what does "I" stand for in OHMS law E=IR.  I know I stand for current, but why not use "C"?
139780
Actually Ohm's law is recorded as I = V/R.  V is for volts.  I is used for current because, believe it or not all great scientists spoke english as their mother tongue.\n\nSome of the worlds most famous scientists also came from countries like France (Coulombe), Germany (Hertz), etc...\n\nI from current comes from the german word Intensität which means intensity.  It was a way of describing the change of charge (Q) with respect to time.
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.
1105216
This question has a simple answer observe\n V=IR now R=V/I\n  so i.e.120/0.6 =200 ohms \n  if you heating effects please see the joule,faradays heating effects of electricity
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
15184
Using Ohm's Law, V=IR, we have 10=I(5).  Solving for I, the current is 2 Amps.
731049
resistance is said to be one ohm if 1 ampere of current flows through the circuit across potential difference of 1volt  becuse resistance =voltage /current\na 1 ohm resistor gives obstruction to the current such that only 1A\ncurrent pass through it when potential difference is1V
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
372456
To measure ohms, DCV, ACV, Amps or what????
31337
It is DC voltage.\n\nhttp://www.ecs.umass.edu/ece/allerton/papers1999/36/
247306
It is Celsius for the Metric System.\n\nFor SI, it is Kelvin.
140428
x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
568661
Let's see if I can remember...\n\na) Find the current flowing through a DC circuit, from a 12V battery and a 3 ohm resistor.\n\nBy Ohm's Law, V = IR, where\n\nResistance = ohms = R\nCurrent = amperes = I\nVoltage = volts = V\n \nYou want to find the current, in amperes, so we switch the equation around a little: if V = IR, that also means that I = V/R (dividing both sides by the same thing gives the same answer).\n\nNow, plug in our values:\nI = (12 volts/ 3 ohms) = 4 amperes\n\nb) what is the current in an AC circuit, with 120 V and 60 W lightbulb.\nIn this case, Watts is a unit of power, and volts is a unit of resistance, and amperes is a unit of current. Electrical power is amperes times volts, or W = I * V\n\nIn this case, W = 60 and V = 120, so we rearrange it:\nW = I * V or \nI = W/V or \nI = (60 watts/120 volts) = 0.5 A\n\nAs for the atom model problem...hmmm! Si is the symbol for silicon. It has an atomic number of 14 (not 16) and an atomic mass of 28.085. According to the Bohr model of an atom, every atom has electrons (with no mass, and a negative charge), protons (with a mass, and a positive charge) and neutrons (with a mass, and no charge). The number of protons and electrons must be equal in order for the atom to have no charge, if one is greater than the other, it results in an ion (positive ion if protons are greater than electrons, negative ion if electrons are greater than protons). Electrons orbit around a nucleus of protons and neutrons.\n\nNow, an element's atomic number (14 in this case) is the same as the number of protons (and electrons, since the element is not an ion, in this case). So a silicon atom has 14 electrons orbiting around its nucleus, and its nucleus also has 14 protons and a number of neutrons.\n\nThe electrons orbit in shells around the nucleus. In the case of silicon, the inner shell has two electrons, the next shell has the maximum number possible or eight, and the third shell has the other electrons, or four. (To check, 2+8+4 = 14.) \n\nFinally, the number of neutrons in the nucleus can be calculated if we know how much a neutron weighs and how much a proton weighs. If N is the weight of a neutron, and P is the weight of a proton, and the number of neutrons is n1, and the number of protons is n2, and we say that the atomic weight is A, we have:\n\nA = (n1 * N) + (n2 * P)\n\nWe want to find n1, so we rearrange this like:\n(n1 * N) + (n2 * P) = A\n(n1 * N) = A - (n2 * P)\nn1 = (A - (n2 * P))/N\n\nNow all we have to do is plug in the relevant values, and we shall get the number of neutrons in an atom of silicon:\nA = 28.085\nn2 = 14\nP = \nN = \n\nSo, n1 = (A - (n2 * P))/N = \n\nSimple! Or at least, simpler! :)\n\nI know what you're going through...getting back into the swing of things after you've been away from school for a long time can be a pain. But keep on plugging away, it's worth it in the end! Good luck!
977033
I happen to care quite a bit about imaginary numbers as a person who studies electrical engineering.  In electrical engineering, we use the imaginary number (denoted i or j) to help represent sinusoidal waveforms, work with linear systems in the frequency domain, solve sinusoidally-excited RLC circuits etc.  It shows up quite a bit in my chosen area of study.  Even something as mundane as the voltage sine wave at a common household outlet can be represented by a phasor 120*e^(i*theta) where theta is the phase shift of the sinusoid.  Also, I hear physicists use imaginary numbers in some applications.\n\nAnyway, one of the interesting properties of the imaginary number is that when you square it you have -1 (since i = sqrt(-1)).\n\nSo:\ni^300 = (i^2)(i^2)(i^2)...(i^2)\ni^300 = (-1)(-1)(-1)...(-1)\n\nThe above set of (i^2) factors will repeat 150 times (the exponents add).  Since (-1)^150 = 1 we have our answer (check with your calculator or realize that an even amount of -1 factors will always give you 1).\n\ni^300 = 1
1079800
in one case you have half an ohm, in the other you have over 500 ohms.  It seems you may be reading one of the meters incorrectly.  One of the meters could be set to auto, while the other is not.
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.
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)
70878
Ohm's Law (named after Georg Ohm, a German physicist) states that the electric current I flowing through a given conducting material (conductor means current can flow through it) is equal to the applied voltage V divided by the resistance R:  I=V/R. Ohm's Law deals with the relationship between voltage and current in an ideal conductor.  The resistance R is sometimes called a "lumped element abstraction."\n\nFor general application to alternating-current circuits where inductances and capacitances as well as resistances may be present, the law must be amended to I=V/Z, where Z is impedance.\n\nOhm's Law is a macro-level, aggregate rule as to how current, voltage, and resistance are related.  But it is not a true "law" of nature in that it really is an approximation of behavior that holds true in most circumstances.\n\nOhm's Law may not hold due to:  1) the kind of conducting material the current is flowing through (some conductors do not follow Ohm's Law)  2) how much voltage is applied to the conducting material; as voltage is increased or decreased to extremes, then the linear Ohm's Law relationship breaks down   3) how small the wavelength of a periodically time-varying voltage applied to the conductor is with respect to the conducting circuit\n\nThe deeper reasons behind why Ohm's Law does describe macro behavior in "ohmic" conducting materials requires diving into a little bit of solid state physics and understanding Maxwell's equations.