Oxford Revise AQA A Level Physics | Chapter P10 answers
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Question |
Answers |
Extra information |
Mark |
AO |
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01.1 |
Solar cell in series with ammeter and variable resistor Voltmeter in parallel with variable resistor (or solar cell) |
1 |
3.5.1.6 |
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01.2 |
Allow sensible suggestions: Control light intensity
Control of temperature
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1 mark for how to control light intensity 1 mark for control of temperature |
1 |
PS1.1 |
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01.3 |
Change the resistance of the circuit using the variable resistance and record a series of potential difference and current readings Plot a graph of p.d. against current The y-intercept is the e.m.f The gradient is the (negative) internal resistance |
1 |
3.5.1.6 |
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01.4 |
The e.m.f depends on the light intensity/temperature so they can only quote a value for average conditions |
Any sensible suggestion |
1 |
3.5.1.6 |
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01.5 |
Use of ε = I (R + r) or recognising ε = 8.2 V ε = IR + Ir = V + Ir 8.2 = 5.5 + (0.1 × r) 2.7 = 0.1r r = 27 Ω |
1 |
3.5.1.6 |
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02.1 |
The current flowing into a junction must equal the current flowing out of the junction (owtte) |
1 |
3.5.1.4 |
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02.2 |
The sum of the p.d.s in a closed loop must equal the sum of e.m.f.s in that loop (owtte) |
1 |
3.5.1.4 |
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02.3 |
Bulbs in series, so same current flows into each \(\begin{array}{l} \textrm{Use of or mention } P = I^2 R (\textrm{or }P = VI\ \rm{or}\ \textit{P} = \frac{V^2}{R})\\ \textrm{Since}\ I\ \textrm{constant}\ P \propto R\ \textrm{so}\ \textbf{A}\ \textrm{must have greater resistance}\\ \end{array} \) |
1 |
3.5.1.4 |
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02.4 |
Bulbs in parallel so this time p.d. the same \(\begin{array}{l} P = \frac{V^2}{R}\\ \textrm{Since }V\textrm{ constant }P \propto \frac{1}{R}\\ \end{array} \) Bulb B brightest |
If they think B has the higher resistance, then allow e.c.f if correct reasoning applied |
1 |
3.5.1.4 |
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03.1 |
1 litre = 1000 cm3 191 litres per hour = 53 cm3 s–1 mass per second = ρV = 1000 × 53 × 10–6 = 0.053 kg s–1 |
1 |
3.4.2.1 |
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03.2 |
\( \begin{array}{l} P = \frac{\Delta w}{\Delta t}\ \rm{or}\ \textit{E}_{p} = \textit{mgh}\\ P = 0.053 \times 9.81 \times 0.3 = 0.16\ \rm{W}\\ \end{array} \) |
1 |
3.4.1.7 |
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03.3 |
\( \begin{array}{l} P = VI\\ I = \frac{1.2}{5} = 0.24\ \rm{A}\\ \end{array} \) |
1 |
3.5.1.4 |
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03.4 |
\( \begin{array}{l} \textrm{Use of } \epsilon = I(R + r)\\ \epsilon = IR + Ir = V + Ir\\ 6 = 5 + (0.24 \times r)\\ 1 = 0.24 r\\ r = 4.2 \Omega\\ \end{array} \) OR \(\begin{array}{l} \textrm{Resistance of pump }R = \frac{V}{I} = \frac{5}{0.24} = 20.8 \Omega\\ \epsilon = I\ (R + r)\\ 6 = 0.24\ (20.8 + r)\\ 6 = 5 + 0.24 r\\ \end{array} \) |
1 |
3.5.1.4 |
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04.1 |
\( \begin{array}{l} P = VI\\ \textbf{A}: I = \frac{0.7}{3.5} = 0.2\ \rm{A}\\ \textbf{B}: I = \frac{1.95}{6.5} = 0.3\ \rm{A}\\ \textbf{C}: I = \frac{0.3}{1.5} = 0.2\ \rm{A}\\ \end{array} \) |
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3.5.1.4 |
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04.2 |
I = 0.2 A + 0.3 A + 0.2 A = 0.7 A |
1 |
3.5.1.4 |
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04.3 |
\( \begin{array}{l} \textrm{p.d. across }R_1 = 9.0 – 6.5 = 2.5\ \textrm{V}\\ R_1 = \frac{V}{I} = \frac{2.5}{0.7} = 3.6\ \Omega\\ \end{array} \) |
Allow e.c.f from answer to 04.2 |
1 |
3.5.1.4 |
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04.4 |
\( \begin{array}{l} \textrm{p.d. across }R_3 = 6.5 – 1.5 = 5.0\ \textrm{V}\\ R_3 = \frac{V}{I} = \frac{0.5}{0.2} = 25\ \Omega\\ \end{array} \) |
1 |
3.5.1.4 |
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05.1 |
In series: ε = 1.5 V + 1.5 V = 3.0 V r = 0.5 Ω + 0.5 Ω = 1.0 Ω |
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3.5.1.4 |
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05.2 |
In parallel: \(\begin{array}{l} \epsilon = 1.5\ \rm{V}\\ r = {\left( {\frac{1}{0.5} + \frac{1}{0.5}} \right)^{-1}} = 0.25\ \Omega\\ \end{array} \) |
