Resistivity and Wire Dimensions: Calculating New Length and Diameter for Reduced Resistance
Understanding the relationship between the physical dimensions of a wire and its electrical resistance is crucial for various engineering and scientific applications. In this article, we delve into the calculation of how changes in wire length and diameter affect its resistance, using a practical example.
Problem Statement
A piece of wire initially has a length of 40 cm and a diameter of 0.2 cm. Its resistance is 0.30 ohms. We need to determine the new dimensions of the wire if its resistance is to be reduced to 0.15 ohms. The goal is to apply the resistivity formula and use the relationships between resistivity, length, and cross-sectional area.
Step-by-Step Solution
Step 1: Calculate Initial Cross-Sectional Area and Resistivity
Initial Length: ( L_1 40 ) cm 0.40 m. Initial Diameter: ( d_1 0.2 ) cm 0.002 m. Initial Resistance: ( R_1 0.30 ) ohms. Calculate the cross-sectional area: [ A_1 frac{pi d_1^2}{4} frac{pi times (0.002)^2}{4} approx 3.14 times 10^{-6} , text{m}^2 ] Calculate the resistivity ( rho ):[ R_1 frac{rho L_1}{A_1} Rightarrow rho frac{R_1 A_1}{L_1} frac{0.30 times 3.14 times 10^{-6}}{0.40} approx 2.355 times 10^{-6} , Omega cdot text{m} ]Step 2: Set Up the Equation for the New Resistance
New Resistance: ( R_2 0.15 ) ohms. Equation for resistivity: [ R_2 frac{rho L_2}{A_2} ] Express ( A_2 ) in terms of ( d_2 ): [ A_2 frac{pi d_2^2}{4} ] Substitute and simplify:[ 0.15 frac{2.355 times 10^{-6} L_2}{frac{pi d_2^2}{4}} Rightarrow L_2 frac{0.15 cdot frac{pi d_2^2}{4}}{2.355 times 10^{-6}} ]Step 3: Express ( L_2 ) in Terms of ( d_2 )
Express ( L_2 ): [ L_2 frac{0.15 cdot frac{pi d_2^2}{4}}{2.355 times 10^{-6}} approx frac{0.15 cdot 3.14 d_2^2}{4 cdot 2.355 times 10^{-6}} approx 1.99 times 10^5 cdot d_2^2 ]Step 4: Use the Relationship Between Length and Diameter
Volume Conservation: [ V_1 A_1 L_1 A_2 L_2 Rightarrow frac{pi d_1^2}{4} cdot 0.40 frac{pi d_2^2}{4} cdot L_2 ] Substitute ( L_2 ):[ 0.40 cdot d_1^2 d_2^2 cdot 1.99 times 10^5 cdot d_2^2 Rightarrow 0.000004 cdot 0.40 1.99 times 10^5 cdot d_2^4 ] Solve for ( d_2 ):[ d_2^4 frac{0.0000016}{1.99 times 10^5} approx 8.04 times 10^{-12} Rightarrow d_2 approx 0.0011 , text{m} 0.11 , text{cm} ]Step 5: Calculate ( L_2 )
Substitute ( d_2 ) back to find ( L_2 ):[ L_2 1.99 times 10^5 cdot (0.0011)^2 approx 1.99 times 10^5 times 1.21 times 10^{-6} approx 0.241 , text{m} 24.1 , text{cm} ]Final Results
The new dimensions of the wire are:
Length ( L_2 ): 24.1 cm Diameter ( d_2 ): 0.11 cmConclusion
This calculation demonstrates the importance of the resistivity formula and the relationship between physical dimensions in determining the resistance of a wire. By manipulating the length or diameter, you can significantly alter the resistance, which is essential in various electrical engineering applications.