Does the length of a wire have an effect on voltage or current?
Thanks A2A. Suppose you have a wire of 10 gm. weight and its length is 10 cm. Now if you increase the length of this wire to 20 cm its resistance R will be twice than it was at 10 cm.If 100 people are running through a 20 ft. broad passage they will easily get across it, but if the same no. of people had to run through a passage of 10 ft. breadth than you can imagine the situation.If the length of the wire is increased the cross section of wire also have be increased accordingly to check the increase in the resistance.Double the length, double the resistance. Cross section is same.Double the cross section, half the resistance. Length is same.Now if the resistance increases, the temperature will also increase because higher the temperature higher will be the vibrations.So if the length is increased keeping the cross section same then there will be an increase in resistance and drop in the voltage and there will also be current loss.So only length is not important cross section of the wire is also an important factor.
Physics: A wire with a weight per unit length of 0.079 N/m is suspended directly above a second wire. The top wire carries a current of 29.8?
A wire with a weight per unit length of 0.079 N/m is suspended directly above a second wire. The top wire carries a current of 29.8 A and the bottom wire carries a current of 60.0 A. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. please show work so i can understand.... answer in mm thank u! (19.9)
A 48 cm long copper wire carries a current of 6.0 A and weighs 0.40 N.?
F = ILB You are trying to find B, so re-arrange the formula: B = F/ IL F = 0.40 N I = 6.0 A L = 0.48 m B = (0.40) / (6.0 * .48) = 0.14 T
The wiring in a house must be thick enough so it does not become so hot as to start a fire. What diameter must a copper wire be if it is to?
The wiring in a house must be thick enough so it does not become so hot as to start a fire. What diameter must a copper wire be if it is to carry a maximum current of 40 A and produce no more than 1.5 W of heat per meter of length? Please help! Thank you soooo much!
A 2 m long wire weighing 0.084 N/m is suspended directly above an infinitely straight wire. The top wire car?
A 2 m long wire weighing 0.084 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of 36 A and the bottom wire carries a current of 55 A. The permeablity of free space is 1.25664 × 10^-6 N/A2. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Answer in units of mm.
A wire suddenly breaks due to a weight being suspended. Will the temperature of the wire change as a result of it?
It depends upon whether it is lying in the air or has come in contact with ground potential (i.e. in contact with tower or earth wire). If it is in air then there would be no current in the wire due to infinite resistance or due to no load present (therefore wire will cool down as compared to before breaking condition).Or if the with has come in contact with the ground then it will carry more current than it was carrying before breaking which will lead to heating of the wire.
Does a charged capacitor weigh more than a discharged capacitor?
This is a very complicated question that illustrates why precision experiments are so time-consuming. First, as others have said, no electrons are added or removed from the system. So no mass change from hereSecond, as others have also said, the energy in the capacitor is [math]CV^2/2,[/math] where [math]C [/math]is the capacitance and [math]V[/math] the voltage, and this corresponds to a mass increase of [math]CV^2/2c^2[/math], where [math]c[/math] is the velocity of light. For a an ultracapacitor (4000F, 2.5V, weighing 0.77 kg), that’s a mass increase [math]1.4\times 10^{-13}[/math] kg, which is almost certainly negligible.That was the easy part.Now for the real thing: when charging a capacitor, the electric field pushes the capacitor plates together. That’s likely squeezing some residual gas out of there, reducing the mass by an amount that I can’t predict.The wires will get warm from the current flow (by a minuscule amount, but that’s what we have to deal with here), which might cause some dirt on the wire surface to evaporate, reducing mass. Or the warm wire starts to oxidize, adding mass. You could avoid that by using noble metal wires.When the current flows through the wire to charge, it will cause a magnetic field that might attract some iron dust from the surroundings.So, if I had to bet, I would say the mass changes by a tiny amount that depends on exactly how the experiment is carried out.
Physics conceptual question!??? Ampere's law and A Current Sheet?
Current Sheet Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current I runs in the -y direction through each wire. There are N/a wires per unit length in the x direction. Write an expression for B(d) , the magnetic field a distance above the xy plane of the sheet. Use Uo for the permeability of free space. Express the magnetic field as a vector in terms of any or all of the following: d,I, N,a ,Uo and the unit vectors i,j ,k
Is there anything that gets heavier when an electric current runs through it?
If you think about it carefully, everything gets heavier when current passes through it, albeit usually only by a very tiny amount.We know that mass and energy are equivalent (remember, E=mc²). So if you add energy to a system, it becomes heavier. For example, one megajoule is equivalent to the mass of 11 nanograms.If current runs to an object, it gains energy. There’s not just the increased momentum of the electrons, the electrons will also collide with other particles and stimulate higher motion in these. In other words, on a macroscopic scale, the resistance of the conductor will result in heat.For example, if the temperature of a 1 kilogram copper (heat capacity of 384J/K, according to WolframAlpha) conductor increases by just 1 Kelvin, this equals to a mass gain of 4.3 picograms, or 0.000000000000043 %.