Understanding Resistance Changes in a Wire When Modified

Resistance in an electrical circuit is a fundamental concept that affects the flow of current. Understanding how changes in the physical properties of a wire—such as its length and cross-sectional area—affect its resistance is crucial for electronic engineers and mathematicians alike. In this article, we'll explore how these changes impact the overall resistance using a specific example and derive a general formula.

Initial Resistance

Consider a wire with a resistance of 10 ohms. The overall resistance, (R), is given by the formula:

[R rho frac{L}{A}]

Symbol Definition Units R Resistance Ohms (Ω) ρ (rho) Resistivity of the material Ohm-meters (Ω·m) L Length of the wire Meters (m) A Cross-sectional area of the wire Square meters (m2)

For a wire made of a material with a constant resistivity, the initial resistance is given as:

[R_1 10 text{ Ω}]

The initial length of the wire is (L) and the initial cross-sectional area is (A).

Modified Wire Characteristics

Now, let's modify the wire's characteristics:

The length of the wire is doubled. The cross-sectional area is halved.

Let's denote the new length and area as (L') and (A'), respectively.

The new length is:

[L' 2L]

The new cross-sectional area is:

[A' frac{A}{2}]

Now, we need to calculate the new resistance, (R_2), using the modified dimensions.

Calculating the New Resistance

The formula for resistance remains:

[R_2 rho frac{L'}{A'}]

Substitute the new length and area:

[R_2 rho frac{2L}{frac{A}{2}}]

Simplify the equation:

[R_2 rho frac{2L times 2}{A} rho frac{4L}{A}]

Using the initial resistance (R_1 rho frac{L}{A} 10 text{ Ω}), we can substitute:

[R_2 4 times R_1 4 times 10 text{ Ω} 40 text{ Ω}]

Conclusion

The new resistance of the wire, when its length is doubled and its cross-sectional area is halved, is:

[R_2 40 text{ Ω}]

This example demonstrates the direct impact of changing wire dimensions on resistance. For a more general formula, we can also consider a scenario where the original wire's resistance is given by:

[R_a rho frac{L}{pi r_a^2}]

And a second wire with:

[R_b rho frac{L_b}{pi r_b^2}]

If the second wire has the same material, double the length, and the radius of the cross-section is halved compared to the original wire, we can derive the new resistance (R_b).

Solving for the resistivity ρ:

[rho frac{R_a pi r_a^2}{L}]

[rho frac{R_b pi r_b^2}{L_b}]

Since (rho) is the same for both wires, we set the two expressions equal:

[frac{R_a pi r_a^2}{L} frac{R_b pi r_b^2}{L_b}]

Solving for (R_b):

[R_b R_a frac{r_a^2}{r_b^2} frac{L_b}{L}]

Given that (r_b frac{r_a}{2}) and (L_b 2L), we can substitute these values:

[R_b R_a frac{r_a^2}{left(frac{r_a}{2}right)^2} frac{2L}{L}]

[R_b R_a frac{r_a^2}{frac{r_a^2}{4}} frac{2}{1}]

[R_b R_a times 4 times 2]

[R_b 8R_a]

For the initial resistance (R_a 10 text{ Ω}), the new resistance (R_b) is:

[R_b 8 times 10 text{ Ω} 80 text{ Ω}]

This example illustrates the principle that changes in the dimensions of a wire directly impact its resistance. Electronic engineers often use these principles to optimize circuit designs. It's essential to keep these calculations in your notes for practical applications.