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.

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