5.2 Electricity - Current-Voltage characteristics – Mark schemes

Q1.A

[1]

Q2.(a) (i)

suitable variable input (variable power supply orvariable resistor) (1)

protective resistor and diode forward biased (1)

correct current and pd measuring devices (1)3

(ii) the mark scheme for this part of the question includes anoverall assessment for the Quality of Written Communication

QWC descriptor mark range

good-excellent

Uses accurately appropriate grammar, spelling, punctuation and legibility.Uses the most appropriate form and style of writing to give an explanation or to present an argument in a well structured piece of extended writing.[May include bullet points and/or formulae or equations].Answer refers to at least 5 of the relevant points listed below.

5-6

modest-adequate

Only a few errors.Some structure to answer, style acceptable, arguments or explanations partially supported by evidence or examples.Answer refers to at least 3 or the relevant points listed below.

3-4

poor-limited

Several significant errors.Answer lacking structure, arguments not supported by

1-2

evidence and contains limited information.Answer refers to no more than 2 of the relevant points.

incorrect,inappropriate

or noresponse

No answer at all or answer refers to unrelated, incorrect or inappropriate physics. 0

The explanation expected in a competent answer shouldinclude a coherent selection of the following physics ideas.

connect circuit up (1)

measure current (I) and pd/voltage (V) (1)

vary resistance/voltage (1)

obtain a range of results (1)

reverse connections to power supply (and repeat) (1)

plot a graph (of pd against current) (1)

mention of significance of 0.6V or disconnect between readingsor change range on meters when doing reverse bias (1)

(b) (i) (use of I = V/R)

I = 12/8 (1)= 1.5A (1)

(ii) I = (12 – 0.65 (1))/4 = 2.8 A (1) sig figs (1)5

[14]

Q3.(a) (i) e.g. filament lamp (1)

(ii) not a straight line [or does not pass through origin, only if applied to example e.g. photodiode] (1)

(2)

(b) (i)

(ii) reverse: high resistance [or constant resistance] (1) forward: low resistance (1)

(max 4)[6]

Q4.(a) resistance / Ω 0.98 1.20 1.50 1.76 2.03 3.00 (1) (1)

[deduct one mark for each incorrect value]2

(b) (i) sensible scales chosen (1)points plotted correctly [deduct one mark for each mistake] (1) (1)line of best fit (1)

(ii) 0.90 Ω (1)

(iii) 0.22 Ω (1)0.38 Ω (1)

(iv) 1.12 W (1)6.0 W (1)

max 8

(c) resistance increases with increasing temperature (1)increase in heat dissipation for 1.0 A to 2.0 A is greater than for 0 to 1.0 A (1)and so a greater corresponding rise in temperature (1)

max 2[12]

Q5.(a)

shape in one quadrant (1)symmetrical (1)(1.0, 1.2) (1)(0, 0) (1)slope at (0, 0) (1)

(max 4)

(b) (i) V is not directly proportional to I [or resistance is constant] (1)

(ii) e.g. semiconductor diode (1)(2)

[6]

Q6.(a)

correct curve in positive quadrant (1)correct curve in negative quadrant (1)passing through origin (1)

(3)

(b) the current heats the filament (1)(temperature rises) resistance increases (1)pd. and current do not increase proportionally (1)some reference to mirror image in negative quadrant (1)

The Quality of Written Communication marks were awarded primarily for the quality of answers to this part

(max 3)

(c) diagram to show:

battery, variable resistance (or variable supply) and filament (1) current sensor in series circuit (1) voltage sensor across filament (1) the two sensor boxes connected to datalogger (1)method: variable resistor or variable supply altered [or choose recording interval] (1) thus changing both V and I (1)

The Quality of Written Communication marks were awarded primarily for the quality of answers to this part

(max 5)[11]

Q7.(a)

forward bias: zero current rising gradually (1) sharp increase at ≈ 0.7 V (1)

reverse bias: zero or slightly less than zero current (1)

sharp negative increase at breakdown (1) breakdown value >50 V indicated (1)

4

(b) forward bias: high resistance (initially gives small current) (1) at ≈ 0.7 V, resistance decreases rapidly (current increases) (1)

reverse bias: high resistance (gives ≈ zero or slightly negative current) (1) at breakdown, resistance ≈ zero (and very large current) (1)

3

The Quality of Written Communication marks are awarded for the quality of answers to this question.

