A group of researchers investigated the electrical properties of a newly developed conduct… — ACT

A group of researchers investigated the electrical properties of a newly developed conducting polymer. They performed two experiments to understand how the physical dimensions of a cylindrical polymer wire affect its electrical resistance, $R$ (in ohms, $\Omega$ ).

In Experiment 1, the researchers measured the resistance of polymer wires of various lengths, $L$ (in meters, $\text{m}$ ), while keeping the cross-sectional area constant at $A = 0.5\text{ mm}^2$ . The results are shown in Table 1.

Table 1

Trial	Wire Length $L$ ( $\text{m}$ )	Resistance $R$ ( $\Omega$ )
1	2.0	0.068
2	4.0	0.136
3	6.0	0.204
4	8.0	0.272

In Experiment 2, the researchers measured the resistance of polymer wires of a constant length $L = 5.0\text{ m}$ while varying the cross-sectional area, $A$ (in square millimeters, $\text{mm}^2$ ). The results are shown in Table 2.

Table 2

Trial	Cross-Sectional Area $A$ ( $\text{mm}^2$ )	Resistance $R$ ( $\Omega$ )
5	0.2	0.425
6	0.4	0.213
7	0.8	0.106
8	1.6	0.053

Based on the results of Experiments 1 and 2, what is the expected electrical resistance of a cylindrical polymer wire with a length of $12.0\text{ m}$ and a cross-sectional area of $0.3\text{ mm}^2$ ?

A
$0.245\ \Omega$
B
$0.408\ \Omega$
$0.680\ \Omega$ Answer
D
$1.133\ \Omega$

Answer

The expected electrical resistance of the polymer wire is

0.680\ \Omega

The correct answer is

0.680\ \Omega

. Experiment 1 shows that resistance

R

is directly proportional to length

L

because the ratio

R/L

is constant (

0.034\ \Omega/\text{m}

A = 0.5\text{ mm}^2

). Experiment 2 shows that resistance

R

is inversely proportional to area

A

because the product

R \times A

is constant (

0.085\ \Omega\cdot\text{mm}^2

L = 5.0\text{ m}

). Combining these gives

R = k \frac{L}{A}

. Solving for the constant gives

k = 0.017\ \Omega\cdot\text{mm}^2/\text{m}

. Applying this to a wire with

L = 12.0\text{ m}

and

A = 0.3\text{ mm}^2

yields

R = 0.017 \times \frac{12.0}{0.3} = 0.680\ \Omega

Step-by-Step Solution

Determine the relationship between resistance (

R

) and wire length (

L

) from Experiment 1.

In Table 1, as the wire length

L

doubles (e.g., from

2.0\text{ m}

4.0\text{ m}

), the resistance

R

also doubles (from

0.068\ \Omega

0.136\ \Omega

). This indicates that resistance is directly proportional to wire length:

R \propto L

Establishing the relationship between length and resistance is necessary to scale the resistance for the new wire length.

Determine the relationship between resistance (

R

) and cross-sectional area (

A

) from Experiment 2.

In Table 2, as the area

A

doubles (e.g., from

0.2\text{ mm}^2

0.4\text{ mm}^2

), the resistance

R

is halved (from

0.425\ \Omega

0.213\ \Omega

). The product

R \times A

remains constant (

0.425 \times 0.2 \approx 0.085

). This indicates that resistance is inversely proportional to cross-sectional area:

R \propto \frac{1}{A}

Establishing the relationship between area and resistance is necessary to scale the resistance for the new cross-sectional area.

Combine the proportional relationships and find the constant of proportionality.

The combined relationship is

R = k \frac{L}{A}

, where

k

is a constant. Using Trial 1 where

L = 2.0\text{ m}

A = 0.5\text{ mm}^2

, and

R = 0.068\ \Omega

0.068 = k \frac{2.0}{0.5} \implies 0.068 = 4k \implies k = 0.017\ \Omega\cdot\text{mm}^2/\text{m}

Finding the formula and constant allows direct calculation of the resistance for any combination of length and area.

Calculate the resistance for a wire with

L = 12.0\text{ m}

and

A = 0.3\text{ mm}^2

R = 0.017 \times \frac{12.0}{0.3} = 0.017 \times 40 = 0.680\ \Omega

Applying the values to the combined formula yields the final expected resistance.

Key Concept

Direct and Inverse Proportionality

Alternative Method

Instead of finding the constant

k

, you can solve this using scaling factors. First, find the resistance for a wire of length

L_2 = 12.0\text{ m}

and the baseline area

A_1 = 0.5\text{ mm}^2

. Since

R

is directly proportional to

L

, scaling the length from

8.0\text{ m}

(Trial 4) to

12.0\text{ m}

scales the resistance by a factor of

\frac{12.0}{8.0} = 1.5

, giving

0.272 \times 1.5 = 0.408\ \Omega

. Next, adjust for the area change from

0.5\text{ mm}^2

0.3\text{ mm}^2

. Since

R

is inversely proportional to

A

, scaling the area by a factor of

\frac{0.3}{0.5} = 0.6

scales the resistance by a factor of

\frac{1}{0.6} = \frac{5}{3}

. This gives

0.408 \times \frac{5}{3} = 0.680\ \Omega

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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