Wallington High School For Girls1 1.C[1] 2.B[1] 3.C[1] 4.C[1] 5.(a)Resistivity is a constant for the material / metalOR resistivity depends on / is a property of the material / metal1Resistance depends on (resistivity and) length / area/dimensionsOR R = l/A with terms defined (do not credit rearrangedequation)1 (b)Correct substitution into the R = l/A formula1R = 0.0085 1[ue applies. Common error is to rearrange eqn and confuse Rand gives answer 3.4 1014 scores zero]Example of calculationR = (1.7 108 m 0.5 m) / 1 106 m2R = 0.0085 [4] 6.(a)(i)Resistance at 20 C = 1250 - 1300 ()1(ii)Converts k [look for 1000 ()]1Use of potential divider formulaOR use of 2300 () to find current1Reading on voltmeter = 2.6 2.7 V(ecf value from (a)(i))1Example of calculationV = (1000 2300 ) 6 VV = 2.6 V (b)(decreasing temp causes) resistance of thermistor to increase1Voltmeter reading decreases1Candidates who think resistance will decrease leading to voltmeterincrease can get 2nd mark.[6] 7.(a)Volt is a Joule coulomb1 or V = J C1 or V = W/Q (not rearranged)1Amp is a Coulomb sec-1 or A = C s1 or I = Q/t (not rearranged)1Show units/symbols cancelling and equating to a watt.This mark can only be given if both the other marks scored. Methodmust be clear, do not allow let t = 1.1 (b)(i)Use of energy = power time1Energy = 2.9 105 J1Example of calculationE = 700 7 60E = 294 000 J (ii)QWC spelling of technical terms must be correct and the answermust be organised in a logical sequenceSee internal resistance / r1Current will be less1Less energy/power is lost in internal resistance OR wastedenergy/power is reduced OR reduced lost volts OR it is more efficient1[8]8.(a)There must be a circuit with a power supply and a labelledwire/identifiable ends of a wire/wavy line/resistor/lamp in order toscore any marksANY TWOAmmeter symbol in series with wire (not in the middle of) (1)Voltmeter symbol in parallel with wire (1)Variable power supply/variable resistor (1)Max 2 (b)Use of P = VI1Rate of work = 4.5 W / J s11 (c)(i)Correct use of I = nqvA with e = 1.6 1019 C1v = 3.0 105 ms11 Example of calculationv = 1.5/ (1.0 1029 1 .6 1019 3.1 106v = 3.02 105 ms1 (ii)Increased lattice/ions/atoms vibrations1(causing) resistance to increase OR increased electron collisions withions/atoms1(This leads to a) reduction in the drift velocity / v1[9] 9.B[1] 10.B[1] 11.C[1]12.C[1] 13.(a)Current (through a conductor) is (directly) proportional to the potentialdifference/voltage (across it)1providing the temperature of conductor remains constant OR externalconditions remain constant.1 (b)Ohmic conductor; fixed resitor horizontal straight line1Filament lamp; graph showing increasing resistance (straight line or curve)1from a non zero resistance start (conditional on 2nd mark)1 (c)Filament lamps work at high temperatures OR as temp of lamp increasesOR as lamp heats up.1Resistance of conductor changes OR the ions vibrate more.1[7] 14.(a)Use of resistors in parallel formula1R = 9.1 1 Example of answer1/R = 1/10 + 1/1001/R = 11/100R = 9.1 (b)Voltmeter is connected in parallel (stated or implied) OR voltmeter drawslittle/no current1Resistors in parallel formula with either RV or large value used11/RV is very small/negligible OR calculated value of 9.1 with comment1[5]15.(a)Use of potential divider formula with 40 in numerator and 120 indenominator1V = 3.0 V1OR ohms law to whole circuit to find current (1)V=IR applied to 40 resistor (1)Example of answerp.d. = 40 9.0 / (40 + 80)p.d. = 3.0 V (b)QOWCWork must be clear and organised in a logical sequenceResistance of parallel combination increases as temperature decreases1Total resistance of circuit increases1e.m.f./p.d. remains constant therefore current decreases.1[5] 16.(a)Use of I = nqvA with e 1.6 1019 C and 8 103 A1v = 2.8 107 m s11 (b)Value for semiconductor is much greater1n for semiconductor (much) less than for conductor1 (c)Its resistance decreases1because (as temperature increases) n increases OR there are more electrons/charge carriers.1[6] 17.(a)(i)e.m.f./total resistance(accept E/(R+r)1 (ii)= D4 * A4 or p.d. = current load resistanceOR(B4 * A4)/(A4+C4) or p.d. = e.m.f. (internal resistance current)ORF4/D4 or p.d.=power/current1 (iii)9.2 (W) R and r in series1(b)R and r in series1Potential divider1r constant so as R increases p.d. also increases1ORV = E - IrE and r constantIdentifies that the term Ir decreasesORE = I(R+r)E and r constantAs R increases I decreases (c)(i)(power) increases and then decreases1Maximum power when load equals 0.8 1 (ii)Similar pattern of power increasing to a max and decreasing1Maximum power when load resistance is 1.6 1[10] 18.C[1] 19.B[1] 20.A[1] 21.Use of V = IR to find total resistance or terminal p.d.1Subtraction of resistance or p.d.s1r = 8.2 (accept 8 )1 OR see E = I(R+r)1Substitution of values into equation1r = 8.2 (accept 8 )1Example of answerTotal R = 1.5 V (17 103 A) = 88.2 r = 88.2 80 = 8.2 [3] 22.Attempt to use I = Q / t1use of e = 1.6 10191I = 2.8 10 6 A [C s1][omit e gives answer 1.73 10 25 scores 1]1 Example of answerI = (2.6 10 26 1.6 1019 C) 15 sI = 2.77 10 6 A[3] 23.(a)Use of V=IR1V = 3.0 V1 (b)pd across 30 resistor = 6.0 V ecf their answer (a)1I2 = 6.0/30 = 0.20 A1 (c)I1 = 0.60 0.20 = 0.40 A1R = 15 full ecf their answer for I2 and their V across30 1[6]24.The answer must be clear and organised in a logical sequenceDifferent currents / current divides in parallel circuit(1)Same potential difference/voltage across each lamp (1)Use of P = V2 / R OR P = VI if identified IA IB (1)Leading to high resistance, smaller power (1)lamp B will be brighter/ lamp A dimmer (1)Each electron loses the same energy (1)There are more electrons/sec in B (1)Hence greater total energy loss /sec in B (1)[5] 25.(a)Effect of stretching wireRefers to R = l/A1Increasing length leads to increase in resistance1Decreasing area leads to increase in resistance [must relatethinner to area]1[last two points may be combined to give single statement,can score both marks] (b)Resistance calculationUse of R = l/A1 81R = 0.22 ()[Omitting 8 gives R = 0.028 scores 1]1Example of answerR = (9.9 108 m) (8 0.025 m) 0.9 107 m 2R = 0.22 (c)Relationship and increase in R(i)Attempts to substitute for A = V/l in R = l/A1R = l2/ V1(ii)Any attempt to relate original resistance of gauge to 2.52(possibly 8, cm or m)1Relates this to resistance associated with increase in length1Change in resistance = 1.76 103 1ORUses V=lA to find new area1Uses this A with new length to find new R1Change in resistance = 1.76 103 1[if candidate assumes A constant and finds new R and R =0.001 , score 1 mark]Example of answerNew R = R = 1.76 103 (d)Zigzag patternEach section of wire increases in length/gives a longertotal length/long wire in small space1Small change in length of gauge leads to larger change inresistance1[13] 26.C[1] 27.A[1] 28.C[1] 29.C[1]30.p.d. is electrical energy(/coulomb) transferred between twopoints/electrical energy transformed/converted to other forms (1)e.m.f is the energy(/coulomb) supplied to a circuit/given to thecharge/energy output of the cell (1)(full credit if wording implies emf as electrical energy source and pdas electrical energy sink)If neither mark scored but reference to energy/charge is made scores1 mark only[2] 31.(a)A coulomb is an Amp sec or As(1)Do not credit current time1 (b)I1 = 10 mA(1)I2 = 5 mA(1)I3 = 30 mA(1)3[4] 32.(a)n; number of charge carriers per unit volume OR number of charge carriersm3OR charge carrier density (1)v; drift velocity (of charge carriers) OR average velocity OR driftspeed (1)(accept free electrons or charge carriers throughout)2 (b)Units of I and qA and A s OR C s1 and C (1)Units of nm3 (1)Units of v and Am s1 and m 2 (1)3[5] 33.(a)Current in A is equal to current in B (1)p.d across A is less than p.d. across B (1)Resistance of A is less than the resistance of lamp B (1)3(b)Resistors in parallel have same p.d (1)Identifies P = V2 /RORP =VI and IA > IB (1)Uses this equation to state PA.> PB.OR bulb A brighter than bulb B. Consequent on 2nd marking point (1)3[6] 34.