List the factors that affect the heat produced in a current-carrying conductor - Leaving Cert Physics - Question c - 2011

To find the maximum resistance per metre (R) of the wire, we can use the formula for power:

$P = I^2 R$

Given:

• Power (P) = 2.7 W
• Current (I) = 20 A

Rearranging for R gives us:

$R = \frac{P}{I^2}$

Substituting values:

$R = \frac{2.7}{20^2} = \frac{2.7}{400} = 0.00675 \; \Omega$

Using the formula for resistivity:

$\rho = R \frac{A}{L}$

where

• $\rho$ = resistivity of copper = $1.7 \times 10^{-8} \; \Omega m$
• $A$ = cross-sectional area
• $L$ = length (1 m in this case)

The area A can be expressed as:

$A = \frac{\pi d^2}{4}$

Substituting this into the resistivity equation:

$\rho = R \frac{\pi d^2/4}{1}$

Rearranging gives:

$d^2 = \frac{4\rho}{\pi R}$

Substituting in the known values:

$d^2 = \frac{4 \times (1.7 \times 10^{-8})}{\pi \times 0.00675}$

Calculating:

$d^2 \approx 8.017 \times 10^{-7}$

Taking the square root yields:

$d \approx 0.9 \times 10^{-4} \text{m} = 0.0009 m = 1.8 mm$