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12 December, 07:52

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).

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  1. 12 December, 08:01
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    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.
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