(a)5% of 60 W (is 3 W)(1)Use of I = P / 4r2(1)ORUses I = P / 4r2 with 60 WFinds 5% of this answerIntensity = 0.038 W m2(1)(accept 0.04 W m2)3 (b)QWC Work must be clear and organised in a logical manner usingtechnical wording where appropriateAny threeFluorescent lamp much more efficient OR filament lamp is lessefficient(1)Sensible attempt to process the values given (1)Indicates that less than 25% of national power used for lighting (1)Reduction in wasted energy as thermal energy (1)Reduction in CO2 emission or preserves fossil fuel resources (1)(Just saying filament lamp is inefficient does not score 1st mark)Max 3[6] 35.(a)Plot of graphCheck points, 4 correct 2 marks, 3 correct 1 markLine of best fit to include 0,0.6 and 0.52,0 (1)3 (b)(i)Use of V = 0.43 V in P = VI (1) ecf values for incorrect best fit lineP = 0.17 W (1)2Example of calculationP = 0.4 A 0.43 VP = 0.172 W (ii)Value of e.m.f. is when the current is zero (1)No lost volts OR no energy loss (1)ORE = V + IrI = 0, E = V2(iii)Identifies current in circuit (1)ecf values for incorrect best fit lineFinds lost volts (1)r = 0.24 (1)3Example of calculationr = (0.52 V 0.40 V) 0.50 Ar = 0.24 (c)Graph of similar shape as in (a) but initially above the first graph (1)ecf values for incorrect best fit lineFinishing at 0.52 V, 0.00 A (1)2[12] 36.D[1] 37.C[1] 38.B[1] 39.B[1] 40.(a)Use of sensorEvent happens very quickly OR cannot take readings fast enough (1)Sampling rate: 50+ samples per second (1)2 (b)Initially the temperature is low so current is highResistance of filament increases as temperature increasesCurrent falls to steady value when temperature is constantMaximum heating is when lamp is switched on / when current is highestFilament breaks due to melting caused by temperature riseMax 4[6]41.The answer must be clear, use an appropriate style and be organised in alogical sequence. (QWC)Reference to I = nqvA (1)For the lampIncreased atomic vibrations reduce the movement of electrons (1)Resistance of lamp increases with temperature (1)For the thermistorIncreased atomic vibrations again reduce movement of electrons (1)But increase in temperature leads to a large increase in n (1)Overall the resistance of the thermistor decreased with increase intemperature. (1)Max 5[5] 42.(a)Voltmeter is across resistor should be across cell (1)1 (b)(i)Plot of graphCheck any three points (award mark if these are correct) (3)Line of best fit3 (ii)e.m.f. = [1.36 1.44 V] (1)1 (iii)Attempt to find gradient (1)Answer [0.38 0.42 W] (1)2 (c)Intercept would twice value above (1) (accept numerical value 2 value(b)(ii))Gradient would be twice value above (1) (accept numerical value 2 value(b)(iii))2[9] 43.(a)Diode or LED (1)1 (b)(i)Use of R = V / I current between 75 and 90 ignoring powers of 10 (1)answer 6.7 8.0 (1)Example of answerR = 0.60 V (85 103) AR = 7.06 2 (ii)Infinite OR very high OR 1(c)ANY ONERectification / AC to DC / DC supply [not DC appliances]Preventing earth leakageStabilising power outputTo protect componentsA named use of LED if linked to LED as component in (a)(egcalculator display / torch)A voltage controlled switch(Allow current in only one direction)1[5] 44.(a)Resistivity definitionResistivity = resistance (1) cross sectional area / length (1)= RA/l with symbols defined scores 2/2equation as above without symbols defined scores equation given as R = l/A with symbols defined scores 1/2(1st mark is for linking resistivity to resistance with some other terms)2 (b)(i)Resistance calculationConverts kW to W (1)Use of P = V2/R OR P = VI and V = IR (1)Resistance = 53 (1)Example of answerR = (230 V)2 1000 WR = 53 3 (ii)Length calculationRecall R = l/A (1)Correct substitution of values (1)Length = 6.3 m (accept 6.2 m) (1)ecf value of RExample of answerl = (52.9 1.3 107 m2) (1.1 106 m)l = 6.3 m3(iii)Proportion methodIdentifies a smaller diameter is needed (1)Diameter = 0.29 mm (1)ORCalculation methodUse of formula with l = half their value in (b)(ii) (1)Diameter = 0.29 mm (1)(Ecf a wrong formula from part ii for full credit)Example of answerdnew = 0.41 mm 2dnew = 0.29 mm2[10] 45.(a)Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)[work done/coulomb 1/2, energy given to a charge 1/2, energygiven to a charge of a coulomb 2/2]ORORE = W/Q (1)E = P/ISymbols defined (1)Symbolsdefined(E = 1 J/C scores 1)(E = 1 W/A scores 1)((Terminal) potential difference when no current is drawn 1/2)2 (b)(i)Internal resistance calculationAttempt to find current (1)Pd across r = 0.2 V (1)r = 0.36 () (1)[You must follow through the working, I have seen incorrectmethods getting 0.36 ]Example of answerI = 2.8 V 5.0 r = (3.0 2.8) V 0.56 A = 0.36 3 (ii)Combined resistanceUse of parallel resistor formula (1)Resistance = 3.3 [accept 3 1/3 but not 10/3] (1)2(iii)Voltmeter reading(ecf bii)Current calculation using 3 V with either 3.3 or 3.7 (1)Total resistance = 3.7 [accept 3.66 to 3.73 ]OR use of V = E Ir (1)Voltmeter reading = 2.7 V (1)ORPotential divider method, ratio of resistors with 3.7 on bottom (1)Multiplied by 3.0 V (1)2.7 V (1)Example of answerRtotal = 3.7 I = 3 V 3.7 = 0.81 AVvoltmeter = 3.3 0.81 A = 2.7 V3 (c)Ideal voltmeterIdeal voltmeter has infinite resistance OR extremely high resistanceOR highest possible R OR much larger resistance than that ofcomponent it is connected across OR quotes value > 1 M (1)Current through voltmeter is zero (negligible) OR doesnt reduce theresistance of the circuit OR doesnt reduce the p.d. it is meant tobe measuring. (1)2[12] 46.(a)Circuit diagramPotentiometer correctly connected i.e potential divider circuit (1)Ammeter in series and voltmeter in parallel with bulb (1)(light bulb in series with resistance can score second mark only)2 (b)(i)Graph+I, +V quadrant; curve through origin with decreasing gradient (1)[do not give this mark if curve becomes flat and then starts goingdown i.e. it has a hook]I, V quadrant reasonably accurate rotation of +I,+V quadrant (1)2(ii)Shape of graphAs current/voltage increases, temperature of the lamp increases /lamp heats up (1)Leading to increase in resistance of lamp (1)Rate of increase in current decreases OR equal increases in Vlead to smaller increases in I (1)Qowc (1)Ecf if a straight line graph is drawn max 3R constant (1)V I (1)Qowc (1)4[8] 47.(a)Calculation of adaptors inputRecall of: power = IV (1)Correct answer [0.01 A] (1)Example of calculation:power = IVI = P/V = 25 W / 230 V= 0.01 A2 (b)(i)Explain why VA is a unit of powerPower = voltage current so unit = volt amp (1)1 (ii)Calculation of efficiency of adaptorUse of efficiency equation (1)Correct answer [24%] (1)Example of calculation:efficiency = (0.6 VA / 2.5 W) 100%= 24 % [0.24]2 (iii)Reason for efficiency less than 100%Resistance (accept explanations beyond spec, e.g. eddy currents) (1)hence heat loss to surroundings (1)2(c)(i)Calculation of chargeRecall of: Q = It (1)Correct answer [4000 C] (1)Example of calculation:Q = It= 0.2 A 6 h= 0.2 A (6 60 60) s= 4000 C (4320 C)2 (ii)Calculation of work doneRecall of: W = QV OR Recall of W = Pt (1)Correct substitution (1)Correct answer [13 000 J] (1)Example of calculation:W = QVW = 4320 C 3 V [ecf]= 13 000 J (12 960 J)ORW = PtW = 0.6 W 6 hW = 0.6 W (6 60 60) s= 13 000 J3[12] 48.(a)Blue light:Wavelength / frequency / (photon) energy1 (b)(i)Frequency:Conversion of either value of eV to JoulesUse of f = E / hCorrect frequency range [4.8 1014 8.2 1014 Hz or range =3.4 1014 Hz][no penalty for rounding errors]eg.2 eV = 2 1.6x 1019 = 3.2 1019 J= 6.63 1034 ff = 4.8 1014 Hz3.4 eV = 3.4 1.6 1019 = 5.4 1019 Jf = 8.2 1014 Hz3(ii)Diagrams:Downward arrow from top to bottom levelOn larger energy gap diagram2 (c)(i)Resistivity drop:Less heating / less energy lost / greater efficiency / lowervoltage needed / less power lost1 (ii)Resistance:Recall of R = L/AUse of R = L/ACorrect answer [80()] [allow 8084 () for rounding errors]Eg.R = (2 102 5.0 103) / (3.0 103 4.0 104)= 83 3[10] 49.(a)n is (number of) charge carriers per unit volume ornumber density or (number of) charge carriers m3 orcharge carrier density(1)[allow electrons]v is drift speed or average velocity or drift velocity(of the charge carriers) (1)[just speed or velocity scores zero]2 (b)/ A and Q A s or / Cs1 and Q C (1)n m3 (1)A m2 and v m s1 (1)[If no equation written assume order is that of equation]3 (c)(i)n 1 and Q Need all three1(i)Ratio vA/ vB less than 1 following sensible calculation (1)Ratio = // 0.25 // 1:4 (1)(ratio 4:1 scores 1)[4vA:1vB scores 1]2[8]50.