Question
A student uses a circuit containing an ammeter, a voltmeter and a cell to take measurements to determine the resistance of a length of nichrome wire.
(a) (i) Define resistance. (1 mark)
____________________________________________
(ii) Draw a circuit diagram to show how the components should be connected. Use the symbol for a resistor to represent the nichrome wire. (2 marks)
____________________________________________
____________________________________________
(b) The student also measures the length and the diameter of the wire. Table 5.1 shows the measurements recorded for each quantity.
| quantity | measurement |
|---|---|
| length | \( (0.864 \pm 0.001)\,\mathrm{m} \) |
| diameter | \( (0.496 \pm 0.002)\,\mathrm{mm} \) |
| voltmeter reading | \( (1.38 \pm 0.02)\,\mathrm{V} \) |
| ammeter reading | \( (0.276 \pm 0.001)\,\mathrm{A} \) |
(i) Show that the resistance of the wire is \(5.00\,\Omega\). (1 mark)
____________________________________________
(ii) Calculate, to three significant figures, the resistivity \(\rho\) of the nichrome. (3 marks)
\(\rho=\) __________________________ \( \Omega\,\mathrm{m} \)
(iii) Calculate the percentage uncertainty in \(\rho\). (2 marks)
percentage uncertainty = __________________________ \( \% \)
(iv) Determine the absolute uncertainty in \(\rho\). (1 mark)
absolute uncertainty = __________________________ \( \Omega\,\mathrm{m} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 10.1: Practical circuits — part (a)(ii)
• 1.3: Errors and uncertainties — parts (b)(ii), (b)(iii) and (b)(iv)
▶️ Answer/Explanation
(a)(i) Definition of resistance [1 mark]
Resistance is the potential difference per unit current.
\(R=\dfrac{V}{I}\)
Answer: \( \boxed{\text{potential difference per unit current}} \)
(a)(ii) Circuit diagram [2 marks]
The ammeter must be connected in series with the nichrome wire so that it measures the current through the wire.
The voltmeter must be connected in parallel across the nichrome wire so that it measures the potential difference across the wire.
The resistor symbol represents the nichrome wire, and the cell is connected in a complete circuit.
Required arrangement: cell, ammeter and resistor in series, with the voltmeter connected in parallel across the resistor.
(b)(i) Resistance of the wire [1 mark]
Using
\(R=\dfrac{V}{I}\)
\(R=\dfrac{1.38}{0.276}\)
\(R=5.00\,\Omega\)
Answer: \( \boxed{5.00\,\Omega} \)
(b)(ii) Resistivity of the nichrome [3 marks]
The resistivity is related to resistance by
\(R=\dfrac{\rho L}{A}\)
Therefore,
\(\rho=\dfrac{RA}{L}\)
The diameter is \(0.496\,\mathrm{mm}\), so the radius is
\(r=\dfrac{0.496\times10^{-3}}{2}=0.248\times10^{-3}\,\mathrm{m}\)
The cross-sectional area is
\(A=\pi r^2\)
Hence,
\(\rho=\dfrac{5.00\times\pi(0.248\times10^{-3})^2}{0.864}\)
\(\rho=1.12\times10^{-6}\,\Omega\,\mathrm{m}\)
Answer: \( \boxed{1.12\times10^{-6}\,\Omega\,\mathrm{m}} \)
(b)(iii) Percentage uncertainty in resistivity [2 marks]
Since
\(\rho=\dfrac{RA}{L}\)
and \(A=\pi d^2/4\), the percentage uncertainty in \(A\) is twice the percentage uncertainty in the diameter.
Therefore,
\(\text{percentage uncertainty}=\left[\dfrac{0.001}{0.864}+\dfrac{2(0.002)}{0.496}+\dfrac{0.02}{1.38}+\dfrac{0.001}{0.276}\right]\times100\)
\(\text{percentage uncertainty}=2.7\%\)
Answer: \( \boxed{2.7\%} \)
(b)(iv) Absolute uncertainty in resistivity [1 mark]
The absolute uncertainty is
\(\Delta\rho=1.12\times10^{-6}\times\dfrac{2.7}{100}\)
\(\Delta\rho=3.0\times10^{-8}\,\Omega\,\mathrm{m}\)
Answer: \( \boxed{3\times10^{-8}\,\Omega\,\mathrm{m}} \)