what is the formula for ohm's law?
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)
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.
621328
ohm, V/m, Amp  <--- with the exception of flux, those units were set as those by the SI
980282
Make sure you have a good, solid connection between your sample and your ohmmeter probes.  Use a scale on your meter that's appropriate for the resistance you are measuring.  If your resistance is 1/2 ohm, then it will read zero on a 0-20Kohm scale, which isn't useful.  It will read half-way on a 0-1ohm scale, where you can get a good reading.\n\nThe resistance of a sample depends not just on the material and temperature, but also the geometry.  The resistance of a uniform wire will be proportional to its length and inversely proportional to its cross-section area.
480559
An ohm-meter measures the resistance by applying a voltage across the component. The current flowing through the component is measured, and a little bit of Computation of the form R=V/I would give the resistance.  \n\nIf a voltage is already supplied, then the ohm-meter may read erroneously or may even get screw**.
222936
Consider a conductor of uniform area of crossection A, length L. Let resistivity of conductor be S and resistance of material to current be R.\n\n  Then,  R = SL/A\n\n     Multiply and divide by mass of conductor say m and L\n\n          R = SL*L*m/L*A*m\n\n But L * A  = Volume = V\nAnd m / V   = Density = d\n\n  Hence,  R =  S*L*L*d/m\n          S =  (m*R/L*L)/d\n\nThus, Resistvity (S) is inversely proportional to density(d) when other quantities are kept constant.
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.
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)
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)
1170181
If the resistors are in series, then to find the equivalent resistance you just add up all the resistances.  6 + 12 + 18 = 36 Ohms\n\nTo find the current, you take V/R\n\n36 Volts / 36 Ohms  =  1 Ampere\n\nGood job on getting the right answer!
279992
Copper's resistance increases by a factor of 0.0068 per degree C (ref.). This resistance increase is linear, not exponential. The formula is Resistance = R(initial) * [1 + alpha * (T(final) - T(initial)], where alpha is the temperature coefficient. Also note that self-heating due to any current flow will add to the ambient temperature effect, so power dissipation in a cable will be higher than what is calculated from resistance based on ambient temperature alone.
971534
Well, u can assume that on every section of wire there is an amount of current flowing say i1,i2.... etc. and give any direction to them.\nKirchoff has two laws:\n1st law(current law) : At any junction the incoming current is equal to the outgoing current. So at any junction u can add all the incoming currents and equate it to the addition of all outgoing currents.\n2nd law (voltage law) : it states that the total voltage in a closed loop or mesh is zero. so if u consider first loop on left and the direction of current in both sections of wire clockwise. Then u can write -i1R1-i1R2+Ub1=0. if u r tracing the loop in the direction of current then voltage on resistance is negative(-iR) or u can take the vice versa it doesn't matter.For voltage source if the current is moving from positive to negative plate then voltage is termed as positive and the other way negative.\n\nSo form that many number of equation, that suffice u in finding the unknowns. And then by applying simple mathematics find the current in each segment and then the voltage across each resistor. U need to consider here the concept of voltage drop too.
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.
1046550
Remember the equation v/i=r, Keep the voltage constant over all the areas. If the resistance goes up, then the current goes down.\nIf you have zero ohms, you have a short. IF you have infinite resistance, you have a break.  This seems general, but this is a great help for most of the electronics applications you will need. \nIf you want more you will need to give us more info.
261339
For series resistances and inductances, and parallel capacitances, simply sum them.\nFor parallel resistances and inductances, and series capacitances, first sum the reciprocals of each one, then take the reciprocal of the sum.\nHow you do this depends on how your calculator expects you to enter numbers and operators. You should consult the calculator manual if in doubt. If you don't have the manual, set up a test case whose answer you know. For instance, assume 2 resistors of 5 and 10 ohms. In parallel they equal 1/(1/5 + 1/10), or 1/0.3, or 3.333... ohms. Try various ways to enter the numbers and operators to get the right answer.
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.
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.