(a)Use of P = IV (1)Current in lamp A 2 A (1)[0.5 A scores zero unless 24 = I 12 seen for 1st mark]2Example of answerI = P V = 24 W 12VI = 2A (b)(i)Voltmeter reading = 12 V (1)1 (ii)2 p.d. across R2 = 6 V or their (b)(i) minus 6V (1)Use of R = V/I (1) conditional on first markR2Answer to this part must be consistent withvoltmeter reading and if voltmeter reading is wrongthis part has a max 2. If (b)(i) = 15 V then need to seeIf (b)(i) = 6V or less they are going to score zero for this section.3 (iii)current through R1 = 5 A (1) ecf answers from (a)1Example of answerCurrent through R1 = 2 A + 3 A = 5 A (iv)p.d. across R1 = 3 V (1) ecf (15V minus their (b)(i))1Example of answerp.d. across R1 = 15 V 12 V = 3 V (v)R11Example of answerR1 = 3 V 5A = 0.6[accept fraction 3/5][9] 51.(a)(i)EI (1)1(ii)I2R (1)1(iii)I2r (1)1 (b)EI = I2R + I2r or E = IR + Irecf Must use values (a)(i)-(iii)1(c)I for circuit given by Imax = E / r or substitution of5000V into the equation (1)(for safety) need I to be as small as possible (1)3[7] 52.(a)(i)Calculate maximum currentRecall of P = IV (1)Correct answer [0.49 A] (1)Example of calculation:P = IVI = 5.9 W / 12.0 V= 0.49 A2 (ii)Show that resistance is about 24 Recall of V = IR (1)Correct answer to 3 s.f. [24.5 ] [no u.e.] (1)Example of calculation:R = 12 V / 0.49 A= 24.5 2 (b)(i)Calculate currentUse of correct circuit resistance (1)Correct answer [0.45 A] (1)Example of calculation:I =V / R= 12 V (24.5 + 2 )= 0.45 A2 (ii)Calculate powerRecall of P = IV and V = IR (accept P = I2R) (1)or P = Correct answer [5.0 W] (1)Example of calculation:P = I2R= (0.45 A)2 24.5 = 5.0 W2(c)Increase in power available to pumpe.g. lower resistance in wire thicker wire, panel nearer to motor (1)(accept relevant answers relating to panels, e.g. more panels)1[9] 53.(i)J C1Potential difference (1)(ii)Product of two quantitiesPotential difference (1)(iii)Rate of changecurrent (1)(iv)Base quantitycurrent (1)(for any part if two answers are given score is zero)4[4] 54.(a)(As temperature of thermistor increases) its resistancedecreases [Do not credit the converse] (1)any TWO(slight) decrease in v (symbol, velocity or drift velocity)Large increase in n increases [accept electrons/charge carriers for n]A, Q and (pd) remain constant (1)(1)[ignore any reference to v staying constant]3(n constant, cant score mark for 3,4) (b)(i)ammeter reading decreases (1)voltmeter reading unaltered (1)(ii)ammeter is used to indicate temperature (1)(iii)Assumption: ammeter; ideal/ has zero/negligible resistance (1)(Reference to meters is zero mark)4[7]55.(a)Tungsten filamentQowc (1)I is not (directly) proportional to VTemperature of filament increases/ filament heats up/gets hotter as current/pd increases[accept bulb or lamp but not wire]Links temperature increase to resistance increasestungsten filament does not obey Ohms law/not anOhmic conductor or resistor. (1)(1)(1)Any THREE4 (b)(i)Reading current from graph 1.5 A (1)answer 5.3 (1)(misread current 0/2)Example of answerV = IRR = 8.0 1.5 = 5.3 2 (ii)Addition of two currents (1)OR use of R = V/I and resistors in parallel formula1.5 + 1.2 = 2.7 A (1)ecf candidates current from above[If you see 2.7 A give 2marks]2[8] 56.(a)(i)Use of P = V2 / R OR P = IV and V = IR (1)Total R = 4.5 (1)2Example of answerR = V2 P = 12 V 12 V 32 WR = 4.5 (ii)Use of 1/R = 1/R1 + 1/ R2 OR R = 1/5R (1)[OR find total current, divide that by 5 and use V = IR]Resistance of strip = 22.5 (1)ecf candidates R.2[common error is to divide by 5 0.9 scores 0/2 butecf to next part gives l = 0.033 m which will then score 3/3](b)R = l / A or correct rearrangement (1)Correct substitution (1)Length = 0.82 m (1)ecf candidates R3Example of answerl = RA/ = (22.5 4.0 108 m2) 1.1 106 ml = 0.82 m (c)See P = V2 / R OR P = IV leading to increase in currentor decrease in resistance (1)more strips in parallel / material of lower resistivity (1)[not greater conductivity]2[9] 57.(a)E.M.F. = work done / charge OR energy transferred / charge (1)OR power / current[There is only one mark here and this is consistent withspecification but it must not be Joules or coulombs]1 (b)(i)Use of V = IR (1)I = 2.0 A (1)2Example of answerI = V / R = 8.0 V / 4.0 I = 2.0 A (ii)Uses p.d. = 4.0 V (1)r = 2.0 ecf their I (1)2Example of answerr = V / I = 4.0 V / 2.0 Ar = 2.0 (iii)Use of P = VI // I2R // V2/R (1)P = 16 W ecf their I (1)2Example of answerP = VI = 8 V 2 AP = 16 W(iv)Uses 4V or 2A 2 or their I r (1)see 5 60 s in an energy equation (1)energy = 2400 J (1)3Example of answerE = VIt = 4 V 2 A 5 60 sE = 2400 J[10] 58.(a)(i)Calculate resistanceRecall of R = V/I (1)Correct answer [8.65 ] (1)2Example of calculation:R = V/IR = 2.68 V 0.31 A= 8.65 (ii)Show that internal resistance is about 0.4 Recall of relevant formula [V = Ir OR lost volts = ( V) (1) OR = I(R + r)] including emfCorrect answer [0.39 ] [no ue] [allow ecf if = I(R + r)] (1)2Example of calculation:V = Irr = ( V)/I= (2.8 V 2.68 V)/0.31 A= 0.39 (iii)Comment on match to maximum powerNot matched [ecf for R in (a) (i) and r in (a)(ii)] (1)Max power when internal resistance = load resistance (1)2 (b)(i)Show that charge is about 14 000 CRecall of Q = It (1)Correct answer [14 400 C] [no ue] (1)2Example of calculation:Q = It= 2 2 A 60 60 s= 14 400 C(ii)Calculate time for which battery maintains currentUse of Q = It OR use of W = Pt (1)Correct answer [46 450 s or 12.9 h] (1)2Example of calculation:t = Q/I= 14 400 C / 0.31 A= 46 450 s (c)Explain effect on efficiencyEfficiency = I2R / I2(r + R) / Efficiency depends on R /(r + R) /more heat dissipated in cells / Efficiency is V/ and V decreases (1)so efficiency is less (1)[Must attempt explanation to get 2nd mark]2[12] 59.(a)Circuit:Potential divider (1)1 (b)Relay potential difference:4 V (1)1Example:5/15 12 = 4V (c)(i)Resistance:Recall of R = L/A (1)Correct substitution of values into formula (1)Correct answer [98()] (1)[allow 97 98 to allow for rounding errors] [no u.e.]3Example:R = (3.4 102 1.44) / (100 0.05)= 98 (ii)Combined resistance:Use of 1/RTot = 1/R1 + 1/R2 (1)Correct answer for R [4.8] (1)[allow 4.7 4.8 to allow for rounding errors]21/R = 1/98 + 1/5 (or = 1/100 + 1/5)R = 4.8 (iii)Relay voltage:P.d. across relay with ballast very similar to p.d acrossthe relay alone / p.d. = 3.9 V / p.d. lower (slightly) (1)1 (iv)Train on track:Relay voltage becomes very small / zero (1)1 (v)Wet ballast:Any twoCombined resistance now small / RT = 0.45 Relay voltage now small / V = 0.52 VRelay voltage too small to trigger green light /signal remains red (1)(1)2[11] 60.Tungsten filament bulb(a)ResistanceUse of P = V2/R or P = VI with V = IR (1)answer 960 (1)2Example of answerR = (240 V 240 V) 60 WR = 960 (b)Drift speedrearrangement of I = nAvQ (1)Use of Q = 1.6 1019 (C) (1)answer 0.15/0.148 m s1 (1)3Example of answerv = 0.25 A (3.4 1028 m3 1.6 10 19 C 3.1 1010 m2) (c)ExplanationQowc (1)Any THREEResistance due to collisions between electrons & ions/atoms/particles(as T increases) ions/atoms/particles have more energy(as T increases) ions/atoms/particles vibrate through largeramplitude /vibrate faster OR amplitude if lattice vibrationincreases.more chance/increased frequency of collision/interactionOR impedes the flow of electrons (1)(1)(1)4[9]61.Emf and Internal resistance(a)DerivationE = I (R + r) OR E = IR + Ir (1)1 (b)(i)Correct working (allow even if evidence of working backwards) (1)Example of answerE/I = R + rRearranging R = E/I r (ii)EmfAttempt to use gradient (1)answer 1.5 V (bald answer 1.5 V scores 0/2) (1)2 (iii)PowerFrom graph find value of 1/I when R = 5 (1)Use of P = I2R (1)answer 0.31 (W) (1)3Example of answer1/I = 4 A1 I = 0.25 AP = 0.25 A 0.25 A 5 = 0.3125 W (c)GraphIntercept at 2 (ohms) (1)Graph steeper than original (1)Gradient is 3.0 V i.e. line passes through [10, 27-29] [no ecf] (1)3[10] 62.Potential divider(a)First circuitMiddle terminal MOuter terminals L and K (any order) (1)1 (b)(i)P.d across lamp.External resistance in circuit is 25 or (20+5) ohms (1)See ratio of resistances (denominator larger) 6.0V (1)OR current = 6/25 Aanswer 4.8 V (1)3 (ii)AssumptionThe resistance of the ammeter is zero/negligible. (1)1(c)Second circuitSee 2 resistors in parallel with supply (1)Supply across ends of variable resistor (10 ) (1)Fixed resistor across one end and slider (consequent mark) (1)3[8] 63.