Question
(a)(i) State and explain the effect, if any, on the resistance of a filament wire in a lamp as the current in the wire decreases. (1 mark)
______________________________________
(ii) On Fig. 5.1, sketch the \(I\)-\(V\) characteristic of a filament lamp. (2 marks)
(b) A battery of electromotive force (e.m.f.) \(E\) and negligible internal resistance is connected in parallel with two filament lamps A and B, as shown in Fig. 5.2.
The current in the battery is \(3.3\,\mathrm{A}\) and the current in lamp A is \(1.5\,\mathrm{A}\). The power dissipated in lamp A is \(18\,\mathrm{W}\).
(i) Calculate the e.m.f. \(E\) of the battery. (2 marks)
\(E=\) ______________________________ \(\mathrm{V}\)
(ii) The filament wire of lamp B has a cross-sectional area of \(1.4\times10^{-9}\,\mathrm{m^2}\). The number density of free (conduction) electrons per unit volume of the metal of the filament wire is \(3.4\times10^{28}\,\mathrm{m^{-3}}\).
Calculate the average drift speed of the free electrons in the filament wire of lamp B. (3 marks)
average drift speed = ______________________________ \(\mathrm{m\,s^{-1}}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 9.2: Potential difference and power — part (b)(i)
• 9.1: Electric current — part (b)(ii)
• 10.2: Kirchhoff’s laws — part (b)(ii), current division in the parallel circuit
▶️ Answer/Explanation
(a)(i) Effect on resistance [1 mark]
As the current decreases, the temperature of the filament decreases.
Therefore, the resistance of the filament decreases.
Answer: \(\boxed{\text{Resistance decreases}}\)
(a)(ii) \(I\)-\(V\) characteristic [2 marks]
The \(I\)-\(V\) characteristic passes through the origin.
As the voltage and current increase, the filament temperature increases and its resistance increases.
Since \(R=\dfrac{V}{I}\), an increasing resistance means that the gradient \(\dfrac{\Delta I}{\Delta V}\) decreases.
Therefore, the curve starts relatively steep near the origin and becomes progressively less steep as \(V\) increases. A similar curve appears in the third quadrant.
Required sketch: a symmetrical non-linear curve through the origin with decreasing gradient in the first and third quadrants.
(b)(i) E.m.f. of the battery [2 marks]
For lamp A,
\(P=VI\)
Since the lamps are connected in parallel, the potential difference across lamp A is equal to the e.m.f. of the battery because the battery has negligible internal resistance.
\(E=\dfrac{P}{I}\)
\(E=\dfrac{18}{1.5}\)
\(E=12\,\mathrm{V}\)
Answer: \(\boxed{12\,\mathrm{V}}\)
(b)(ii) Average drift speed of electrons [3 marks]
The total current divides between the two parallel branches.
Therefore, the current in lamp B is
\(I_B=3.3-1.5\)
\(I_B=1.8\,\mathrm{A}\)
For a current-carrying conductor,
\(I=Anvq\)
where \(A\) is the cross-sectional area, \(n\) is the number density of charge carriers, \(v\) is the average drift speed and \(q\) is the charge of an electron.
Rearranging,
\(v=\dfrac{I}{Anq}\)
\(v=\dfrac{1.8}{(1.4\times10^{-9})(3.4\times10^{28})(1.60\times10^{-19})}\)
\(v=0.24\,\mathrm{m\,s^{-1}}\)
Answer: \(\boxed{0.24\,\mathrm{m\,s^{-1}}}\)