How to convert Resistance to Conductivity.?
815817
You must have the CELL CONSTANT to do this properly.\nResistance and Conductance are reciprocals.   But to go from\nConductance to ConductIVITY, you need the Cell Constant.\nIt's different for each probe, and should be easy to find on the probe you're using.   Find that and get back to me.
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!
388634
Inductive reactance can be calculated using this formula: XL = 2πfL\n\nGood Luck
1102715
There is always some resistance, unless you are dealing with a superconductor.  You have to have resistance to develop voltage.
925581
Resistivity is independent of length or diameter.  It's 49 × 10^-8 ohm*m for constantan.
763289
You need to post some details - then I can post an answer...\n\nIn general, P = V*I = V^2/Z\nso calculate the complex impedance Z (= R+jX)\nthen take the real component of the complex power.
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
837447
PROBLEM ONE\nThe electric resistance R of a wire varies directly as its length L and inversely as the square of its diameter.  So you mean R=cte*L/d2 where the constant depends on the metal used.\nfor that alloy, C=Rd2/L=36*0.6*0.6/20=0.648\nIn the second wire, you use the same constant\nR=0.648*60/(1.2*1.2)=27 ohms\n\nPROBLEM TWO\nhere, period=T=cte*sqrt(L)\ncte=T/sqrt(L)=2/sqrt 6=0.816\nL=(T/cte)2=(1/0.816)2=1.22*1.22=1.5
953475
Specific resistance, or resistivity, is the resistance in ohms offered by a unit volume (the circular-mil- foot or the centimeter cube) of a substance to the flow of electric current. Resistivity is the reciprocal of conductivity. A substance that has a high resistivity will have a low conductivity, and vice versa. Thus, the specific resistance of a substance is the resistance of a unit volume of that substance. Many tables of specific resistance are based on the resistance in ohms of a volume of a substance 1 foot in length and 1 circular mil in cross-sectional area. The temperature at which the resistance measurement is made is also specified. If you know the kind of metal a conductor is made of, you can obtain the specific resistance of the metal from a table. \n\nThe electrical resistivity ρ (rho) of a material is usually defined by the following:\n\nρ=R*A/l\n \nwhere\n\nρ is the static resistivity (measured in ohm metres) \nR is the electrical resistance of a uniform specimen of the material (measured in ohms) \nl is the length of the specimen (measured in metres) \nA is the cross-sectional area of the specimen (measured in square metres) \nElectrical resistivity can also be defined as:\nρ=E/J\n \nwhere\n\nE is the magnitude of the electric field (measured in volts per metre) \nJ is the magnitude of the current density (measured in amperes per square metre)
840883
For two resistors in parallel, take the product over the sum.\n\n(100*100) / (100+100) = 10000/200 = 50.\n\nWhenever you put two resistors of the same value in parallel, the effective resistance is 50% of the actual value of one resistor.
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.
315095
When resistors are in series both have same current, i.e V/(r1+r2) = 12/(2+4) = 2 A\n\np.d across each resistor will be I*R\nSo for 2 ohm resistor p.d is 2 * 2 = 4 V\nFor 4 ohm resistor p.d is 4*2 = 8 V\n\nWhen the reistors are in series the p.d across both is the same, i.e 12 V\n\nCurrent across each resistor is given by V/R\nSo across 2 ohm resistor current = 12/2 = 6A\nAnd across 4 ohm resistor current = 12/4 = 3A\n\nResultant resistance = r1*r2/(r1+r2) = 8/6 = 4/3 ohm
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.
65822
You'll have to describe the circuit a little better \n\nAre you adding either the 3.00 ohm or 2.50 ohm (but not both) in parallel with the 4.00 and 3.75 ohm, and need to know the total resistance with either the 3.00 or 2.50 ohm resistor connected?\n\nOr are BOTH the 3.00 AND the 2.5 ohm being placed in parallel with the 4.00 and 3.75 ohm?\n\nAnd without knowing the applied voltage, your first question (current through a 3.00, 4.00, 3.75, 2.50 ohm resistor) can't be answered.
597402
let the resistances be r1 and r2.\nwe know the equation for power'p' is p=V*I\nV-Voltage applied,\nI-current in the circuit,\nby ohms law,\nV=IR\nsubstitute the value of V in power,\nwe get P=I^2*R  \nif the resistances  are in series then the value of R is,\nR=r1+r2.\nso the power dissipated when the resistance in series is,\np=I^2*(r1+r2),\nsimillarly power dissipated when the resistances are in parallel is\np=I^2*((r1*r2)/(r1+r2))
1100598
voltage = current x resistance\nE = I x R  It IS Ohm's law. An easy way to remember the formula and it's configurations is to draw a circle and divide it in half horizontally. Above the line write "E". In the lower half divide the semicircle vertically and write "I" on the left, "R" on the right. It should look something like this:\n                                                     E\n                                                   I    R\nNow just cover up the one you need. and either divide, or multiply.
402553
Your circuit diagram looks strange - some resistance are denoted in Volts. Hmm, anyway, otherwise it's rather simple. You got to use your formulaes for calculating resistance in a series and ohm's law V=IR. I hope I'm right - I last did physics in 1999.\n------------------------------------------------\n\nThe resistors are all in series, therefore:\nTotal Resistance = 33 + 23 + 2/I + 1/I + 43\n                 = 99 + 3/I Ohms\n\nUsing Ohms law, V = IR. Therefore:\n=> 100 = I . Total Resistance\n=> 100 = I (99 + 3/I)\n=> 100 = 99 I + 3\n=> I = 97/99 Amperes  --- ans\n\nSub. value for I into Total Resistance:\n=> Total Resistance = 99 + 3/(97/99)\n                    = 9900/97 Ohms (or 102.06 Ohms) --- ans \n\nUsing V=IR\n\nV(3) = potential difference from 100 V to when current I passes through the 33 Ohm resistor\n     = 100 - [(97/99 x 33)] \n     = 202/3 V or 67.33 V (The reading of V (3)'s galvanometer) --- ans.\n\nV(2) = potential difference from V(3) to when current I passes through the 23 Ohm resistor\n     = 202/3 - [(97/99 x 23)] \n     = 4435/99 V or 44.80 V (The reading of V (2)'s galvanometer) --- ans.\n\nV(1) = potential difference from V(2) to when current I passes through the 2V resistor\n     = 4435/99 - 2\n     = 4237/99 V or 42.80 V (The reading of V (1) galvanometer) --- ans.