(a)Circuit diagram and explanationammeter and voltmeter shown in series and parallel respectively (1)current measured with ammeter and voltage / p.d. with voltmeter (1)2 (b)Calculation of resistanceRecall of R = V/I (1)Correct answer [25.0 ] (1)2Example of calculation:R = V/IR = 3.00 V 0.12 A= 25.0 (c)Calculation of resistanceRecall of P = V2/R (1)Correct answer [29.4 ] (1)2Example of calculation:P = V2/RR = (230 V)2 1800 WR = 29.4 [Accept calculation of I = 7.8 A (1), calculation of R = 29.4 (1)] (d)Explanation of difference in values of resistanceAt higher voltage value element is at a higher temperature (1)(resistance higher because) increased lattice ion vibrations impedecharge flow (more) (1)2[8]64.(a)(i)Critical angle calculation:Use of sin C = 1/da (1)Correct answer [24.4 only acceptable answer] [no u.e] (1)2eg.SinC = 1/da = 1/2.42C = 24.4 (ii)Ray diagram:Small angle ray shown passing into air, away from the normal (1)Large angle ray showing T.I.R. with angles equal [by eye] (1)2 (iii)Labelling of angles:An incident angle correctly labelled between normal and rayin diamond (1)An angle of refraction correctly labelled between normaland ray in air (1)2 (iv)Amount of trapped light:Any 3 of the following:The higher the refractive index the greater the amount oftrapped lightThe higher the refractive index the lower the critical angleT.I.R occurs at angles greater than the critical angleSo, if critical angle is smaller, more light is reflected (1)(1)(1)Max 3 (b)Comment on angle:Lower critical angle so more sparkle (1)1[10] 65.(a)Graph scale:Log scale (1)1 (b)(i)Choice of material:Any 2 of the following:(almost) all of the voltage is dropped across the carbon rodgives the greatest speed variationothers need to be longer (to have same resistance as carbon)others need to be thinner (to have same resistance ascarbon) (1)(1)Max 2(ii)Resistance calculation:Use of R = L/A (1)Correct units used for all terms [all in cm or all in m] (1)Correct answer [1.9 ] (1)3[allow 1.8 for rounding errors no u.e]eg. R = 1.4 105 0.4 / 3.0 106 = 1.9 (iii)Available voltage:X 12 VY 0 V (1)1 (iv)Effect of connecting wires:Less voltage available for train set as some wasted across wires (1)0.5 is (relatively) large % of total resistance, so effect ishigh / not negligible (1)orCalculation of potential difference available now (1)[9.5 V] [allow 9.5 9.6 V]Significant drop from 12 V (1)Vxy = (Rxy / RTotal) Vsupply = (1.9/ (1.9 + 0.5)) 12 = 9.5 V2[9] 66.(a)(i)Potential difference = work (done)/(unit) chargeOR Potential difference = Power/current (1)1 (ii)J = kg m 2 s 2 (1)C = A s or W = J s1 (1)V = kg m2 A1 s3 (1)3 (b)Converts 2 minutes to 120 seconds (1)Multiplication of VIDt or VDQ (1)Energy = 1440 J (1)3Example of answer:Energy = 6.0 V 2.0 A 120 s= 1440 J[7] 67.(a)n = number of charge carriers per unit volumeORn = number of charge carriers m3ORn = charge carrier density (1)v = drift speed/average velocity/drift velocity (of the charge carriers) (1)2(b)n is greater in conductors / n less in insulators. (1)[There must be some comparison]larger current flows in a conductor. Dependant on havingreferred to n (1)2(statement that n large in conductor and so current large max1) (c)(In series), so same current and same n and Q (1)vB greater vA (1)vA/vB = // 0.25 (1)3[7] 68.(a)pd = 3.6 V (1)1Example of answer;p.d. = 0.24 A 15 = 3.6 V (b)Calculation of pd across the resistor (1)[6.0 3.6 = 2.4 V]Recall V = IR (1)I1 calculated from their pd / 4 (1)[correct answer is 0.60 A. Common ecf is 6V/4 gives 1.5 A]3Example of answer:I1 = 2.4 V / 4.0 = 0.6 A (c)Calculation of I2 from I1 0.24 [0.36 A] (1)[allow ecf of their I1. common value = 1.26 A]Substitution V = 3.6 V (1)R = 10 (1)3[7] 69.(a)(i)( gradient =) r = 1.95 2 (1)E = 8.9 9 V (1)2 (ii)I = 2.15 2.17 A (1)1 (iii)Use of V = IR (1)R = 2.1 2.2 (1)2 (b)(i)Battery or cell with one or more resistive component (1)Correct placement of voltmeter and ammeter (1)2(ii)Vary R e.g. variable resistor, lamps in parallel (1)Record valid readings of current and pd (consequent mark) (1)2[Do not give these marks if the candidate varies the voltage as well][9] 70.Area of wire:Use of A = r2 (1)Correct answer [1.9 107 (m2). Allow 1.9 107 and 2.0 107 (m2)](1)2[No u.e.]e.g.A = r2 = (2.5 104)2= 1.96 107 m2 Table + graph:Length / Area / 106 m10.00.51.01.52.02.53.13.6(1)4.0 4.1First 2 points plotted correctly to within 1 mm (1)Rest of points in straight line with origin by eye (1)3 Resistivity calculation:Drawn through the origin, ignoring first 2 points (1)Recall = R /(L/A) [in any form] (1)Large triangle drawn on graph OR accept the use of a pair of values (1)read from the line[ x> 3 10-6 m1)is required in both cases][x-axis allowed as bottom of triangle]Correct answer [1.2 10-7 W m)] (1)[allow 1.1 1.3 10-7( m)] [no u.e.]4e.g.0.4 / 3.4 106 = 1.2 107 mAnomalous results:Any two of the following:Higher current/lower resistance for shorter lengths/at these pointsAt shorter lengths/at these points wire gets hotterNon-uniform area/diameterCable / contact resistanceSensitivity of metersEffect on resistance of any of the above (2)max 2[11] 71.(a)(i)Lamp brightnessLamp A (1)Larger current through it (at 9.0 V)/greater power (1)2(at 9.0 V)/smaller resistance (at 9.0 V) (ii)Battery currentAddition of currents (1)Current = 1.88 1.92 A (1)2 (iii)Total resistanceR = 9 V/1.9 A or use of parallel formula (1)R = 4.6 4.9 W (1)2[full ecf for their current] (b)Lamps in seriesCurrent same in both lamps/current in A reduced from original value (1)Pd across A less than pd across B (1)Lamp A has a lower resistance than lamp B (1)P = VI or P = RI2 (1)Any 2Lamp A will be dimmer than B [conditional on scoring ONE of (1)1the above marks][9] 72.(a)(i)ResistanceUse of V/I [ignore 10x] (1)3800 W (3784 W) (1)2(ii)Resistance of thermistorUse OR 9V/.74mA R OR (1)6.2 V = 0.74 mA RTH8400 W [8378 W] [substituting 4000 W gives 8857 W ie 8900 W] (1)[method 2 substituting 3800 W gives 8362 W: substituting 4000 Wgives8162 W]2 (b)Suggestion and ExplanationThe milliammeter reading increases (1)Thermistor resistance becomes zero /Short circuit (1)Since supply voltage is constant / I = 9.0 V/R (1)ORCircuit resistance reduced3[7] 73.(a)Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)ORE = W/Q (1)Symbols defined (1)[E = 1J/C scores 1]ORE = P/I (1)Symbols defined (1)[terminal pd when no current drawn or open circuit scores max 1]2 (b)Voltmeter calculationAny attempt to find any current (1)Attempt to calculate pd across 10 W resistor (1)5.77 V2ORPotential divider method; ratio of resistors with 10.4 W on the bottom (1)Multiplied by 6.0 V (1)5.77 V (1)3[For either method, an answer of 0.23 V scores max 1](c)Second battery addedVoltmeter reading increased (1)Any two of:EMF unchangedTotal resistance reducedcurrent increases or lost volts decreases (2)3[8] 74.Explanation increase of resistance with temperatureTemperature increase leads to increased lattice vibrations (1)scattering flowing electrons / increased collisions of electrons. (1)2 Calculation of resistance at 200 CR = V/I [stated or implied] (1)= 7.4 V 0.19 A= 39 W (1)2 Discuss whether results support hypothesisNo. Resistance is not increasing with temperature. (1)1 Calculation of mains voltageP = V2 R (1)V2 = PR= 1200 W 41 W [Mark for rearrangement OR substitution] (1)[Accept 39 41 W] [ecf]V = 220 V (1)[Allow P = I2R (1),3calculate I = 5.4 A and use in 1200 W = 5.4 A V (1),V= 220 V (1)][8] 75.Resistance calculationUse of R = L/A (1)Substitution R = 1.6 104 0.02/(5 (103) 0.02 (103)) (1)= 32 W (1)3Total resistanceEither Section 2 = R1 (16 W) OR Section 3 = R1 (10.7 W) (1)Use of RTotal =R1 + R2 + R3 (1)RTotal = 58.7 W [55 W if 30 W used as starting point] (1)3[ecf if section 3 calculated as R1 = 56 W OR 52.5 W if 30 W usedas starting point] Why thermochromic ink becomes warmCurrent produces heat / reference to I2RORThermal conduction from conductive ink (1)1[Mark for identifying that the heating effect originates in the conductive ink] Why only thin section transparentThinner / section 1 has more resistance (1)So even a small current will heat it/Power (heating effect) given byI2R / current will heat it more (1)2[OR opposite argument explaining why thicker section is harder to heat][9] 76.CircuitsBase unit:ampere OR amperes OR amp OR amps (1)Derived quantity:charge OR resistance (1)Derived unit:volt OR volts OR ohm OR ohms (1)Base quantity:current (1)4[If two answers are given to any of the above, both must be correct to gain the mark][4] 77.