what is Ohm's Law i just don't get it?
349738
i=v/r\n\ni=current\nv=voltage\nr=resistance\n\nThink of it as a hose full of water. The current is the amount of water going through the hose. The voltage would be the pressure of the water, and the resistance would be the diameter and the length of the hose. So obviously if you have a huge pressure on a huge diameter hose, you are going to have a huge current. But if your resistance is very high, like say the hose is just a little straw, then even with alot of pressure the current will be pretty small. And same thing the other way around. If you have a really big pipe, then you only need a little pressure to move alot of water.\n\nElecricity works the same way, except that resistance is not so much the wire as is the appliance in the circuit. So you might say that a light bulb has a certain resistance that causes a certain current when hooked up to a 12 volt power source.
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.
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
706823
It doesn't VIOLATE ohms law in any sense at all. The formula V=IR is independently applicable to the 2 coils (Primary and secondary) where R is the resistance of the coil. The resistance of the 2 coils is different as in this case due to the soft iron core and winding, a electromagnetic resistance (inductance) is introduced in the circuit, taking that into account every thing falls in place.
822878
It's impossible to generate a perfect impulse voltage.\n\nIt's a mathematical thing to help engineers make the math work.
946535
The standard for voltage drop in a circuit is 2%.  Remember that the current must go back to the source so if you are running 3 Phase you must multiply 750ft by 1.732 to get the proper length.\n  So, using ohms law -voltage drop=current (in amps) X resistance (in ohms).  #000 gage copper wire has .00073 ohms/ ft.  Using 200 amps in this formula, If you work out the math you will find that a 30 amp service is about all you can safely install at that distance.  Even at 30 amps you are running the risk of over heating and shortening the life of your equipment. \nexample:  750ft X 1.732 X .00073 ohms per ft = .9 ohms total\n\nSo:  .9 ohms X 30 amps = 27 volts dropped\n\nThus:  if your service is 240 volts your voltage drop will be a little over 10 % or 213 volts at the point of use..  Some equipment may endure this.  I would not chance it.
724085
Its substitution.\n\nYou are given p=i*v\n\nand you need to express it in terms of one variable using ohms law v=i*r\n\nYou want p, so you need to trade out either the i or the v from p=i*v\n\nif you trade out the v using ohms law as (i*r) then your power equation becomes p=i*(i*r) = i^2 *r\n\nif you trade out the i using ohms law, first you solve for i getting i=v/r then substitute into the p-equation.  \nThe substitution yields p=(v/r)*v  = v^2 /r\n\nThis is straightforward algebra.
896490
Ohm's law:   V=RI\nBy changing I you will have V adjust to keep R constant.\n\nIf I goes up, so will V, because it's a direct relation.\n\nTry it with 100 ohm.  You can have\n100=R=V / I=500 volt / 5 amps\n100=R=V / I=700 volt / 7 amps
211231
It depends on both the type of current and the circuit.  For example, AC and DC currents passing through a circuit with reactive elements (capacitors and inductors) will much diffreently from one another.... Ohm's law always applies however...
821458
Use Ohm's Law.\n\nMake the network, put 10 volts across it, find how much current is being drawn across the 10 volts.\n\nE = iR so R = E/i, where E is 10 volts, and i is the current, amps, then R is the resistance in Ohms.
349215
mechanical power, P = Work done per unit time = W/t\n\nElectical Power P = current times potential difference = IV\n\nfor ohmic devices we can also use the substition from Ohm's law V = IR in the power formula above.
248482
Ohm's Law is always obeyed.\nIf a conductor is non-linear it has a coefficient of resistance - thermal, pressure, tensile, even voltage-dependent.\nMeasure the current thru it and the voltage across it and you have the instantaneous resistance at the time of measurement.\nThe law stands, but the result may not be static.
151570
ohm's law states that current throught a conductor is directly proportional to the voltage applied to it.\n\nso,  i oc V\n=> i = R*V where R is proportionality constant called resistance of the conductor. \n\nResistance of a material is the electrical property which opposes the flow of electrons or current.
832719
Not really here to give you the lesson you should have had!\nBut, in short:\nA resistor LIMITS the current passing through, as Ohms law, V = R x I (Volts = Resistance x Current)\nA capacitor keeps an electric charge (like a battery), but does NOT let the current passing through.  When C is empty, and a voltage is put across, a current flows through C until the cap is charged with the same voltage as the voltage across.  Once this is reached, the cap does not let any current through.
543060
I'd say that you already know.  If my memory serves (at 6 AM) it is an equation where you can determine the unknown if you have two values.  The equation can be expresssed multiple ways.  Ohms are given in impedance or interference/resistance, and I think the other values are volts(or watts) and amperage (amps).  Whether it's home, commercial or electronics, the law applies to the flow  of electron in any circuit.
216438
By Ohms law for a fixed resistance, high volt means high current and low voltage means low current.\n\nWhen ever we touch a point of any voltage, there will be certainly a current through our body to the ground. \n\nThe ground is considered to be at zero potential.\n\nSuppose we wear a shoe, then the shoe acts as an insulator and hence no current flows through our body. \n\nThough wearing a shoe, if we touch some wall then our body is short circuited to the ground and a current will flow.\n\nTherefore unless we are well insulated from ground the voltage is the cause to send the current through our body and it may be fatal.\n\nIn power lines if we hang ourselves with out touching any other wire then no current will flow. But accidentally if we touch the other wire then we will be killed.\n\nBirds sit on one wire. They are safe as long as they do not touch the other wire\nBirds die if they touch the other wire.\n\n Though it is technically correct to say volts burn current kill, one must remember the cause is the voltage and not the current.\n\nIt is like saying man shoots bullets kill.\n\nThere is also another reason that some times we are not killed even if we feel the shock of the current.\n\nThe current grows from zero to a value in a fraction of a second. But reflex action of our muscles pulls us away sooner from the point of contact.  The time of rise of current depends on the voltage. It is the reason for the burn.\n\nI write all these because any body reading the answers may thing that voltage is harmless. \n\nIt is always dangerous to deal with current with out fully knowing its properties.
53767
Inductance is the "definite answer" that I was given 30 years ago. I don't suppose that my confirmation is any better than yours, but it's a vote anyway.