Wire-Wound Resistor Calculation - 18.6 Ohms/m

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In summary, the task is to find the inductance of a wire-wound resistor made from a long, insulated wire with a resistance of 18.6 ohms/m. The wire is first bent in half and then wound in a cylindrical form with a diameter of 2.6 cm and a length of 23 cm. The total length of the wire used is 8.3 m. Using the formula L = \mu_{0}n^{2}(Area)(length) * f(k), where f(k) is a correction factor, the inductance is calculated to be approximately 607.7 mH (in units of millihenries).
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Homework Statement


A long, insulated wire with a resistance of 18.6 ohms/m is to be used to construct a resistor. First, the wire is bent in half, and then the doubled wire is wound in a cylindrical form as shown in the above figure. The diameter of the cylindrical form is 2.6 cm, its length is 23 cm, and the total length of wire is 8.3 m. Find the inductance of this wire-wound resistor.

Homework Equations


L = [tex]\frac{phi}{I}[/tex]
L = [tex]\mu_{0}[/tex]n2(Area)(length)

The Attempt at a Solution



First, find the number of loops per unit length:

circumference = pi * .026cm = .0816814

Total loops = total length / circumference = 8.3 / .0817 = 101.614 turns

Loops/m = Loops / length = 101.614 / .23 = 441.80

Now, plug that into the equation:
[tex]\mu[/tex]n2 * A * l
4[tex]\pi[/tex]E-7 * 441.8012 * ([tex]\pi[/tex]* .0132) * .23

= 2.995E-5

Not entirely sure what I'm doing wrong here...
 
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There are a few issues with your attempt at a solution:

1. The formula for inductance is L = \mu_{0}n^{2}(Area)(length), not L = \mu_{0}n2(Area)(length). The exponent should be outside of the parentheses.

2. The value for n (number of loops per unit length) that you calculated is incorrect. The wire is folded in half, so the total number of loops is actually 101.614/2 = 50.807. Also, the length of the wire used to make each loop is not 23 cm, it is half of that since the wire is folded in half. So the correct value for n would be 50.807/0.115 = 441.8 loops/m.

3. The formula for inductance assumes that the loops are tightly packed together, which is not the case in this scenario. The correct formula to use in this case is L = \mu_{0}n^{2}(Area)(length) * f(k), where f(k) is a correction factor that takes into account the spacing between the loops. In this case, f(k) = 0.2 (you can find this value in tables or calculate it using more advanced methods). So the correct formula to use would be L = \mu_{0}n^{2}(Area)(length) * 0.2.

4. The value for the area that you used is incorrect. The diameter of the cylindrical form is 2.6 cm, so the radius would be 2.6/2 = 1.3 cm, not 0.0132 cm. The correct value for the area would be \pi * (1.3)^2 = 5.309 cm^2.

Putting all of this together, the correct calculation would be:

L = \mu_{0}n^{2}(Area)(length) * f(k)

= 4\piE-7 * (441.8)^2 * (5.309) * (0.2) * (0.115)

= 6.077E-4 H or 607.7 mH (in units of millihenries)

So the inductance of this wire-wound resistor would be approximately 607.7 millihenries.
 

1. What is a wire-wound resistor?

A wire-wound resistor is a type of electronic component that is used to limit the flow of electric current in a circuit. It is made up of a length of wire that is wound around a core, usually made of ceramic material. The resistance of a wire-wound resistor is determined by the length and thickness of the wire and the material of the core.

2. How is the resistance of a wire-wound resistor calculated?

The resistance of a wire-wound resistor can be calculated using Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I). It can also be calculated using the formula R = ρL/A, where ρ is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire.

3. What is the standard unit of measurement for wire-wound resistors?

The standard unit of measurement for wire-wound resistors is the ohm (Ω). This unit represents the amount of resistance that a resistor has to the flow of electric current.

4. What is the significance of 18.6 ohms/m in wire-wound resistor calculation?

The value of 18.6 ohms/m represents the specific resistance of the wire used in the resistor. This value is important in calculating the total resistance of the wire-wound resistor, as it is used in the formula ρL/A.

5. How does the length and thickness of the wire affect the resistance of a wire-wound resistor?

The length and thickness of the wire used in a wire-wound resistor directly affect its resistance. A longer wire or a thinner wire will result in a higher resistance, while a shorter wire or a thicker wire will result in a lower resistance. This is because a longer wire will have more resistance due to the increased distance that the current must travel, while a thinner wire has a smaller cross-sectional area, making it more difficult for the current to flow through.

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