(a)Io and Jupiter: Time taken for electrons to reach Jupitert = s/ = (4.2 108 m)/(2.9 107 m s1) = 14.48 sCorrect substitution in = s/t (ignore powers of ten) (1)Answer: 14.48 s, 14.5 s [no ue] (1)2(b)Estimate of number of electronsQ = ne = Itn = It/en = (3.0 106 A) (1s)/(1.6 1019 C)Use of ne = It (1)(1.8 2.0) 1025 (1)2 (c)Current directionFrom Jupiter (to Io) / to Io / to the moon (1)1[5] 78.(a)p.d. across 4 resistor1.5 (A) 4 ()= 6 V (1)1 (b)Resistance R2Current through R2 = 0.5 A (1)R2 = R2 = 12 (1)2[allow ecf their pd across 4 ](c)Resistance R1p.d. across R1 = 12 6 4= 2 V (1)Current through R1 = 2 A (1)R1 = = 1 (1)[allow ecf of pd from (a) if less than 12 V]Alternative methodParallel combination = 3 (1)Circuit resistance = 12(V)/2 (A) = 6 (1)R1 = 6 (3 + 2) = 1 (1)3[allow ecf of pd from (a) and R from (b)][6] 79.(a)Current in filament lampP = VI or correct rearrangement (1)2 A (1)2 (b)(i)Sketch graphCorrect shape for their axes (1)IV quadrant showing fair rotational symmetry (1)2(ii)Explanation of shape(As the voltage/p.d. increases), current also increases (1)(As the current increases), temperature of lamp increases (1)(This leads to) an increase in resistance of lamp (1)so equal increases in V lead to smaller increases in I OR rate ofincrease in current decreases OR correct reference to their correct (1)4gradient[8][If a straight line graph was drawn though the origin then (1)(0)(0)(1) forthe following:V is proportional to Rtherefore the graph has a constant gradient] 80.(a)(i)ReplacementV1 (1)1 (ii)Explanation[ONE pair of marks]Resistance: resistance of V1 [not just the voltmeter] is much largerthan 100 OR combined resistance of parallel combination is (1)approximately 100 Voltage: p.d. across V1 is much greater than p.d. across 100 OR (1)all 9 V is across V1ORCurrent: no current is flowing in the circuit / very small current (1)Resistance: because V1 has infinite/very large resistance (1)OR(Correct current calculation 0.9 x 10 6 A and) correct pd calculation90 x 10 6 A (1)This is a very small/negligible pd (1)2 (b)Circuit diagram(i) or equivalent resistor symbol labelled 10 M (1) or equivalent resistor symbol labelled 10 M (1)2[They must be shown in a correct arrangement with R](ii)Value of R6 (V): 3 (V) = 10 (M): 5 (M) / Rtotal of parallel combination is 5 (1)M1/5 (M) = 1/10 (M) + 1/R OR some equivalent correct (1)substitution to show workingR = 10 M (1)3[8] 81.Explain zeroing of meterNo resistance when leads touched together/short circuit/calibration forzero error (1)1Show that resistance is about 70 R = V I (1)= 0.54 V 0.0081 A= 67 [no ue] (1)2Explain section from passageOther currents/voltages/resistances present (1)change in current changes reading for resistance (1)2Explain changes in meter reading with temperature increaseIncreased lattice vibrations/vibration of atoms/molecules (1)scattering flowing electrons/more collisions (1)increased resistance/increase meter reading (1)3[8] 82.Circuit diagramAmmeter and power source in series (1)Voltmeter in parallel with electrodes (1)2[Allow both marks if diagram shows an ohmmeter without a powerpack 1 if power pack]Calculation of resistanceUse of area = r2 (1)R = 2.7 103 m 5.0 104 m/A (1)= 172 (171.9 ) (1)3Plotting graphAxis drawn with R on y-axis and labelled with units (1)Points plotted correctly [1 for each incorrect] (1)Sensible scale (1)Curve added passing through a minimum of 4 points (1)4Diameter of holeCorrect reading from graph = 0.23 mm [Allow 0.22 0.26 mm] (1)1[10] 83.Temperature calculationCurrent = 4.5 103 A (1)p.d. across thermistor is 4.2 V (1)Rthermistor = 930 W ecf their current and pd subtraction error (1)Temperature = 32 C - 34 C [Allow ecf for accurate reading] (1)4 Supply doubledAny two from:Current would increase / thermistor warms up / temperature increasesResistance of thermistor would decrease (1) (1)Ratio of p.d.s would changeNo OR voltmeter reading / pd across R more than doubles (1)3[This mark only awarded if one of the previous two is also given][7] 84.DiagramLabelled wire and a supply (1)Ammeter in series and voltmeter in parallel (1)ORLabelled wire with no supply (1)Ohmmeter across wire (1)2ReadingsCurrent and potential difference OR resistance ( consistent with diagram) (1)Length of wire (1)Diameter of wire (1)3Use of readingsR = V/I OR r = RA/l (1)Awareness that A is crosssectional area (may be seen above and credited here) (1)Repetition of calculation OR graphical method (1)3PrecautionAny two from:Readings of diameter at various places /different orientationsContact errorsZeroing instrumentsWire straight when measuring lengthWire not heating up / temperature kept constant (1) (1)2[10] 85.Conductor resistanceR = rl/A (1)Correct substitution of data (1)R = 4.3 102 W (1)3 Manufacturers recommendationLarger A has a lower R (1)Energy loss depends on I2R / reduces overheating in wires (1)2[5] 86.Car batteryVoltmeter reading: 12.2 (V) (1)1EquationTerminal p.d. = 12 V + (5.0 A 0.04 W)See 12V (1)See 5.0 A 0.04 W (1)Addition of terms (1)3Wasted powerSee 0.04 + 0.56 OR 2.8 V + 0.2 V OR 5 x (15 12) W (1)Power = 15 W (1)2Efficiency(same current) 12 V / 15 V OR POUT/PIN = 60 W/75 W (1)Efficiency = 0.8/80% Efficiency = 0.8/80% (1)2 ExplanationAny two from:Starter motor / to start car needs (very) large currentI = (E and R fixed) rmin Imax (1) (1) (1)2[10] 87.Circuit diagramVariable voltage (1)Includes ammeter and voltmeter (1). in series and parallel respectively (1)3[No penalty for LED bias] Description of current variation in LEDsInitially, increasing voltage still gives zero currentORCurrent doesnt flow until a specific minimum voltage (1)Current then increases (1).with an increasing rate of increase (1)3 Discussion of whether LEDs obey Ohms lawNo (1)I not proportional to VORR not constant / V/I not constant / R decreases (1)2 Calculation of resistance of green LED at 1.9 VR = V/I [Stated or implied] (1)= 1.9 V 1.46 103 A= 1300 W (1)2Calculation of power dissipated by red LED at 1.7 VP = IV [Stated or implied] (1)= 3.89 103 A 1.7 V [do not penalise mA twice]= 6.6 103 W (1)2[12] 88.ChargeCharge is the current time (1)1Potential differenceWork done per unit charge [flowing] (1)1Energy9 V 20 C (1)= 180 J (1)2[4] 89.Number of electrons(64 109 C) / (1.6 1019 C) = 4.0 1011 electronsUse of n = Q/e (1)Seeing 1.6 1019 C (1)Answer of 4.0 1011 (electrons) (1)3[Use of a unit is a ue][ve answer: 2/3] Rate of flow(6.4 108 C)/3.8 s = 16.8/17 [nC s1] OR 16.8/17 109 [C s1](6.4) / 3.8 s i.e. use of I = Q/t [Ignore powers of 10] (1)Correct answer [No e.c.f.] [1.7 or 1.68 x 108 or 1.6 108] (1)2UnitAmp(ere)/A (1)1[6]90.Explanation of observationAny two from:LED on reverse bias/R in LED infinite/ LED wrong way roundso current is zero /LED does not conduct / very small reversebias currentsince V = IRV = 0 1K = 0 V (1) (1)2 Explanation of dimnessRV very large / RV much greater than RLED (1)Current very low / pd across LED very small (not zero) (1)2Circuit diagramLED in forward bias (1)Variation of pd across LED (1)Voltmeter in parallel with LED alone (1)3[LED in series with voltmeter 0/3][7] 91.Circuit diagramAmmeter in series with cell and variable resistor (correct symbol) (1)Voltmeter in parallel with cell OR variable resistor (1)2Power output at point XPower = voltage current (1)= 0.45 V 0.6 A= 0.27 W (1)2 Description of power outputAny three from:Current zero; power output zero/small/lowAs current increases power output also increasesThen (after X ) power decreasesMaximum