determine one from the other, but you can't convert one into the other.\n\nHaving said that, here's OHM's law: \n\nV=I x R, where V is volts, I is Amps and R is resistance.
154630
Voltage=Current*Resistance (V=I*R) \n\nV=voltage in volts, I=current in amps, R=resistance in ohms\n\nVariations: I=V/R when you know Voltage(V) and Resistance (R); R=V/I when you know Voltage(V) and Current(I)
304043
Because algebra skills are very lacking in most people. It's just to make sure that they know what the rearranged formula's look like. But you are right, those 2 laws are probably all you need to remember. P=IE is always true, but remember than Ohm's Law is only an approximation for materials that have a linear E vs R graph, such as metals at room temperature. Its not true for semiconductors and gas discharges.
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
455603
If it's for a simple hobbyist application they need to know only the resistance value and its power rating.\n\nA full specification would also include:\n\ntolerance,  thermal coefficient,  maximum voltage.
8775
E = hf applies to light, not electricity.\n\nThe power consumed by AC current is assessed by the root mean square, which is equivilent to the peak amplitude divided by radical 2.  The frequency doesn't factor into the matter.
1088099
insulators ; conduction