[7]

Q8.(a) circuit diagram to show:

ammeter in series, voltmeter in parallel (1)variable source (e.g. battery + rheostat or potential divider) (1)

2

(b) (i) RX = = 56 Ω (1)

(ii) RX = (e.g.) = 23 (Ω) (1)

RX depends on current (or voltage) non-ohmic3

(c) (i) col C col D0.15 2.530.20 2.830.25 3.090.30 3.370.35 3.660.40 3.94

four pairs of values correct (1)all six pairs correct and col D to no more than 4 s.f (1)

(ii) axes labelled (1)suitable scales chosen (1)at least five points plotted correctly (1)acceptable straight line (1)

(iii) k = gradient (1)

gradient = = 5.7 (V–1) (1)

intercept on y-axis =ln A (1)

(intercept = 1.68 gives) A = e1.68 = 5.4 (mA) (1)

unit for k or A correct (1)

(iv) the points define a straight line (1)valid over given range (1)

max 10[15]

Q9.(a) Ohm’s law obeyed (or straight line graph) initially (1)

at a given voltage) current heats filament (to certain temperature) (1) resistance constant at that temperature (1) increase in voltage gives increase in current (1) temperature of filament increases and resistance increases (1)

rate of increase of current less than if resistance was constant (1) negative voltage and current produces same effect (1)

5

(b) P = I2R (1) 20 = (90 × 10 -3 ) 2 R and R = 2.5 × 10 3 Ω (1) (2470 Ω)

2[7]

Q10.(a)

straight line in both quadrants,through origin for A and B (1)greater gradient for B (1)

2

(b)

characteristic to show:positive current increasing slowly and then rapidly (1)at ≈ 0.6V (1)negative current either zero or just < zero (1)

3

(c) as voltage increases, current increases (1)current heats filament (1)therefore resistance increases (1)correct argument to explain curvature(1)mirror image in negative quadrant (1)

max 4[9]

Q11.(a) a non-ohmic conductor does not have a constant resistance (1)

1

(b) (i) curve of decreasing gradient with increasing V (1)

attempt to make graph symmetric in two opposite quadrants (1)2

(ii) resistance increases as pd increases/current increases (1)1

(c) (i) (use of P = V2/R)

36 = 144/R (1)

R = 4.0 (Ω) (1)2

(ii) reference to temperature change (1)

(resulting in) a lower resistance (1)

(hence) power rating would be greater (1)3

[9]

Q12.(a) ratio of voltage (across component) to current (through

component) or R = V/I with terms defined and R as subject

B11

(b) (i) correct curve

B11

(ii) resistance increases / increase in resistivity

B1

energy transfer increases lattice vibration/ temperature riseincreases lattice vibration / electron collisions increaseslattice vibration

B1

more frequent collisions/ ions now a larger target for electrons

B13

[5]

Q13.

(a)

first mark for linear at origin and decreasing gradient in either quadrant (linear region can be very small)second mark for symmetry plus no dip at end or extended horizontal section at endstraight line scores zero

2

(b) (i) resistance (of filament lamp) increases1

(ii) filament lamp is a non-ohmic conductor as current is not (directly) proportional to voltage / resistance is not constant

proportionality can be shown using graph1

(c) eithercircuit / total resistance increases(hence) current decreases and pd / voltage across R decreasesORresistance of PQ combination increases(hence) greater share of pd / voltage across lamp P

implication that current is different in different parts of series circuits scores 0implication that new total current is greater scores zerovoltage flowing loses second mark

2

(d) (i) (use of energy = VIt)(energy converted by X = 60 × 120 × 3600 =) 2.59 × 107 J (energy converted by Y = 11 × 120 × 3600 =) 4.75 × 106 J

Accept answers to 1 sig. fig.2

(ii) in lamps energy is wasted as heat / thermal energyspecific lamp considered e.g. in lamp, X / filament lamp more energy is wasted OR in X / filament lamp less energy is converted to light / luminosity

2[10]