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3.5.1.4 |
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05.3 |
Two cells in parallel with one cell in series |
1 |
3.5.1.4 |
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05.4 |
\( \begin{array}{l} \textrm{Use of }\epsilon = I(R + r)\ \rm{or}\ r = \frac{3}{4} = 0.75\ \Omega\\ 3.0 = I\ (2.0 + 0.75)\\ I = 1.1\ \rm{A}\\ \end{array} \) |
1 |
3.5.1.6 |
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05.5 |
\( \begin{array}{l} P = I^2 R = 1.1^2 \times 2\\ = 2.4\ \rm{W}\\ \end{array} \) |
1 |
3.5.1.4 |
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06.1 |
Diagram showing 3 resistors connected in series |
1 |
3.5.1.4 |
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06.2 |
Total resistance = 33 Ω + 110 Ω + 67 Ω = 210 Ω |
1 |
3.5.1.4 |
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06.3 |
Diagram showing three resistors connected in parallel |
1 |
3.5.1.4 |
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06.4 |
\( \begin{array}{l} \textrm{Total resistance } = {\left( {\frac{1}{110} + \frac{1}{67} + \frac{1}{33}} \right)^{-1}}\\ R = 18\ \Omega\\ \end{array} \) |
1 |
3.5.1.4 |
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06.5 |
Diagram showing two 33 Ω resistors in series and also in parallel with two other 33 Ω resistors in series |
1 |
3.5.1.4 |
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06.6 |
Smaller current through each resistor/less power transferred by each resistor/still resistance if 1 resistor breaks |
Accept any sensible suggestion |
1 |
3.5.1.4 |
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07.1 |
\( \begin{array}{l} P = \frac{V^2}{R}\\ R = \frac{V^2}{P} = \frac{12^2}{50} = 2.9\ \Omega\\ \end{array} \) |
1 |
3.5.1.4 |
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07.2 |
\( \begin{array}{l} \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots\\ \frac{1}{2.9} = \frac{8}{R}\\ R = 8 \times 22.9 = 23.2\ \Omega\ (23.0\ \textrm{if you use unrounded number})\\ \end{array} \) |
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3.5.1.4 |
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07.3 |
\( \begin{array}{l} R = \frac{\rho l}{A}\\ A = \frac{\rho l}{R}\\ A = d \times 0.003\ \rm{m}\\ d = 1.1 \times 10^{-5} \times \frac{0.75}{23} \div 0.003 = 0.12\ \rm{mm}\\ \end{array} \) |
1 |
3.5.1.3 |
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07.4 |
5 μΩ cm = 5 × 10–6 cm = 5 × 10–8 Ω m |
1 |
3.1.1 |
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07.5 |
This 1 cm strip will have a much lower resistance, \(R \propto p\) But the strip is 75 cm long so will not affect overall resistance |
owtte |
1 |
3.1.3 |
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08.1 |
LDR and resistor drawn in series, correct symbols Some indication that VOUT is across the fixed resistor |
1 |
3.5.1.5 |
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08.2 |
The resistance of the LDR decreases so current increases The p.d. across the fixed resistor increases since current increases, V = IR/has a greater share of the total resistance so p.d. increases |
This first mark maybe given even if VOUT wrongly labelled in 08.1 |
1 |
3.5.1.5 |
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08.3 |
Examples of calculation: \(\begin{array}{l} 1\ \rm{M}\Omega:\\ \rm{Dark}\ V_{\rm{out}} = \frac{1 \times 10^6}{1 \times 10^6 + 1 \times 10^6} \times 6 = 3\ \rm{V}\\ \rm{Light}\ V_{\rm{out}} = \frac{1 \times 10^6}{1 \times 10^6 + 5400} \times 6 = 6\ \rm{V}\\ 10\ \rm{k}\ \Omega:\\ \rm{Dark}\ V_{\rm{out}} = \frac{1 \times 10^4}{1 \times 10^4 + 1 \times 10^6} \times 6 = 0.06\ \rm{V}\\ \rm{Light}\ V_{\rm{out}} = \frac{1 \times 10^4}{1 \times 10^4 + 5400} \times 6 = 4\ \rm{V}\\ 1\ \rm{k}\ \Omega:\\ \rm{Dark}\ V_{\rm{out}} = \frac{1 \times 10^3}{1 \times 10^3 + 1 \times 10^6} \times 6 = 0.006\ V\\ \rm{Light}\ V_{\rm{out}} = \frac{1 \times 10^3}{1 \times 10^3 + 5400} \times 6 = 0.9\ V\\ 10\ \rm{k}\ \Omega\textrm{ has the greatest range}\\ \end{array} \) |
3 marks max for examples of calculations 1 mark for correct deduction with explanation |
max 4 |
3.5.1.5 |
Skills box answers
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Question |
Answer |
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1 |
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2 |
Circuit is connected as briefly as possible to prevent the cell heating up, which would alter the resistance. |
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3 |
e.m.f = 3.15 V, resistance = 15 Ω (The gradient is in \(\frac{\rm{V}}{\rm{mA}}\) so need to multiply by 103 to obtain Ω.) |
