A 24-gauge copper wire has a diameter of 0.51 mm and is used to connect a speaker to an amplifier. The speaker is located 8.5 m away from the amplifier.

Part A: What is the minimum resistance of the connecting speaker wires at 20 ˚C?

Part B: Compare the resistance of the wire to the resistance of the speaker (RSP = 8 Ω).

Given Information:

Diameter of the wire = d = 0.51 mm

Length of the wire = L = 8.5 m

Resistance of the speaker = Rsp = 8 Ω

Required Information:

Part A: Resistance of the connecting wire = R = ?

Part B: Ratio of the resistance of connecting wire and speaker = ?

Answer:

Part A: R = 1.43 Ω

Part B: R is 17.8% of Rsp

Explanation:

R = ρL/A

where ρ is the resistivity of the copper wire 1.72x10⁻⁸ Ω/m (at 20 ˚C)

A is the area of the copper wire A = πr²

r = d/2 = 0.51/2 = 0.255 = 0.0255 cm 0.000255 m

Part A: What is the minimum resistance of the connecting speaker wires at 20 ˚C?

R = ρL/πr

R = 1.72x10⁻⁸ (8.5) / π * (0.000255) ²

R = 0.715 Ω

Since two connecting wires are needed so total resistance becomes

R = 2*715 = 1.43 Ω

Part B: Compare the resistance of the wire to the resistance of the speaker (RSP = 8 Ω)

Resistance of connecting wires = 1.43 Ω

Resistance of the speaker = 8 Ω

1.43/8 = 0.178 = 17.8 %

Therefore, resistance of the connecting wires is 17.8% of the speaker resistance.