Q14.(a) (i) Use of P = VI with pair of valid coordinates from graph

C1

0.52 (W)Allow 1sf if within 0.49 to 0.52

A12

(ii) Correct general shape

M1

Linear rise between 0.0 ‒ 0.5 V and falls to zero at 0.71 V

A12

(iii) Use of efficiency =

C1

Use of I =

C1

Their (i) / 67.5 (m2) (7.7 × 10−3 if correct)

A13

(b) (i) 0.7 J of work done (by cell) per 1 C of charge (when moved round circuit)OR(Terminal) pd across (solar) cell with no load / current is 0.7 V

Not “per unit charge”

B11

(ii) 20 cells in series (to produce 14 V)

B1

Series arrangement has internal resistance of 15.6 Ω

B1

Cells in parallel (needed to reduce total internal resistance of array)

B1

80 cells / 4 parallel sets of 20 cells in series

B14

(c) The marking scheme for this question includes an overall assessment for the quality of written communication (QWC). There are no discrete marks for the assessment of QWC but the candidate’s QWC in this answer will be one of the criteria used to assign a level and award the marks for this question.

Descriptor ߜ an answer will be expected to meet most of the criteria in the level descriptor.

Level 3 ‒ good-claims supported by an appropriate range of evidence;-good use of information or ideas about physics, going beyond those given in the question;-argument is well structured with minimal repetition or irrelevant points;-accurate and clear expression of ideas with only minor errors of grammar, punctuation and spelling.

Level 2 ‒ modest-claims partly supported by evidence;-good use of information or ideas about physics given in the question but limited beyond this;-the argument shows some attempt at structure;-the ideas are expressed with reasonable clarity but with a few errors of grammar, punctuation and spelling.

Level 1 ‒ limited-valid points but not clearly linked to an argument structure;-limited use of information about physics;-unstructured;-errors in spelling, punctuation and grammar or lack of fluency.

Level 0-incorrect, inappropriate or no response.

Some points:Use on communication satellite:Continuous supply of energy from SunNo need for fuel (for power purposes)Large area of solar cells not needed (but possible)Low massCan be unfolded (after launch)No environmental hazardReliable/no moving partsContinuous operation:Arrays need to track sun (to maximise absorption)Shielding required as can be damaged by meteors or cosmic raysNeed storage system (rechargeable batteries / capacitors) for back up (if in shadow)Limit use of energy-intensive operationsUse on space probe:

Light intensity / energy too low at large distanceIntensity falls as inverse-squareArea of array would be too largeSolar cells will have degenerated too much over this time

B66

[18]

Q15.A

[1]

Q16.(a) emf is the work done / energy transferred by a voltage source / battery /

cell per unit chargeORelectrical energy transferred / converted / delivered / producedper unit chargeORpd across terminals when no current flowing / open circuit

not in batteryaccept word equation OR symbol equation with symbols defined if done then must explain energy / work in equation for first mark

2

(b) (i) by altering the (variable) resistor1

(ii) reference to correct internal resistancee.g. resistance of potato (cell)

terminal pd = emf ߜ pd across internal resistance / lost voltspd / lost volts increases as current increases OR as (variable)resistance decreases greater proportion / share of emf across internal resistance

accept voltage for pd3

(iii) draws best fit straight line and attempts to use gradientuses triangle with base at least 6 cmvalue in range 2600 ‒ 2800 (Ω)

3

stand-alone last mark

(c) total emf is above 1.6 Vbut will not work as current not high enough / less than 20 mA

2[11]

Q17.(a) voltmeter, ammeter and lamp connected correctly to measure

V and I must include a cell

B1

cell and potentiometer correctly connected with othercomponents across the output terminals

B12

(b) 0.24 A

B11

(c)

general shapeLine through origin and correct curvature clear(condone levelling off)

B11

[4]

Q18.(a) V I [allow proportional]

M1

physical condition constant

A12

(b) (i) Line goes through (12, 2) [within one square]

B1

Straight line at origin aimed at (1,0.5) and smooth curve(correct shape) beyond (1,0.5)

B1

Calculation clearly supporting second mark[V=IR, I = 0.5, so V = 1]

B13

(ii) Correct shape for V +ve

M1

Non-zero, positive breakaway from V-axis,V <= 1V; line not > 1V]

A1

Zero current for reverse bias explicit

B13

[8]

Q19.

