Electricity: Question 4

Worked solution

Part (a): Cross-sectional area

The radius is half the diameter, converted to metres: $r = \frac{0.460\times10^{-3}}{2} = 2.30\times10^{-4}\text{ m}$

The cross-section is a circle, so: $A = \pi r^2 = \pi \times (2.30\times10^{-4})^2 = \pi \times 5.29\times10^{-8}$ $A = 1.66\times10^{-7}\text{ m}^2 \ (3\text{ s.f.})$

Part (b): Resistance from resistivity

Using $R = \dfrac{\rho L}{A}$, and keeping an extra figure in $A$ ($1.6619\times10^{-7}\text{ m}^2$) to avoid rounding error: $R = \frac{4.90\times10^{-7} \times 2.50}{1.6619\times10^{-7}} = \frac{1.225\times10^{-6}}{1.6619\times10^{-7}}$ $R = 7.37\,\Omega \ (3\text{ s.f.})$

Part (c): Doubling the diameter

For a circular cross-section, $A = \pi r^2 \propto d^2$: doubling the diameter quadruples the cross-sectional area. Since $R = \rho L/A$ with $\rho$ and $L$ unchanged, resistance is inversely proportional to area, so quadrupling the area means the resistance falls to one quarter of its original value: $R_2 = \frac{R}{4} = \frac{7.371}{4} = 1.84\,\Omega \ (3\text{ s.f.})$

The thicker wire offers a wider path for the charge carriers, so it opposes the current less and has a lower resistance.

Part (d): Power dissipated

Using $P = I^2R$ with the original resistance $R = 7.371\,\Omega$ (unrounded) and $I = 0.600\text{ A}$: $P = I^2R = (0.600)^2 \times 7.371 = 0.360 \times 7.371$ $P = 2.65\text{ W} \ (3\text{ s.f.})$

Final answers

• (a) Cross-sectional area $=$ $1.66\times10^{-7}\text{ m}^2$
• (b) Resistance $=$ $7.37\,\Omega$
• (c) Resistance of the thicker wire $=$ $\approx 1.84\,\Omega$ (a quarter of the value in (b), since $R \propto 1/A$ and $A \propto d^2$)
• (d) Power dissipated $=$ $2.65\text{ W}$