current; power output zero (1) (1) (1)3[Accept reverse order]e.m.f. of cell0.58 V (1)1Internal resistanceAttempt to use OR = V+ IR (1)= = 0.217 / 0.2 W (1)2[ecf an emf greater than 0.45 V][10] 92.Statement 1Statement is false (1)Wires in series have same current (1)Use of I = nAeu with n and e constant (1)3[The latter two marks are independent] Statement 2Statement is true (1)Resistors in parallel have same p.d. (1)Use of Power = V2/R leading to R , power (1)3OR as R , I leading to a lower value of VI 3rd mark consequenton second[6] 93.Explanation of assumption that voltmeter does not affect valuesVoltmeter has very high resistance/takes very small current (1)1Current through X4.8 A 6 = 0.8 AOR 48 V 60 W = 0.8A (1)1Value missing from E7P =IVP = 4.4 A 53 V = 233 W (1)1Description of appearance of lamp X as lamps switched onGets dimmerfrom table, voltage decreasing / current in X decreasing / power per lamp decreasing (1)So P decreases (1)3 Formula for cell C6I = / Rtot (1)I = 120 / (15 + B6) (1)2Effect of internal resistance on powerPower has a maximum value (1)when external resistance = internal resistance (1)2[10] 94.Measurement neededAny three from:ResistanceDistance between probesEffective area/cross sectional areaR = r (1) (1) (1)3 Equation of line AIntercept = 3.5 (W m) (+/ 0.3) (1)Gradient = 1.5 (W mm1) (+/ 0.05) (1)So equation is r = 1.5 d 3.5 [Or equivalent, e.c.f. allowed] (1)3Addition of linePoints correctly plotted (1 for each error, allow square tolerance) (1) (1)Line of best fit drawn (1)3Best distanceBetween 1.90 and 1.99 km (1)1[10] 95.Resistance calculationsEvidence of 20 W for one arm (1) (1)R = 10 W (1)3 CommentThis combination used instead of a single 10 W resistor [or samevalue as before] (1)because a smaller current flows through each resistor/reduce heatingin any one resistor/average out errors in individual resistors (1)2[5] 96.GraphsDiode:RH quadrant: any curve through origin (1)Graph correct relative to labelled axes (1)LH side: any horizontal line close to axes (1)3Filament lampRH quadrant:Any curve through origin (1)Curve correct relative to axes (1)LH quadrant:Curve correct relative to RH quadrant (1)3[Ohmic conductor scores 0/3][6] 97.CircuitAmmeters and two resistors in series (1)1[1 mark circuit penalty for line through cell or resistor]Cell e.m.fE= 150 x 106 (A) x 40 x 103 (W) total R (1)Powers of 10 (1)2E = 6.0 (V) New circuitVoltmeter in parallel with 25 (kW) resistor (1)1Resistance of voltmeter(Total resistance) = = (35.3 kW)(Resistance of ll combination) = 35 15 kW= (20 W) [e.c.f. their total resistance]RV = 100 kW [108 kW if RT calculated correctly](1)(1)(1)Alternative route 1:3p.d. across 15 kW = 2.55 V(\p.d. across ll combination = 3.45 V)resistance combination = 20 kW RV = 100 kW(1)(1)(1)[7] Alternative route 2:3p.d. across parallel combination = 3.45 VI through 25 kW = 138 mA RV = 100 kW(1)(1)(1) 98.Resistance of strain gaugeState R = (1)Use of formula (1)x 6 (1)R = 0.13 W [ecf their l] (1)4 Change in resistanceDR = 0.13 W 0.001DR = 1.3 104 (W) [no e.c.f.]ORDR = 0.02 0.001DR = 2.0 105 W0.1% 0.001 (1)Correct number for DR (1)2Drift velocityStretching causes R to increase (1)Any two from: Current will decrease I = nAuQ Drift velocity u decreases nAe constant (1) (1)3[9] [For R decreasing, max 1:Any one from: I will increase I = nAuQ u will increase nAe constant] 99.Resistance in darknessIn the dark R = 4 kW (1)so resistance per mm = 4000 W/40 mm = 100 W(mm1) (1)2Resistance of 8 mm lengthIn the light R = 200 W (1)so resistance of 8 mm strip = (8 mm/40 mm) 200 W [= 40 W] (1)2 CalculationsResistance of remainder = 32 mm 100 W mm1 = 3200 W (1)1(i)Total resistance = 3240 W (1)Current = V/R = 1.2 V/3240 W = 3.7 104 A (1)(ii)Across 8 mm, p.d. = IR = 3.7 104 A 40 W (1)= 0.015 V (1)4 Explanation of why current decreasesAny two points from:more of strip is now in the darkgreater total resistanceI = V/R where V is constant (1) (1)Max 2[11]100.Resistance of lampsP = OR I= 60/12 = (5 A)1R = R = V/I1R = 2.4 W1 Resistance variationLamp A: resistance of A decreases with current increase1Lamp B: resistance of B increases with current increase1 Dim filamentLamps are dim because p.d. across each bulb is less than 12 V1 Why filament of lamp A is brighterBulbs have the same current1p.d. across A > p.d. across B/resistance A> Resistance B1ORpower in A > power in B2[8] 101.Current in heating elementp = VII=I = 2.2 Ap = R = I = 2.2 A 1 1 1 Drift velocityDrift velocity greater in the thinner wire / toaster filament1ExplanationQuality of written communication1See I = nAQu1I is the same (at all points )1(probably) n (and Q) is the same in both wires1[8] 102.Resistance of filmsR = 1R = or A = wt [consequent on first mark]1[i.e. product = wt] Resistance calculationR = ORR = R = 160 WCorrect substitution except powers of 101Correct powers of 101Answer1 Resistance of square filml = w1R = 1[7]103.Definition of e.m.f. of a cellWork/energy (conversion) per unit charge1for the whole circuit / refer to total (energy)1ORWork/energy per unit charge1converted from chemical to electrical (energy)1 ORE = for whole circuit1All symbols defined1ORE = for whole circuit1All symbols defined1[Terminal p.d. when no current drawn scores 1 mark only] Circuit diagramR1R (can be variable)12A in series1A and V correct1V as shownOr across R + AOr across battery [2nd mark is consequent on R(fixed, variable) or lamp]Sketch graphGraph correctly drawn with axes appropriately labelled andconsistent with circuit drawn1Intercept on R axes Gradient ()r [Gradient mark consequent1 ()r on graph mark][Gradient may be indicated on graph][6] 104.Total e.m.f of cells in seriese.m.f. in series add up / 6000 40 10-3 V (1)= 240 V (1)2 Internal resistance of cells in series6000 0.70 W = 4200 W (1)1Calculation of currentI= V R (1)= 240 V 4200 W = 0.057 A (1)2 Calculation of total current20 0.057 A = 1.1 A (1)1Explanation of voltmeter readingSince V = IR (1)and R = 0 W (1)V = 0 V e.m.f.ORV= E Ir (1)r 0 W (1)so V < EORLost volts (1)across internal resistance (1)2 Voltmeter reading0 V [No u.e.] (1)1[9] 105.Resistor(i)A = pr2 = p (4.0 10-4)2 (1)= 5.03 10-7 m2 (no u.e) (1)2(ii)Recall of R = rl/A (1)Length l = RA/r= 0.12 5.0 10-7 / 1.8 10-8 [substitutions]= 3.3 m (1)3Advantage of using iron wire of same diameterShorter piece of wire needed (if iron chosen) (1)1[6] 106.Unit of currentAmps/ampere (1)1 Base units of p.d.For V = IR methodAny three from:V = J ClC = A sJ = N mN = kg m s2[kg m2 s3 A1][See J = kg, m2 s2 (1) (1)]ORFor P = VI methodWatt is J s-1 / J/sV = J s1 A1J = NmN = kg m2 s2 (1) (1)] (1) (1) (1)[See kg m2 s2 (1) (1)]3[4] 107.Show that resistance is approximately 45 W = 44.7 W [No u.e.] (1)[Must see this value and not 45] TableSwitch XSwitch YResistance of heater/WOpenClosed22.5/22.35(1)ClosedOpen45/44.7(1)ClosedClosed15/14.9(1)[No u.e.]3 Calculation of maximum power Use of equation with 15 W OR their minimum value (1) = 3526 W,3500 W [full ecf] (1)2 Explanation of power output fallSince V is constant P = OR P = VI and V = IR[Then P as R ] (1)2OR P [so P as R][10] 108.Explanation of greater drift velocity(Electrons have greater drift velocity) in the thinner wire (1) Any two from:Same current in both wiresReference to I = nAQunQ same in both wires (1) (1)3 Explanation of higher dissipation of power(Higher power is dissipated) by the smaller(er)/ low resistor (1) Any two from:Resistors have same p.d. across themThe small resistor has the largest current [or reverse]Power = voltage current, OR voltage2 resistance [NOT I2R] (1) (1)3[6] 109.Circuit diagramResistor with another variable resistor/potential divider/variable powerpack (1)Ammeter reading current through resistor (1)Voltmeter in parallel with resistor (1)3 Graph labels (1)1Potential differenceAt 0.5 A p.d.