(a) V = IR seen or used (condone = for this mark only)C1

I = 1.8 A and V = 12 VC1

6.7 ΩA1

(b) current = 0 for negative voltages (allow no mark seen on negative voltage axis if graph drawn for positive)

B1

characteristic curve with steep rise: voltage not exceeding 1V allow higher turn on voltages if candidates specify that they have drawn graph for LED

B1[5]

Q20.(a) Diode or LED

B11

(b) Use of V/I

C1

= 1.03 OR 1.04 OR 1.0 Ωcorrect numerical answer only

A12

(c) rectification/description such as “a.c. to d.c”/demodulation/protection against current surges

B11

[4]

Q21.(a) indication of complete circuit with correct components

must show variable resistor / pot divider / variable power supply circuit to be completethermistor symbol or resistor labelled ‘thermistor’can omit ammeter / voltmeter here

B1voltmeter correctly placed

B1ammeter correctly placed

B1(3)

(b) (i) V = (6 / 1000) × 2000C1

= 12 VA1

(2)

(ii) correct readoffB1

power = I × VM1

= 0.34 × 10–3 * 12 = 4.1 mW [e.c.f. from I]A1

(3)

(c) decrease in resistanceB1

more charge carriers released at high temperatureB1

(2)[10]

Q22.(a) R = ρL/A

C1

area = 3.14 × 10−8 m2 or 12.6 × 10−8 or π(1 × 10−4)2

or π(2 × 10−4)2

A1

1.7 (1.74 or 1.75) m (not 1.8 m)

A13

(b) correct curvature for positive V (decreasing gradient)

B1

consistent graph for negative V through origin

B12

[5]

Q23.(a) (i)

correct diode bias for variable supply, must have some attempt to vary pd

Condone variable resistor (condone missing arrow) don’t allow thermistor symbol

correct symbols and positions for voltmeter, ammeter:voltmeter in parallel with diode onlyammeter in series with diode

Allow mA symbol instead of A symbol for ammeterAllow symbols for diode without line through triangle and / or with a circleDiode symbol must consist of a triangle and a straight line at nose perpendicular to wiring in circuit.

allow voltmeter across ammeter and diode

2

(ii) The candidate’s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear.The candidate’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marksThe information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question.Candidate explains how to obtain sufficient values of I and V. They

mention the need to limit the current through the diode and give an indication of the range and frequency of measurements. They discuss an advantage of using a data logger. Voltage does not exceed 1.0V, diode is forward biased

Intermediate Level (Modest to adequate): 3 or 4 marksThe information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate.Candidate explains how to obtain sufficient values of I and V. Includes mention of diode is forward biased or suitable voltage for switch on mentioned or advantage of data logger

Low Level (Poor to limited): 1 or 2 marksThe information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.vary pd obtain several readings of I and Vor an advantage of using data loggeror forward biasedlow level safety may include switch off / avoid overheating type arguments / don't touch

The explanation expected in a competent answer should include a coherent selection of the following points concerning the physical principles involved and their consequences in this case.means of controlling pd across diodeindication of range and frequency of measurementmention of limiting current to avoid damage to diodea consideration of the advantages of a datalogger e.g. many readings, computer display of resultsuse of potential divider instead of series resistor

All signs of quality that could lift markLower bandvary pd obtain several readings of I and Vor an advantage of using data logger or low level safety and action to minimise riskMiddle bandvary pd and obtain several readings of I and V, at least 6 different values including an advantage of using data logger or mention of forward bias or mention of switch on voltage (0.6V) or safetyTop BandMention of how to vary pd (seen in viable circuit) obtain several readings of I and V, at least 6 different values (range given where maximum value of pd does not exceed 1.0V)mention of limiting current through diode using protective resistor

consider advantage of data loggermention forward biasmust include potentiometer for 6 marksmust have voltage as independent, no current led arguments in Top bandData logger advantages:Not more accurateNot removes human error

6

(iii) reverse connections to the power supply / battery / cell / reverse diode

not switch wires around (need clear link to reversing connections at supply's terminals)

1

(b) (i) divide V by I for a reading from graph or uses R = V ⁄ I for a reading from graph

Treat gradient = as TOrepeat for different values of V and I

Must score 1st mark to achieve 2nd2

(ii) (Resistance) decreases Or resistance starts off very high and then becomes much lower

1[12]

Q24.(a) correct general shape

accurate plotting to within square 2

(b) 12 (V) , 30 (W) 2

(c) R = = 3.2 (Ω) 1

(d) Resistance increases

Temperature increases

More collisions / interaction of electrons with lattice ions Condone ‘atoms’, ‘molecule’.

