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? view available hint(s) 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 $20^{\circ}C$, the resistivity $
ho= 1.72\times10^{-8}\ \Omega\cdot m$. The wire is used to connect the speaker to the amplifier, and since it's a two - way connection (positive and negative), the length of the wire $L = 2\times6\ m=12\ m$. The cross - sectional area of a circular wire $A=\pi r^{2}$, with $r=\frac{d}{2}=\frac{0.51\times10^{-3}}{2}\ m$.

Step2: Calculate cross - sectional area

$A=\pi r^{2}=\pi(\frac{0.51\times10^{-3}}{2})^{2}\ m^{2}\approx2.04\times10^{-7}\ m^{2}$

Step3: Calculate resistance

$R=
ho\frac{L}{A}=(1.72\times 10^{-8}\ \Omega\cdot m)\frac{12\ m}{2.04\times10^{-7}\ m^{2}}\approx1.01\ \Omega$

Step4: For part B

We want to find $\frac{R_{wire}}{R_{sp}}$, where $R_{sp} = 8\ \Omega$ and $R_{wire}\approx1.01\ \Omega$. So $\frac{R_{wire}}{R_{sp}}=\frac{1.01\ \Omega}{8\ \Omega}\approx0.126$

Answer:

Part A: $1.01\ \Omega$
Part B: $0.126$