Electrical Resistance?
953683
The entire reason that every house and business and car/truck has breaker boxes is because of resistance. Current flowing where it shouldn't, and causes somehting to heat up is because of resistance!\n\nComputers....same thing. Most circuit boards have relayes that will 'trip' if a short occurs, and usually this results in things getting overheated.......because of resistance.\n\nResistance is like friction for electrons. If you rub two sticks together they get hot (resistance)....same for electrons flowing thru a wire. The electrons 'bump' into various things as they go, and it gives rise to heat to put it simply.\n\nToasters rely on resistance to work. Ever look inside when one is on? Ever see the glowing wires in there?.....resistance causes the wires to heat and get red hot! This is what 'cooks' your toast.\n\nSame for an electric oven.......electric heating in your home.\n\nDoes this help?\n\nGood luck to you!\n\n:)
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.
363388
The total resistance is 12 + 8 + 6 = 26 ohms. In series the total resistance is the sum of individual resistors. Apply ohm's law V=IR, I = 20/26 Amperes. The voltage drop across each resistor is found by multiplying the current with the resistance of each resistor.
117733
That is a function of\n\nH = I^2 R t {a quantity of heat generated by a current applied for a time across a resistance}\n\nH=Heat in joules {1 joule = .24 calories\nI=Current in amperes\nR=resistance in ohms\nt=time in seconds\n\nIf you wanted to make the equation a rate\nJ/s=I^2•R \n\nSo what you need to know essentially is the resistance of your wire and its heat dissipating capacity\nthen you can balance the impedance with heat loss\n\nAs for a cigarette lighter in a car look at the fuse size and multiply by 12 volts to get wattage {a rate}\n\nA thinner wire will have a greater resistance\nA longer wire will have a greater resistance\nand both will lead to greater heat loss through surface area\n\nThe best thing you could do is place a rheostat and an ammeter in series with your wire and experiment\n\nAlso remember that the resistance of your element will increase slightly as it grows hot this principle is called impedance
576166
the electrical circuit?
910217
There are two issues to consider, cable heating and voltage drop.  Of the two, voltage drop is by far the easiest.  You need to determine what the allowable voltage drop in you cable is.  The voltage drop is just I x R where I is the current and R is teh round trip cable resistance.  The specification of 100 kW isn't enough.  If its a high tension 100 kV line, the current is only 1 amp and a run of 16 AWG wire would deliver about 99,999 volts out of the original 100,000 volts to your load.  If its a 100 volt system then a 4 AWG wire would only deliver only 20 volts to the load.  \n\nThe formula for wire resistance (in AWG) is that 10 AWG has 0.001 ohm/ft.  Each increase of 1 in the gauge increases resistance by the cube root of 2 (about 25%).  You calculate by picking what voltage drop you can live with, for the known current figure the resistance, knowing the length figure the resistance per foot, and then pick the AWG using the above.\n\nThe second concern is wire heating.  Although you could do the heat transfer calculations, instead look up the allowable ampacity for any wire in an electrical engineers or NEC handbook.
616731
The equivalent resistance of the parallel resistances R3, R4, R5 is given by\n\n1/Re = 1/5 + 1/10 + 1/ 15. =  55/150\nRe     = 2.73 ohm.\nThe equivalent resistance of the series resistances =R1+R2+Re =32.73 ohm\nTotal resistance =32.73 ohm.\n\nCurrent total = voltage /total resistance = 10 /32.73 ohm = 0.305 A.\n\nVolts of R1 = current x resistance = 0.305 x 10 = 3.05 V.\nVolts of R2 = current x resistance = 0.305 x 20 = 6.11 V.\nVolts of Re   = current x resistance = 0.305 x 2.73 = 0.83265 V.\n\nVolts of R3= Volts of R4= Volts of R5= Volts of Re =0.83265 V.\n\nCurrent through R1 = Current through R2 = Current through Re =  0.305 A\n\nCurrent through R3 = volts of R3/ R3 = 0.83265/5 =0.16653A\nCurrent through R4 = volts of R4/ R4 = 0.83265/10 =0.083265A\nCurrent through R5  = volts of R5/ R5 = 0.83265/15 =0.05551A
1079575
By Ohm's Law, current = voltage / resistance\n\nTherefore the current is 120 / 2 = 60 amps (answer b)
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
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.
471231
If the bundle diameter is the same and the individual wires are round in cross-section, then obviously the area of wire metal in the cross section is less in the bundle than in the solid wire so the bundle has a higher resistance. But let's say you mean that the bundle is enough larger in diameter that the metal areas of both cross sections are equal as assumed by answer 2. Suppose there are N strands in the bundle and the solid wire has a resistance R. Then each strand's resistance is N*R. Contrary to answer 2 (if I am reading it correctly), it makes no difference if the individual wires are connected side-by-side or not, as long as they are connected together at each end. Connecting them thus in parallel makes the bundle's resistance equal to R. Since the strands are identical in resistance and therefore carry the same current, the voltage at any distance along one strand is exactly the same as at the same distance along any other strand. Connect them together at that point and what happens? No change in current flow because no change in voltage, and therefore no change in resistance.
1139339
The best way is to get it from the manufacturer.  If you don't know who made it, you can look up resistances from a bunch of places on the net if you know the wire gauge, but remember that copper resistance increases as it gets hot.  \n\nMost magnet wire can handle a few hundred volts with no troubles, but if you're doing some high voltage app, better be sure about the voltage rating.
888374
Conductors allow electricity to flow through them with little resistance.  Resistors resist the flow of electricity.
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.
1011938
Resistance does not depend upon the ductility of the metal used in the wire.\nDuctility is the property of the substance to be easily drawn into wires.  This does not effect electrical resistance.
844258
um, basically, there is no connection between resistance and frequency. there are however some circuits that output a frequency based on a time constant that can be set by a resistor though. if that is what you mean, then we need to know what circuit you are talking about.
230873
Volt = I * R \nmeasure V and I or use the resistance  setting on an ohm meter. Use a fine mechanical pencil refill and subtract out the resistance on the meter leads. If more accuracy is needed there are milli ohms meters or make a resistor bridge.