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 7 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).

a) R_c = 1.18 ohms

b the resistance of copper wire is 14.7 % of resistance of speakers.

Explanation:

Given:

- 24-Gauge copper wire resistivity p = 1.72*10^-8

- The Length of the single wire L = 7 m

- The diameter of single copper wire d = 0.51 mm

Find:

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).

Solution:

- We are given details of as single copper wire. However, it takes two copper wire to connect a speaker to the amplifier. After establishing that fact we can use the relation between the dimensions of the wire and the resistance as follows:

R = p*L / A

Where,

R is the resistance of the wire in ohms

A is the cross sectional area of the wire

- Now for two wires the resistance would be twice:

R_c = 2*p*L / A

- plug in the values:

R_c = (2*1.72*10^-8 * 7) / (pi * (0.00051) ^2 / 4)

- Evaluate:

R_c = 1.18 ohms

- The wire resistance calculated can be compared with that of speaker by taking a ratio of the two:

R_c / R_s

1.18 / 8 * 100 = 14.7 %

- Hence, the resistance of copper wire is 14.7 % of resistance of speakers.