5.2 Resistance and resistivity — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Write the unit at every step.

Warm-up: remember

A component carries a current of $0.20\,\text{A}$ with a potential difference of $4.0\,\text{V}$ across it. Find the power.
Write $2.5\,\text{mm}^2$ in $\text{m}^2$ .
Rearrange $R = \dfrac{\rho L}{A}$ to make $A$ the subject.

Answers

$P = VI = 4.0 \times 0.20 = 0.80\,\text{W}$ .
$2.5 \times 10^{-6}\,\text{m}^2$ . The millimetre is squared, so the factor is $10^{-6}$ .
$A = \dfrac{\rho L}{R}$ .

If question 1 goes wrong, look again at Section 5.1 and the relation $P = VI$ . If question 2 goes wrong, remember that the power applies to the prefix as well as to the unit. If question 3 goes wrong, multiply both sides by $A$ and then divide by $R$ .

Practise with help

1. A resistor has a potential difference of $9.0\,\text{V}$ across it and a current of $0.30\,\text{A}$ through it. Find its resistance.

Hint

Resistance is the potential difference divided by the current.

Next step

Write $R = \dfrac{V}{I}$ with $V = 9.0$ and $I = 0.30$ .

Full working

R = \frac{V}{I} = \frac{9.0}{0.30} = 30\,\Omega

The resistance is $30\,\Omega$ .

2. A wire of resistivity $1.7 \times 10^{-8}\,\Omega\,\text{m}$ has a length of $3.0\,\text{m}$ and a cross-sectional area of $5.0 \times 10^{-7}\,\text{m}^2$ . Find its resistance.

Hint

Use the resistivity equation. The length goes on the top and the area goes on the bottom.

Next step

Substitute $\rho = 1.7 \times 10^{-8}$ , $L = 3.0$ and $A = 5.0 \times 10^{-7}$ into $R = \dfrac{\rho L}{A}$ .

Full working

\begin{aligned} R &= \frac{\rho L}{A} \\ &= \frac{1.7 \times 10^{-8} \times 3.0}{5.0 \times 10^{-7}} \\ &= 1.0 \times 10^{-1}\,\Omega \end{aligned}

The resistance is about $0.10\,\Omega$ .

3. A wire of length $2.0\,\text{m}$ and cross-sectional area $4.0 \times 10^{-7}\,\text{m}^2$ has a resistance of $6.0\,\Omega$ . Find the resistivity of the material.

Hint

Rearrange the resistivity equation to make $\rho$ the subject.

Next step

Start from $\rho = \dfrac{RA}{L}$ , then substitute the three values.

Full working

\begin{aligned} \rho &= \frac{RA}{L} \\ &= \frac{6.0 \times 4.0 \times 10^{-7}}{2.0} \\ &= 1.2 \times 10^{-6}\,\Omega\,\text{m} \end{aligned}

The resistivity is $1.2 \times 10^{-6}\,\Omega\,\text{m}$ .

4. A filament lamp has a current of $0.25\,\text{A}$ when the pd across it is $3.0\,\text{V}$ , and a current of $0.40\,\text{A}$ when the pd is $6.0\,\text{V}$ . Find the resistance at each pd, and explain the change.

Hint

Find each resistance with $R = \dfrac{V}{I}$ , then compare them.

Next step

Work out $\dfrac{3.0}{0.25}$ and $\dfrac{6.0}{0.40}$ , and note which is larger.

Full working

R_1 = \frac{3.0}{0.25} = 12\,\Omega \qquad R_2 = \frac{6.0}{0.40} = 15\,\Omega

At the higher pd the filament is hotter. The ions in the metal lattice vibrate more, so they scatter the drifting electrons more often. The current is smaller than it would be at a constant resistance, so the resistance has risen from $12\,\Omega$ to $15\,\Omega$ .

5. A copper wire has a diameter of $0.50\,\text{mm}$ and a length of $2.0\,\text{m}$ . The resistivity of copper is $1.7 \times 10^{-8}\,\Omega\,\text{m}$ . Find its resistance.

Hint

Find the cross-sectional area first. It is the area of a circle, so you need the radius, not the diameter.

Next step

Halve the diameter to get $0.25\,\text{mm}$ , change it to metres, and use $A = \pi r^2$ .

Full working

\begin{aligned} r &= 0.25\,\text{mm} = 2.5 \times 10^{-4}\,\text{m} \\ A &= \pi r^2 = \pi \times (2.5 \times 10^{-4})^2 = 1.96 \times 10^{-7}\,\text{m}^2 \\ R &= \frac{\rho L}{A} = \frac{1.7 \times 10^{-8} \times 2.0}{1.96 \times 10^{-7}} \\ &= 0.17\,\Omega \end{aligned}

The resistance is about $0.17\,\Omega$ , to two significant figures.

6. Two wires are made of the same metal. One has twice the length and twice the diameter of the other. Explain what happens to the resistance of the longer, thicker wire.

Hint

Doubling the diameter makes the area four times larger. The resistance is proportional to the length and inversely proportional to the area.

Next step

Compare the two effects. The length doubles, so $R$ doubles. The area becomes four times larger, so $R$ becomes one quarter. Find the combined factor.

Full working

The resistance is proportional to $\dfrac{L}{A}$ . Doubling the length doubles $R$ . Doubling the diameter makes the area four times larger, so it divides $R$ by four. The combined change is a factor of $\dfrac{2}{4} = \dfrac{1}{2}$ . The longer, thicker wire has half the resistance of the shorter, thinner wire.