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The thin wire has more resistance to the flow of electric current than the thick wire. If you connect the wires to a battery the battery will supply electrical pressure (voltage) and the wires serve similar to pipes that conduct water under pressure. A small pipe exhibits more resistance to the flow of water and a thin wire exhibits more resistance to the flow of electrons. However, as you point out different wire materials exhibit different resistances for equal sizes (silver conducts better than copper, etc.).
Assume that the increase in length is achieved by uniform reduction in the cross-sectional area of the wire. Then an increase in length by 4 times will result in the cross sectional area being reduced to a fifth of it original value. This will increase the resistance to five times its previous value.
1. The potential difference given to the conductor 2. The resistance of the conductor 3. The change in magnetic field linked with the conductor 4. The temperature of the conductor 5. The material that we have chosen as the wire
1.Resistance is dependent on the material.Like wood is insulator(ALMOST infinite resistance). 2.Resistance of a wire having more cross sectional area is less and less cross sectional area is more(i.e. it is inversely propotional to the cross sectional area.) 3.It is more for more length and less for less length. 4. Resistance varies with temprature.For metals like platinum it increses with temprature.
Actually resistance is directly proportional to the length provided area remains constant. But as we stretch the wire only its volume would remain constant. So its area is to be decreased as length increases. V = pi r^2 * L Now we have R = K * L / pi r^2 Multiplying numerator and denominator by L we get R = K/V * L^2 So resistance is found to be proportional to square of length Hence as length gets increased by 2 times, its resistance value would increase by 4 times.
We know that, Circumference of the wire = 2πr Thus, resistance per unit volume of the wire = 4/2πr = 2/πr So, resistance of the specimen = 2/πr × 2r = 4/π And resistance of the halves of the wire = 2/πr × πr = 2 Now, Equivalent resistance will be decided as 4/(4+π) as 4/π,2,2 are in parallel combination.
resistance is directly proportional to wire length and inversely proportional to wire cross-sectional area. In other words, If the wire length is doubled, the resistance is doubled too. If the wire diameter is doubled, the resistance will reduce to 1/4 of the original resistance.
The thin wire has more resistance to the flow of electric current than the thick wire. If you connect the wires to a battery the battery will supply electrical pressure (voltage) and the wires serve similar to pipes that conduct water under pressure. A small pipe exhibits more resistance to the flow of water and a thin wire exhibits more resistance to the flow of electrons. However, as you point out different wire materials exhibit different resistances for equal sizes (silver conducts better than copper, etc.).
Length, cross section, material, temperature.AnswerWithout wishing to sound pedantic, there are only threefactors that affect resistance. These are the length, cross-sectional area, and resistivity of a material. Temperature affects resistivity.
I would check the maximum current produced by the alternator then check charts to see what gauge wire is needed. As a general rule, I would start with #4 AWG copper. As wire size decreases, the resistance of the wire increases and more heat is generated in the wire. Also as resistance increases, the voltage at the other end decreases due to voltage drop in the wire.
I = E/R = 9/4 = 2.25 AmpereThe wire is dissipating 20.25 watts, and is getting pretty hot.
You can, but you may damage your amp. If you wire them in parallel (both wires from each connected to the amp) then you will have an effective resistance less than 2 ohms. If you wire them in series then you will have an effective resistance of 6 ohms. Resistance that is too low draws more power than the output is rated for. Resistance that is too high reduces the volume and may distort the sound.
Assume that the increase in length is achieved by uniform reduction in the cross-sectional area of the wire. Then an increase in length by 4 times will result in the cross sectional area being reduced to a fifth of it original value. This will increase the resistance to five times its previous value.
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The main difference is in the price. Oxygen-free copper is sold to people with plenty of money for loudspeaker wire, but provided the wire has less resistance than one tenth of the speaker resistance (usually 4 or 8 ohms), the resistance of the wire is immaterial, and ordinary copper wire is perfectly all right. Highly refined copper has about 1% better conductivity than the usual variety. That difference in conductivity is insignificant for audio use and is also produced by a 3 degree C temperature rise in copper.
10 AWG wire will have less resistance per foot and therefore you can have longer runs with 10 AWG than 12 AWG without as much loss of signal. Additional to loss of signal there is a need to maintain a low impedance driving the typical speaker with its 4 ohm or 8 ohm coil resistance, to avoid frequency-distortion. A guide rule is that the speaker cable should have a total resistance around 1% of the speaker resistance (or less). On that basis, for 5 yards of speaker cable for a 4 ohm speaker the wire size would be 4 sq. mm (11 AWG) or for an 8 ohm speaker 2 sq. mm (14 AWG). <<>> Using the above formula of 1% of the speaker resistance the above answer is not correct. For a 4 ohm speaker at 1% is .04 ohms. The resistance of #12 wire is .001588 ohms per foot. This will allow you to run 25 feet and still stay within the parameters. For a 4 ohm speaker at 1% is .04 ohms. The resistance of #10 wire is .00100 ohms per foot. This will allow you to run 40 feet and still stay within the parameters. You can see, what the first answer states is correct.
An RTD or Pt100 sensor is connected with two, three or four wires to the measuring device.we learned that we are in fact measuring resistance to determine the temperature. Now when measuring the resistance of the sensing element, we also measure the resistance of the leads and cables used. This gives an error! To compensate for this, the three wire type (bridge) is used, giving enough accuracy in most industrial applications. Even better accuracy is possible with a four wire Pt100 (laboratory applications). Our Pt100 panel mounted indicators have an offset compensation when using two wire sensors.