Oxford Revise AQA A Level Physics | Chapter P9 answers
P9: Current electricity
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Question |
Answers |
Extra information |
Mark |
AO |
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01.1 |
\(R = \frac{V}{I}\)
7.00 to be seen either on the table or by the question |
Must be written to 3 s.f. |
1 |
3.5.1.1 |
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01.2 |
±0.01 A |
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3.1.2 |
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01.3 |
Point plotted to within \(\frac{1}{2}\) a small square AND suitable line of best fit drawn |
Both needed for mark |
1 |
MS 3.2 |
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01.4 |
Systematic error Resistance of connecting wires or Error in measuring length introduced by crocodile clips |
Allow any sensible source of systematic error |
1 |
3.1.2 |
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01.5 |
Large triangle seen or suitable data points from line of best fit Gradient = 12 ± 0.5 |
MUST NOT be data from table |
1 |
MS 3.4 |
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01.6 |
Cross-sectional area of wire = π (0.11 × 10–3)2 = 3.8 × 10–8 m2 \(\textrm{Use of }R = \frac{\rho l}{A} \textrm{ to give gradient } = \frac{\rho}{A}\)ρ = 12 × 3.8 × 10–8 = 4.6 × 10–7 (Ω m) |
Ignore errors in powers of 10 for this mark possible error carried forward from gradient |
1 |
3.5.1.3 |
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01.7 |
\(\% \textrm{ difference } = \frac{4.9 \times 10^{-7} – 4.6 \times 10^{-7}}{4.9 \times 10^{-7}}\)
% difference = 6% (or 7% if you use unrounded numbers) This data is accurate as below 10% difference or not accurate as above 5% |
Allow justification for any sensible comment Allow any valid comment |
1 |
3.1.2 |
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02.1 |
\( \begin{array}{l} I = \frac{\Delta Q}{\Delta t}\\ \Delta t = \frac{\Delta Q}{I} = \frac{15\,{\rm{C}}}{30\,000\,{\rm{A}}} = 5 \times 10^{-4}\,{\rm{s}} \end{array} \) |
1 |
3.5.1.1 |
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02.2 |
\( \begin{array}{l} \textrm{number of electrons } = \frac{\Delta Q}{\textrm{charge on 1 electron}}\\ \textrm{number of electrons } = \frac{15\,{\rm{C}}}{1.6 \times {10}^{-19}\,{\rm{C}}} = 9.4 \times 10^{19} \end{array} \) |
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3.5.1.1 |
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02.3 |
W = VQ = 40 × 106 × 15 C = 6.0 × 108 J |
1 |
3.5.1.1 |
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02.4 |
Use of E = mcΔθ or E = mL E = (0.58 × 830 × 1800) + (0.58 × 156 000) = 9.6 × 105 J Yes, as 6 × 108 J >> 9.6 × 105 J |
valid comment consistent with their calculation |
1 |
3.6.2.1 |
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03.1 |
X is a resistor The resistance is constant so the voltage and current are directly proportional Y is a filament lamp The resistance increases with increasing voltage/current/as temperature increases resistance increases |
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3.5.1.2 |
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03.2 |
\( \begin{array}{l} R = \frac{V}{I}\\ R = \frac{5.0}{0.3} = 16.7\ \Omega \end{array}\) |
Must not draw a tangent here |
1 |
3.5.1.1 |
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03.3 |
p.d. across Y = 2.5 V (read from graph) p.d. across X = 5.0 V (read from graph) e.m.f = 7.5 V |
Can also solve by determining resistance of each component with that current and calculating V by multiplying resistance by current |
1 |
3.5.1.2 |
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03.4 |
Current in Y = 0.30 A (read from graph) Current in X = 0.20 A (read from graph) Total current = 0.50 A |
1 |
3.5.1.2 |
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04.1 |
Area of 1 strand of cable = πr2 = π (1.665 × 10–3)2 \(\begin{array}{l} \textrm{For 1 strand } R = \frac{\rho l}{A} = \frac{{2.82 \times {10}^{-8} \times 1000}}{{\pi {{\left( {1.665 \times {10}^{-3}} \right)}^2}}}\\ R = 3.2 \Omega\\ \textrm{Therefore, for cable}\\ \frac{1}{R_T} = 27 \times \left( {\frac{1}{R}} \right)\\ R_T = 0.12\ \Omega\\ \end{array} \) |
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3.5.1.3 |
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04.2 |
\( \begin{array}{l} R \textrm{ of } 1 \textrm{ m } = \frac{0.12}{1000} = 1.2 \times 10^{-4}\ \Omega\ \textrm{or use of } P = I^2 R\\ I^2 = \frac{P}{R} = 500\ \textrm{A}\\ \end{array} \) |
Allow e.c.f from answer to 04.1 |
1 |
3.5.1.1 |
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04.3 |
A superconductor has zero resistivity at or below the critical temperature A high temperature superconductor has a higher critical temperature |
1 |
3.5.1.3 |
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04.4 |
Allow any sensible suggestion here for one mark:
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3.5.1.3 |
