Question
(a) Define the potential difference across a component. [1 mark]
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(b) The variation with potential difference \(V\) of the current \(I\) in a semiconductor diode is shown in Fig. 6.1.
Use Fig. 6.1 to describe qualitatively:
(i) the resistance of the diode in the range \(V=0\) to \(V=0.25\,\mathrm{V}\). [1 mark]
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(ii) the variation, if any, in the resistance of the diode as \(V\) changes from \(V=0.75\,\mathrm{V}\) to \(V=1.0\,\mathrm{V}\). [1 mark]
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(c) A battery of electromotive force (e.m.f.) \(12\,\mathrm{V}\) and negligible internal resistance is connected to a uniform resistance wire \(XY\), a fixed resistor and a variable resistor, as shown in Fig. 6.2.
The fixed resistor has a resistance of \(5.0\,\Omega\). The current in the battery is \(2.7\,\mathrm{A}\) and the current in the fixed resistor is \(1.5\,\mathrm{A}\).
(i) Calculate the current in the resistance wire. [1 mark]
current = ____________________ \( \mathrm{A} \)
(ii) Determine the resistance of the variable resistor. [2 marks]
resistance = ____________________ \( \Omega \)
(iii) Wire \(XY\) has a length of \(2.0\,\mathrm{m}\). Point \(Z\) on the wire is a distance of \(1.6\,\mathrm{m}\) from point \(X\).
The fixed resistor is connected to the variable resistor at point \(W\). Determine the potential difference between points \(W\) and \(Z\). [3 marks]
potential difference = ____________________ \( \mathrm{V} \)
(iv) The resistance of the variable resistor is now increased.
By considering the currents in every part of the circuit, state and explain whether the total power produced by the battery decreases, increases or stays the same. [3 marks]
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 9.3: Resistance and resistivity — part (b)
• 10.2: Kirchhoff’s laws — parts (c)(i) and (c)(ii)
• 10.1: Practical circuits — parts (c)(iii) and (c)(iv)
▶️ Answer/Explanation
(a) Definition of potential difference [1 mark]
Potential difference is the energy transferred per unit charge when charge passes through a component.
The relationship is
\(V=\frac{W}{Q}\)
Answer: \(\boxed{\text{energy transferred per unit charge}}\)
(b)(i) Resistance from \(0\) to \(0.25\,\mathrm{V}\) [1 mark]
From the graph, the current is approximately zero while the potential difference increases from \(0\) to \(0.25\,\mathrm{V}\).
Since
\(R=\frac{V}{I}\)
a very small current for a finite potential difference corresponds to a very large resistance.
Answer: \(\boxed{\text{The resistance is very large.}}\)
(b)(ii) Resistance from \(0.75\,\mathrm{V}\) to \(1.0\,\mathrm{V}\) [1 mark]
Over this range, the graph becomes progressively steeper. Therefore, a small increase in \(V\) produces a relatively large increase in \(I\).
Since \(R=\frac{V}{I}\), the resistance decreases as \(V\) increases.
Answer: \(\boxed{\text{The resistance decreases.}}\)
(c)(i) Current in the resistance wire [1 mark]
The current from the battery splits between the resistance wire and the lower branch containing the fixed and variable resistors.
Using Kirchhoff’s first law,
\(I_{\mathrm{battery}}=I_{\mathrm{wire}}+I_{\mathrm{fixed}}\)
\(2.7=I_{\mathrm{wire}}+1.5\)
\(I_{\mathrm{wire}}=1.2\,\mathrm{A}\)
Answer: \(\boxed{1.2\,\mathrm{A}}\)
(c)(ii) Resistance of the variable resistor [2 marks]
The fixed resistor and variable resistor are in series in the lower branch.
The p.d. across this branch is \(12\,\mathrm{V}\), and the current is \(1.5\,\mathrm{A}\).
Therefore, the total resistance of the lower branch is
\(R_{\mathrm{total}}=\frac{V}{I}\)
\(R_{\mathrm{total}}=\frac{12}{1.5}=8.0\,\Omega\)
Hence,
\(R_{\mathrm{variable}}=8.0-5.0\)
\(R_{\mathrm{variable}}=3.0\,\Omega\)
Answer: \(\boxed{3.0\,\Omega}\)
(c)(iii) Potential difference between \(W\) and \(Z\) [3 marks]
The wire \(XY\) is uniform, so the potential drop along it is proportional to length.
The current in the wire is \(1.2\,\mathrm{A}\), and the wire is connected directly across the \(12\,\mathrm{V}\) battery. Therefore, the total p.d. across \(XY\) is \(12\,\mathrm{V}\).
The p.d. across \(XZ\) is therefore
\(V_{XZ}=12\times\frac{1.6}{2.0}\)
\(V_{XZ}=9.6\,\mathrm{V}\)
The p.d. across the fixed resistor is
\(V=IR\)
\(V_{XW}=(1.5)(5.0)=7.5\,\mathrm{V}\)
Hence, the potential difference between \(W\) and \(Z\) is
\(V_{WZ}=9.6-7.5\)
\(V_{WZ}=2.1\,\mathrm{V}\)
Answer: \(\boxed{2.1\,\mathrm{V}}\)
(c)(iv) Effect of increasing the variable resistance [3 marks]
Increasing the variable resistance increases the resistance of the lower branch.
Therefore, the current in the lower branch decreases.
The resistance wire is connected directly across the ideal \(12\,\mathrm{V}\) battery, so its current remains unchanged at \(1.2\,\mathrm{A}\).
Hence, the total current supplied by the battery decreases.
Since the battery voltage remains constant,
\(P=VI\)
a decrease in total current means that the total power produced by the battery decreases.
Answer: \(\boxed{\text{The total power produced by the battery decreases.}}\)