Answer:
Step-by-step explanation:
The resistance of each wire can be calculated using the formula:
R = ρl/A
where ρ is the resistivity of copper (1.68 × 10^-8 Ω∙m), l is the length of the wire (2 m), and A is the cross-sectional area of the wire.
A = π(d/2)^2 = π((28 AWG)/2)^2 = 2.37 × 10^-7 m^2
R = (1.68 × 10^-8 Ω∙m)(2 m)/(2.37 × 10^-7 m^2) = 0.141 Ω (for each wire)
The internal inductance of each wire can be approximated using the formula:
L = µrµ0l/π ln(d/2r)
where µr is the relative permeability of copper (approximately 1), µ0 is the permeability of free space (4π × 10^-7 H/m), l is the length of the wire (2 m), d is the diameter of the wire (28 AWG = 0.321 mm), and r is the radius of the wire (0.321 mm/2 = 0.1605 mm).
L = (4π × 10^-7 H/m)(2 m)/π ln(0.321 mm/0.1605 mm) = 3.74 nH (for each wire)
The external inductance of the cable can be calculated using the formula:
L = µ0l/π ln(2h/d)
where h is the separation between the wires (50 mils = 0.050 inches = 1.27 mm).
L = (4π × 10^-7 H/m)(2 m)/π ln(2 × 1.27 mm/0.321 mm) = 1.518 µH
The capacitance between the wires can be approximated using the formula:
C = εrε0A/h
where εr is the relative permittivity of the insulating material between the wires (approximately 1 for air), ε0 is the permittivity of free space (8.854 × 10^-12 F/m), A is the area of overlap between the wires, and h is the separation between the wires.
A = π(d/2)^2 - πh^2/4 = π((28 AWG)/2)^2 - π(1.27 mm/2)^2 = 6.945 × 10^-8 m^2
C = (8.854 × 10^-12 F/m)(6.945 × 10^-8 m^2)/(1.27 mm) = 29.28 pF
The characteristic impedance of the cable can be approximated using the formula:
Z0 = √(L/C)
Z0 = √((1.518 µH + 5.95 nH)(29.28 pF)) = 227.7 Ω (approximately)
Therefore, the resistance, internal inductance, external inductance, capacitance, and characteristic impedance of the cable are approximately 0.141 Ω, 3.74 nH, 1.518 µH, 29.28 pF, and 227.7 Ω, respectively.