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05.1 |
(As temperature increases) resistance of thermistor decreases p.d. across thermistor decreases or p.d. across fixed resistor R increases Vout increases, (which means it will switch something on when the temperature increases) |
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3.5.1.3 |
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05.2 |
This is because the number of charge carriers increases in the thermistor |
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3.5.1.3 |
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05.3 |
60°C: 280 Ω 100°C: 190 Ω (allow ±5 Ω) |
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3.5.1.3 |
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05.4 |
Variable resistor needed Range: \(\begin{array}{l} \frac{R}{5} = \frac{280}{4},\ R = 350 \Omega\\ \frac{R}{5} = \frac{190}{4},\ R = 238 \Omega\\ \end{array} \) |
1 |
3.5.1.5 |
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06.1 |
Axes labelled – resistance on y-axis and length/cm on x-axis R changing and not constant Graph will be an inverse of shape of paper – when area is large, resistance is small and vice versa |
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3.5.1.3 |
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06.2 |
Resistivity is constant for a material Resistance depends on the length/cross-sectional area/resistivity of the sample |
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3.5.1.3 |
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06.3 |
Measurements of p.d. and current can be used to determine resistance or \(R = \frac{V}{I}\) |
1 |
3.5.1.1 |
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06.4 |
Any sensible suggestions: Wall would have higher resistance than surrounding soil so would show up/well would have concentration of water lower resistance/changes in water content would show up/broken crockery may change resistivity of soil |
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3.5.1.3 |
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07.1 |
Diode/LED The component only conducts once you are above the threshold voltage/a certain voltage/2.6 V |
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3.5.1.2 |
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07.2 |
Circuit diagram using either a potential divider arrangement or a variable resistor Voltmeter in parallel and ammeter in series with component Correct diode symbol used with correct orientation |
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ATg |
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07.3 |
Infinite/allow very large |
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3.5.1.2 |
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07.4 |
\( \begin{array}{l} \textrm{Use of } R = \frac{V}{I}\\ R = \frac{V}{I}\\ R = \frac{4}{0.020\ \rm{A}} = 200\ \Omega\\ \end{array} \) |
Ignore powers of 10 for this mark |
1 |
3.5.1.1 |
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08.1 |
\( \begin{array}{l} E_K = \frac{1}{2}\ mv^{2}: 20\ \textrm{keV} = 20\ 000 \times 1.6 \times 10^{-19}\ \textrm{C} = 3.2 \times 10^{-15}\ \textrm{J}\\ \frac{1}{2}\ mv^{2} = 3.2 \times 10^{-15}\\ v^2 = \frac{{2 \times 3.2 \times {10}^{-15}}}{9.11 \times {10}^{-31}}\\ v = 8.4 \times 10^{7}\ \rm{m\ s}^{-1}\\ \end{array} \) |
Either statement for this mark |
1 |
3.4.1.8 |
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08.2 |
\( \begin{array}{l} v = \frac{d}{t}\textrm{ and } d = 4.22 \times 10^{8}\ \rm{m}\\ t = \frac{4.22 \times 10^8}{8.4 \times 10^7} = 5.0\ \rm{s}\\ \end{array} \) |
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3.4.1.3 |
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08.3 |
\( \begin{array}{l} Q = It\\ ne = 3 \times 10^{6}\ \textrm{A} \times 1\\ n = \frac{3 \times 10^6}{1.6 \times 10^{-19}}\\ n = 1.9 \times 10^{25}\\ \end{array} \) |
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3.5.1.1 |
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08.4 |
From Jupiter to Io |
1 |
AO1 |
Skills box answers
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Question |
Answer |
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1(a) |
0.34 mm |
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1(b) |
0.37 mm |
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2(a) |
The reading of 0.351 mm is anomalous, and may have been written incorrectly. Ideally this measurement should be retaken. However, to calculate the mean, this value is discarded. \({\rm{The\ mean\ is}}\ \frac{{\left( {0.314 + 0.315 + 0.316 + 0.315} \right)}}{4} = 0.315\ \rm{mm}.\) |
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2(b) |
\(\rm{area}\ = \pi \frac{{{{\left( {0.315 \times {{10}^{-3}}} \right)}^2}}}{4} = 7.79 \times {10^{-8}}\ \rm{m}^2.\) |
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3 |
Turn the barrel of the micrometer until the jaws are closed. At this point the scale should be on zero. If not, then there is a zero error. |