Please help, Voltage vs. Current question.?
967090
Voltage is the power potential.  Current is the speed and power at which that voltage goes through a given medium.  \n\nThink about this, Hoover dam.  It has inordinate amounts of potential.  If it sprung a small leak at the top and shot out watter, the force of that wouldn't be much at all to hurt you.  But, if you blew an 8 foot wide hole at the bottem of the dam, the force of the water would disentegrate you in an instant.\n\nElectricity was a very difficult concept for me to get.  However, when you start comparing it to something that is easy to see, touch, and feel, then it helps a lot.  Water is a perfect analogy in many ways.
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
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.
734859
Divide by the voltage and multiply by 1000.\n\nIf you are working with 3 phase power, divide again by sqrt(3)
698161
No, the voltages are different
210547
power=voltage^2/resistance\nhigher power means lower resistance.
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
214344
Voltage across parallel circuts is always the same, so 7V. The current, however will be split (if the lamps are exactly identical, split in half between them).\n\nSeries circuits are the other way. Voltage is different, current is the same.\n\nThis is a consequence of Kirchoff's laws. Kirchoff's law says that all the current going into a node must equal the current going out of the node (conservation of energy, really). And also, the voltage drops around a loop must equal. Common sense really. There can't be any extra (or deficit) current or voltage.
752024
When you copy and paste these, why don't you convert the "W"s into "ohms"?\n\nHey!  The guy who did your homework for you ("The Great One") gave three WRONG answers (as of 1930 MST 11-14.  He'll probably redo it after he reads this)...heh, heh, heh...
450422
Bah! I thought it was AMD based on the voltage alone...\nchecking Intel now...\n\nBased on every 3.4GHz CPU I could find on Intel's website, your voltage is within range. Intel specifies an operating range of 1.25-1.40v for most, and up to 1.525v for some.
238179
Resistance=Voltage/Amperes
702226
Its fault with the socket or the wiring. Check them out. You have not mentioned the voltage. Wattage matter in performance. Its only the Voltage & Ampere that matters for working.
756481
Normally this is a "Restive" not voltage question/answer, and the proper neutral to ground = 2 Megg ohms (min.) this being incorporated in a residential application.\nHigher voltage in let us say a 3-phase utilization this # is on the low side. (assuming a.c. source. d.c. is a completely different animal!
158260
Need more information. Why are you people bellow suggesting you can just increase current?, you cannot if your load (resistance/impeadance) and voltage remain the same. Current is the result, Voltage and resistance are the only two forces that can be manipulated directly. Just because you read something, doesn't mean it can be applied in the real world.
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)
501507
Here is the equation :  \n  \nI0 = Is*(exp(V0/nUT) - 1)\n\nwhere :  - I0 is the current (function of V0, the applied\n                              voltage)\n         - The others parameters should be given by the   \n           constructor:\n            -    1 < n < 2\n            -    UT = 26mV at 27°C\n            -    Is is the reverse saturation current\n\nPS : I'm not sure I'm using the right english vocabulary as it is not my mothertongue. Sorry for that...
833466
Volts times amps = watts.