Question

a single 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 6 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). $\frac{r_{wire}}{r_{sp}}=$

Explanation:

Step1: Recall resistance formula

The resistance formula is $R =
ho\frac{L}{A}$, where $
ho$ is the resistivity of the material, $L$ is the length of the wire, and $A$ is the cross - sectional area of the wire. For copper at $20^{\circ}C$, $
ho= 1.72\times10^{-8}\Omega\cdot m$. The length of the wire $L = 2\times6m$ (since there are two wires from amplifier to speaker), and the cross - sectional area $A=\pi r^{2}$, with $r=\frac{d}{2}=\frac{0.51\times10^{-3}}{2}m$.

Step2: Calculate cross - sectional area

$A=\pi r^{2}=\pi(\frac{0.51\times10^{-3}}{2})^{2}\approx 2.04\times10^{-8}m^{2}$

Step3: Calculate resistance

$R=
ho\frac{L}{A}=1.72\times 10^{-8}\frac{2\times6}{2.04\times10^{-8}}\approx 1.01\Omega$

Step4: For part B

To find the ratio $\frac{R_{wire}}{R_{sp}}$, where $R_{sp} = 8\Omega$ and $R_{wire}\approx1.01\Omega$. So $\frac{R_{wire}}{R_{sp}}=\frac{1.01}{8}\approx0.126 = 12.6\%$

Answer:

Part A: $1.01\Omega$
Part B: $12.6\%$