Do not allow electron−electron collisions.3

(e) Can attain neither maximum nor minimum voltage

Explanation of either maximum OR minimum 2

[10]

Q25.(a) An increase in current / voltage leads to an

increase in temperature (more heat generated) Ignore 'of particles' in first markDo not condone ‘particles’ in second mark

This causes an increase in the movement of the lattice/ions/atoms

And therefore an increase in the rate of collisions with electrons

Allow more frequent collisions

So the resistance increases as shown by V / I changing/V not proportional to I (on the graph)

Allow correct reference to gradient of I / V curve unless the answer suggests that this is the resistance or inverse of resistance.

Max 4

(b) 14.3 (Ω)Allow range 14 to 15but calculated answer must lie between 14 and 15

1

(c) Determination of pd across either filament or resistor from graph

Pd across resistor can be calculated from resistance value in (b)Eg V = 0.18 × 14.3 = 2.6

Determination of pd across the other component, and values added

Use of V = IR to give 3.4 (V)Allow ecf if either value is wrong allow 2 max

Or

Clear attempt to determine total resistance and multiply by 0.18

Condone small rounding error

(Resistance of lamp at 0.18A = 4.4 Ω)

Total resistance = 18.7 Ω ecf from 2,2

3.4 V (ecf from 2.2) Allow for small rounding errors (eg allow range 3.3 to 3.5)

3

(d) Determination of current through either filament or resistor from graph Allow calculation of resistor current using 4/(answer to 2.2)

Determination of current through the other component, and values added

(Current through resistor = 0.28 A

Current through filament = 0.36 A)

R = V/I = 4/ (0.28 + 0.36) = 6.25 (Ω)If either value wrong allow 2 maxCondone small rounding errors.

Or

Calculation of filament resistance or statement of resistor resistance Resistance of filament = 11.1 (Ω)

Calculation of other resistance and use of parallel formula (allow ecf from part b)

Either resistance gets the first mark

6.2 -6.3 (Ω) 3

(e) Calculation of area, ignoring power of ten errors.A = 8.0 × 10-10 m2

Correct resistivity 3.1 × 10-8 Allow ecf for A (for example use of d for r gives 3.2 × 10-11 for A and 1.2 × 10-7 for answer)

Ω m Some working must be shown for award of unit mark.

3[14]

Q26.planning

(a) sensible key factor e.g. p.d. across paper, that, when varied, leads to the determination of resistance: candidate then goes on to estimate

the thickness of the paint layer on strip [only allow direct measurement of resistance if the investigation is of how either width or length of a rectangular strip affects the resistance of the paper] (1)

(b) correct measuring instrument given [allow circuit diagram] (1)

(c) dimensions of paper constant when resistance measured [to see how a certain dimension influences the resistance, width (if length varied)/length (if width varied)] (1)

(d) check that current through paper does not exceed 200 mA (1)

(e) sensible qualitative prediction given: thickness can only be estimated due to uncertainty in resistivity (1)

(f) thickness of layer (assuming uniform coating) in range 10–7 to 10–11 m (1)

[or (e) sensible qualitative prediction given: R ∝ l or R ∝ w–1]

(g) reasonable physics reasoning given in support: similarity with behaviour of a metallic conductor (1)

(h) use of VlI to find R [use of repeated readings to reduce uncertainty in measurement of dimension] (1)

(i) calculating possible range of thickness using limiting values of resistivity / assessing the uncertainty in result [plotting graph of results to check relationship] (1)

(j) any other sensible measure, e.g. maintain steady temperature (1)[max 8]

Q27.C

[1]

Q28.C

[1]