Question

you want to connect two 1.00 ω resistors with 1.00 m of copper wire which has a diameter of 2.00 mm. what percentage does the wire contribute to the total resistance of the circuit? o 35.0 % o 0.270 % o 0.001 % o 1.30 % 26.3 resistance and resistivity save for later submit answer

Explanation:

Step1: Recall resistance formula

The resistance formula is $R =
ho\frac{L}{A}$, where $
ho$ is the resistivity of copper ($
ho_{copper}= 1.72\times10^{-8}\ \Omega\cdot m$), $L$ is the length of the wire, and $A$ is the cross - sectional area of the wire. The cross - sectional area of a circular wire $A=\pi r^{2}$, with $r = \frac{d}{2}$, where $d = 2.00\ mm=2\times10^{- 3}\ m$, so $r = 1\times10^{-3}\ m$ and $A=\pi(1\times10^{-3})^{2}\ m^{2}=\pi\times10^{-6}\ m^{2}$, and $L = 1.00\ m$.

Step2: Calculate wire resistance

$R_{wire}=
ho\frac{L}{A}=1.72\times10^{-8}\times\frac{1}{\pi\times10^{-6}}\ \Omega=\frac{1.72\times10^{-8}}{\pi\times10^{-6}}\ \Omega\approx5.48\times10^{-3}\ \Omega$.

Step3: Calculate total resistance of circuit

The total resistance of the circuit with two $1.00\ \Omega$ resistors and the wire is $R_{total}=1.00 + 1.00+R_{wire}=2.00 + 5.48\times10^{-3}\ \Omega\approx2.00548\ \Omega$.

Step4: Calculate percentage contribution

The percentage contribution of the wire to the total resistance is $\frac{R_{wire}}{R_{total}}\times100\%=\frac{5.48\times10^{-3}}{2.00548}\times100\%\approx0.270\%$.

Answer:

$0.270\%$