= 3.5 V / 3.4 V + 7.8 V / idea of adding p.d. [for same current] (1)= 11.2 V/11.3 V (1)2[Accept 11.0 11.5 V] Resistance of lamp [OR their value of p.d. across lamp 0.5 A] (1)= 7.0 W (1)2[e.c.f. their value][8] 110.Meaning of m 103 (1)1 Calculation of resistance for reading 3R = V/I OR R = 74 103 V 150 109 A [ecf for milli] (1)R = 4.9 105 W (1)2 Calculation of power for reading 4P = I V OR P = OR P = I 2R (1)= 210 109 A 57 103 (1)2= 1.2 108 W Plotting points on graphTwo correct points (1)Third correct point (1)Best fit straight line for points as they appear on students graph (1)3 Predicting short-circuit currentCorrect from graph, e.g 450 nA (1)1 Suggested e.m.fCorrect from graph, or table, 110 mV (1)1Explanation of why voltage fallsCell has internal resistance/ lost volts (1)Lost volts = Ir, so lost volts increase as current increasesORV = E Ir, so V decreases as I increases (1)2[12] 111.Equation to define resistivity(1)All symbols defined (resistivity, resistance, length, cross-sectional area) (1) (1)[3 symbols only defined (1)]3 Resistance meterAny two from:the resistance between the two probes is measured, not the resistivitybecause you cannot measure the cross-sectional area of skin between the probesA and 1 both vary; cannot calculate resistivity (1) (1)2[5] 112.Word EquationQuantity DefinedVoltage CurrentResistance(1)Voltage CurrentPower(1)Charge TimeCurrent(1)Work done ChargeVoltage/p.d./e.m.f(1)[4] 113.Charge calculationQ = 20 000 4.0 10-4 s [substitution]Q = 8.0 C/A s2Resistance calculationR = = R = 8.5 10-4 W Formula (1)Correct substitution (1)Answer (1)3Potential difference calculationV = IR= (20 000 A) (85 10-5 W) [or their value] (1)= 17 V [Allow full e.c.f] (1)2ExplanationFor the tree: R or p is larger (1)1[8] 114.NetworksFirst network:2.5(W) (1)Second network:25 (W) (1)Third network:10 (W) (1)3 Meter readingsAmmeter:25 (mA) (1)Voltmeter V1:25 10 OR 50 5 [ignore powers of 10] (1)= 0.25 V (1)Voltmeter V2:50 25 [ignore powers of 10] (1)= 1.25 V (1)5[Allow full e.c.f. for their resistance for 2nd network OR their V1 answer][8] 115.Potential difference across resistors2.0 MW:6.0 V5.99998 V (1)OR4.0 W:0V1.2 10-5 V (1)2Second potential divider circuitp.d. across 45 W:( 6.0 V) = 5.4 V (1)p.d. across diode:(6.0 V 5.4 V) = 0.6 V (1)2 [Allow e.c.f. for 2nd mark if candidate uses 6.0 V = 0.7 V for diodethen6.0 V 0.7 V5.3 V for 45 W] Graph (1)1[5] 116.Calculation of currentP = IVI = P/V (stated or implied) (1)= 0.78 W 6 V= 0.13 A (1)2Calculation of resistanceP = V2/ROR R = V/IOR R = (6 V)2 0.78 WOR R = 6 V 0.13 A [ecf] (1)= 46 W (1)2 Explanation of operation from mainsIn series (1)240 V 40 lamps = 6 V per lamp (1)2Explanation of constant brightness of lampsCurrent equal justified, e.g. in series or same V/R or same P/V1 Statement and explanation of different resistance with ohmmeterLower resistance with ohmmeter (1)Identify lower temperature with ohmmeter [may be implied] (1)(Lattice) ions/atoms vibrations impede electrons/current (1)[Require interaction][Allow converse argument]3[10] 117.Explanation of amp hoursCharge = current time or Q = It (1)Amp current and hour time (so amp hour charge) (1)2 Show that charge about 5000 CCharge = 1.5 A 1 60 60 s= 5400 C1 Calculation of energy storedW = QV OR W = I v t (1)= 5400 C 3 V [may use 5000 C]= 16 200 J [up] (1)2Show that energy is about 20 000 JEnergy = Ivt (1)= 0.3 A 3.1 V 6 60 60 s= 20 100 J (20 088 J) (1)2 Calculation of efficiencyEfficiency = (stored energy/input energy) 100%= 16 200 J [allow ecf from 3rd part] (1) 20 088 J [allow 20 000 J from 4th part] (1) 100%= 80.6% [Accept fractional answers. Allow ecf, but check nos.] (1)3[10] 118.DevicePotential divider or potentiometer1 Voltmeter readingA 9.0 V (1)B 0 V (1)2DiagramLabel X two thirds of the way down from A [Allow e.c.f.]1 ExplanationAny 3 points from the following:lamp in parallel with lowest 1/3 of ABwhen resistors in parallel, resistance decreasesp.d. across lamp reduced to below 3 Vcurrent divides no longer enough current to light lamp3[7]119.Circuit diagram(Variable) resistor symbol (1)Voltmeter in parallel with cell/resistor (1)Ammeter in series [even if R missing] (1)3 Maximum power availableUse of P = IV (1)Any pair of values which round to 1.4 W (1)2 Analysis of data1000 W m2 P = 1.4 W100 W m2 P = 0.11 W OR at least one further value of P (1)Inspection of ratio[e.g. 100 1000, 0.11 1.4, 1000 1.4] (1)Comment based on candidates result[e.g. Yes as ratio is similar] (1)3 GraphE/V + scale : 2 large squares = 0.05 V (1)Points (1)Straight line good fit (1)t/C + scale: 1 large square = 10 (1)[No penalty if t vs E]4 Determination of mathematical relationshipIntercept = 0.640 0.655 (1)A gradient evaluated (1)= 2.1 2.3 103 (1)E = 2.2 103t + 0.65 [e.c.f.] (1)4Other axes:Intercept 290 310 (1)Gradient (1)440 460 (1)t = 450E + 300 (1) Determination of light power from the sunAttempted evaluation of an area (1)= 0.13 0.17 [1 if 106] (2)[0.10 0.20 (1)]Their answer 4.0 = .(W) [no u.e.] (1)4[20] 120.Base unitDerivedunitBasequantityDerivedquantityMassChargeJouleAmpereVolt(1)(1)(1)(1)(1)5[5] 121.Explanation:As the temperature rises, the resistance decreases (1)As the resistance decreases, so the ammeter reading/current increases (1)[No mention of resistance 0/2][Current controls temperature controls R is wrong physics 0/2][If T changes so R changes OR vice versa so I changes 1 mark only][Correct static relationship (extremes) 1 mark only]Reading on milliammeter:At 20 C R = 1.4 (kW) (1)Substitute correctly in V = IR i.e. 6 V = I 1400 W (1)[Allow their incorrect R; ignore 10x] (1)Milliammeter reading = 0.0043 A OR 4.3 mA [no e.c.f.] (1)[Accept 4 mA/4.2 mA]5[5] 122.Current:Conversion, i.e. 0.94 103 m s1 (1)Use of 1.6 1019 C (1)Answer 3.0 A1.0 1029 m3 0.20 106 m2 1.6 1019 C 0.94 103 mm s1 (1)Current = 3.0 A [Accept 2.8 A if 0.9 103 used.] 3Resistance:Recall R = (1) Substitution:R = (1)Resistance = 0.34 W (1)3Potential difference:Potential difference = 3.0 A 0.34 W (1)= 1.0 V (1.02 V)[Mark for correct substitution of their values or for the answer of 1.0 V]1 Explanation:(Increasing resistivity) increases resistance (1)Leads to a smaller current (1)2Comparison:Drift velocity decreases (in second wire) (1) 1[Allow V1/V2 = I1/I2][Allow e.c.f. answer consistent with their current answer][Resistivity up, current down r up, I down / 2 (2nd mark)][10]123.E.m.f.Use of intercept mentioned/indicated on graph/when I = 0 (1)e.m.f. = 1.5 V (1)2Use of graph:Internal resistance: mention use of gradient/use of numbers/triangle on graph (1)Internal resistance = 0.5 W (1) 2[Finds r and/or V by substitution, can score answer mark, but NOT method mark] [Gradient = They might write gradient = W OR gradient = - ignore signs] Graph:Negative gradient of a straight line starting anywhere (1)from (0.0, 3.0) [No e.c.f.] (1)heading for (1.0, 2.0[1.9 2.1])/gradient of -1 [Consequent mark] 13 Filament lamp: any two ofif the variable resistor is set to zero [NOT, as RVR down] (1)the lamp prevents I from becoming too large (1)and overloading/damaging the ammeter (1)bulb acting like a fuse OR prevents short circuit (1)bulb means there is still resistance in circuit (1)Max 2[9] 124.Completion of circuit:Ammeter and voltmeter used [correct symbols required] (1)Ammeter in series, voltmeter in parallel (1)[Do not penalise variable resistor in series]2 Explanation of difference voltages:Any two from:Internal resistance of cell/batteryWhen current flows, energy transferred to / lost by internal resistance/heating in cellHence voltage across internal resistance/ lost voltsReduced terminal p.d. / V = E Ir / E = V + Ir2Show that internal resistance is about 0.6 W:R = V/I= (1.5 1.25) (1) 0.4 A= 0.63 W [No u.e.] (1)2 Calculation of resistance of bulb:R = V/I (1)= 1.25 V 0.4 A= 3.1 W (1)2 Explanation of lower resistance with ohmmeter:Identify lower temperature with ohmmeter (1)Lattice ions/atoms vibrations impede electrons (1)[Allow converse argument]2[10] 125.Explanation of why it is a good approximation:Resistance of connecting lead is (very) small(1)So I R(very) small = (very) small p.d./e1s do little work so p.d. small/r small (1)compared with rest of the circuit so p.d. small2 Circumstances where approximation might break down:If current is large OR resistance of rest of circuit is small(1)[Not high voltage/long lead/thin lead/high resistivity lead/hot lead]1Calculation:Use of R = with A attempted sectional area(1)Correct use of 16(1)Use of V = IR(1)0.036 V(1)4[10] 126.Number of carriers or electrons per unit