Question
(a) A battery of electromotive force (e.m.f.) \(6.0\,\mathrm{V}\) and negligible internal resistance is connected in series with a variable resistor and a uniform resistance wire XY, as shown in Fig. 6.1.
Wire XY has length \(2.00\,\mathrm{m}\) and resistance \(8.0\,\Omega\). The resistance \(R\) of the variable resistor is adjusted so that the potential difference across wire XY is \(2.4\,\mathrm{V}\).
(a) Determine \(R\). (2 marks)
\(R=\) ______________________________ \(\Omega\)
(b) Explain why the potential difference \(V\) between any two points on wire XY is proportional to the distance \(L\) between those points. (2 marks)
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(c) A cell of e.m.f. \(E\) and internal resistance \(r\) is connected to the circuit, as shown in Fig. 6.2.
Resistance \(R\) is unchanged.
The movable connection P is positioned on wire XY so that the galvanometer reading is zero. Distance XP is \(1.24\,\mathrm{m}\).
(i) Calculate \(E\). (2 marks)
\(E=\) ______________________________ \(\mathrm{V}\)
(ii) The value of \(R\) is now decreased.
State and explain the change that must be made to the position of P on wire XY so that the galvanometer reads zero again. (2 marks)
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 10.3: Potential dividers – part (c): potentiometer principle, potential comparison and galvanometer null method.
▶️ Answer/Explanation
(a) Resistance of the variable resistor [2 marks]
The current is the same through the series combination of \(R\) and the \(8.0\,\Omega\) wire.
Using the potential-divider relationship,
\(\dfrac{2.4}{6.0}=\dfrac{8.0}{R+8.0}\)
\(\dfrac{2.4}{6.0}=0.40\)
\(0.40=\dfrac{8.0}{R+8.0}\)
\(R+8.0=20\)
\(R=12\,\Omega\)
Answer: \(\boxed{12\,\Omega}\)
(b) Potential difference and distance along a uniform wire [2 marks]
For a uniform wire,
\(R=\dfrac{\rho L}{A}\)
The resistivity \(\rho\), cross-sectional area \(A\), and current \(I\) are constant along the wire.
Therefore, \(R\propto L\).
Using \(V=IR\), with \(I\) constant,
\(V\propto R\)
Hence,
\(\boxed{V\propto L}\)
(c)(i) E.m.f. of cell [2 marks]
At the null point, the galvanometer reads zero, so the p.d. across \(XP\) is equal to the e.m.f. \(E\) of the cell.
Since the wire is uniform,
\(\dfrac{V_{XP}}{V_{XY}}=\dfrac{L_{XP}}{L_{XY}}\)
\(\dfrac{E}{2.4}=\dfrac{1.24}{2.00}\)
\(E=2.4\times\dfrac{1.24}{2.00}\)
\(E=1.488\,\mathrm{V}\)
Answer: \(\boxed{1.5\,\mathrm{V}}\)
(c)(ii) Effect of decreasing \(R\) [2 marks]
When \(R\) is decreased, the total resistance of the circuit decreases.
Therefore, the current in the circuit increases.
Since the wire XY has constant resistance, the potential difference across XY increases.
The potential gradient along XY therefore increases. To obtain the same p.d. \(E\) at the new null point, a shorter length of wire is required.
Hence P must move towards X, away from Y.
Answer: \(\boxed{\text{P moves towards X}}\)