volume / per m3 /carrier density/electron density (1)[Not charge density / concentration]Drift velocity OR drift speed OR average/mean/net/overall velocity (1)2[Not just velocity; not speed unless drift]m3 (1)m2 As m s1 (1)Multiply and reduce to A (1)3 [Base units not needed][Mixed units and symbols could get the third mark][mA = m1 loses 1 mark]Metal:2[Ignore anything about v. Allow e.g. electron density for n][7]127.Use R = rl/A OR correct rearrangement OR plot R l gradient = r /A (1)[Symbols or words]With A = tw (1)2l = RA/r [Rearrangement mark symbols or numbers] (1)Use of A = tw (1)[Correct physical quantities substituted but ignoring unit errors, powers of 10]= 110 m[111 m] (1)3 Reduce width/w of strip OR use thinner/t foil [Not reduce A; not increase T, V, I] (1)Smaller w/t/A will be less accurate OR have larger error OR larger R will be more accurate (1)2[Increase w or t, could give e.c.f. to increased accuracy][7] 128.I 2 R / (eI I 2 r) / (1)I 2 r / (eI I 2 r) (1)eI OR I 2 R + I 2 r / e 2 / (R + r) (1)eI = I 2 R + I 2 r OR (It = I 2 RT + I 2 rt / their (iii) = their (i) + their (ii) (1)Cancel I (OR I and t) and arrange [only if energy equation is correct] (1)5 Maximum current occurs when R = 0 (1)Imax = e/r (1)2OR larger r means smaller I (1 mark)1 MW [Could be underlined OR circled] (1)It gives the smallest current (1)[If 100 kW this reason: 1 only]2[9]129.No, because V is not proportional to I OR not straight line through origin / (1)only conducts above 0.5 V / resistance changes1Use of R = 0.74 / current from graph (1)= 9.25 W [9.0 9.5 W] [Minimum 2 significant figures] (1)2Calculation of p.d. across R [8.26]Calculation of total resistance[109 115]Ratio R: ratio VE = SIR (1) I diode resistance [9]Correct substitutionsCorrect substitutions (1)103 W [100 106] (1)3 [If not vertical line, 0/2] (1)(1)2[Otherwise 0 0 ][8] 130.Potential difference = OR OR in words: work done in moving 1 coulomb of charge between two points. (1)1Unit: volt OR J C1 OR V (1)1Base units: kg m2 A1 s3 (1)(1)2 [2/2 possible even if final answers wrong for recognising that As = C J = Nm][4]131.Current in motor:= 300 000 W / 420 V= 714A [allow710][no u.e.] (1)1Problem:Overheating in wires OR circuit/motor becomes hotOR Need thick/large/heavy cablesOR other sensible comment (1)1 Why e.m.f. of battery must be more than 420 V:Mention of internal resistance (1)Detail e.g. loss of p.d. inside battery when current delivered/ lost volts (1)OR equations used correctly 2 marks2Overall efficiency of motor:K.E. gained = mu 2= 1160 107 2 J= 6.64 M J (1)Energy input = P t= 300 000 100 J= 30 M J (1) efficiency = out/in 100= 6.64/30 100 =22% (1)3 Reasons for energy losses: (1)Work (done) against air resistance (1)Work (done) against friction (1)Heating in wires of circuit (1)Heating, in battery (1)Heating in motor coils (1)Max 2OR other sensible comments e.g. sound[Friction or heat loss scores zero unless detailed e.g. heat due to friction/airresistance but heat to surroundings ][9]132.Why resistance changes:Wire lengthens OR crosssectional area OR diameter reduces (1)Use of R = rl/A to explain [R and l, R 1/A (1)2Advantage:A long length of wire OR small area OR multiple stretching (1)1Diagram: Circuit with ammeter in series (1)voltmeter in parallel (with strain gauge) (1)OR multimeter across strain gauge (1)(1)[Multimeter with power supply 1 only]Max 2 Resistance:R = r l /A= 4.9 107 W m 0.2 m/p (2 10 4 m /2)2 (1)[i.e. area = (1)] (1)= 3.1W (1)3[8] 133.Threshold wave:Electron requires certain amount of energy to escape from surface (1)This energy comes from one photon (1)Use of E = hf (1)(So photon needs) minimum frequency (1)Hence maximum wavelengthOR use of E = hc/l (1)Max 4 Work function:f = c/l = 3.0 108 / 700 109 m (1)= 4.28 10 14 Hz (1)E = hf = 6.63 1034 J s 4.28 10 14 Hz = 2.84 1019 (J) [Allow e.c.f.] (1)3Circuit :Circuit showing resistors only in series (1)Potentials labelled (1)[Use of potential divider allowed]Resistor values 1: 1: 1 OR 1:2 (1)Max 2Suggestion:Cosmic rays travel more slowly than light (1)1[10] 134.Definition of symbols:n=number of electrons/carriers per unit volume (per m3)ORelectron (or carrier) density (1)u=average (OR drift) velocity (OR speed) (1)2RatioValueExplanation1Same material (1) (1)1Connected in series/Kirchoffs 1st law/conservation of charge/current is the same (1) (1)2A is halved so n double[Accept qualitative, e.g. A so v , or goodanalogy] (1) (1)6 [Accept e.g. ny = nx.....][No e.c.f ][NB Mark value first, without looking at explanation. If value correct, mark explanation. If value wrong, dont mark explanation except: if uy/ux = or 1:2, see if explanation is correct physics, and if so give (1). No e.c.f.][8]135.Demonstration that resistance is 0.085 W:R =rl/A (1)=1.7 108 Wm 20 m / (4.0 106 m2) (1)2Calculation of voltage drop:V=37 A 0.085 W (1)=3.1 V 2 = 6.3 V [Not if Vshower then found] (1)2[Only one conductor, leading to 3.1 V, gets 1st mark][Nothing if wires in parallel] Explanation:Lower resistance/R = 0.057 W/less voltage drop/new V=old V (1)Power dissipated in cable/energy wasted/wire not so hotOR more p.d/current/power to showerOR system more efficient (1)2[6] 136.Explanation: .I= E/r + R (1)1 Appropriate formula for cell E9:C9 * D9 OR RI OR 1 4 A (1)1Appropriate formula for cell F 11D11 *E11 OR VI OR 3A 6V OR C11 * D11 *D11OR RI2 OR 2 (3 A)2 (1)1Short circuit current:6 A (1)1 Explanation:r and R in series potential division (1)1as R , r constant R has greater share of 12 V (1)1OR other valid argumentSketch graph of power against resistance: 18 (1)2 (1)Shape including asymptote (1)3 Comment:Maximum when R = r (1)in accordance with maximum power theorem (1)OR P 0 as R (1)Max 2[11]137.(a)Mark the method before marking the circuitSuitable circuitShort circuit optionWhat is measured Set of readings of V and IV and ITwo sets of V and IV and IWhat is then donePlot V against IRecord V for open circuitSubstitute inV = E IrRecord V for open circuitFinding E and rE = interceptr = gradientE = open circuit voltager from V = EIrSolve simultaneous equationsE=open circuit voltage r from r=EIr-Suitable circuitPotentiometerWhat is measuredV for known RI for known RTwo sets of I and Rl for known RWhat is then doneRecord V for open circuitRecord V for open circuitSubstitute in E = I(R + r)l for open circuitFinding E and rE = open circuit voltage r fromE/V = (R + r)/RE = open circuitvoltager fromE = I(R + r)Solve simultaneous equationsE from l (calibrated)Mark other procedures in a similar way4 [Mark text, then tick for circuit if it does the job described.If diagram alone, ask if it can do the job and give mark if yes](b)(i)p.d. across battery:V = E - Ir= 12.0 V - 3.0 A 3.0 W (substitution)= 3.0 V2 (ii)Straight line from (0,12) to (3,3) (e.c.f.)1Current: 2.05 to 2.10 A1[8][Allow correct intersection of their line (ignore shape), 0.05 A, of the characteristic with their graph, even if theirs is wrong. A line MUST be drawn for the last mark.] 138.Explanation of variation shown on the graph:More electrons set free. Any one from: as temperature increases; thermal energy/vibration increases/resistance decreases/current increases2 Resistance of thermistor:V (across thermistor) = 1.20 VResistance ratio = voltage ratioR = 495 WorI = 0.80 V/330 W (substitution)= 0.002424 AV across thermistor = 1.20 VR = 1.20 V/0.002424 A= 495 WorI = 0.80 V/330 W= 0.002424 AR(total) = 2.0 V/0.002424 A= 825 WR = 825 W - 330 W= 495 W3Explanation:Thermistor resistance lowWhy: thermistor hotter/more current, power, charge carriersWhy v. small: thermistor takes smaller fraction of p.d. or ratio of p.d.3[8] 139.Completion of a correct circuit diagram:Ammeter in series with lamp and supply [Ignore voltmeter position]Voltmeter across lamp and ammeter [and maybe with ammeter2 Measurements:Record voltmeter readingRecord corresponding ammeter reading [corresponding may beimplied]Repeat for range of supply voltage settings [or currents]3Labelled sketch:Label axes I and V [with or without units]Graph line with correct curvature [overlook any tendency of the currentvalue to saturate]Show 12 V, 2 A correctly [Allow 12 and 2 if units are labelled on axes]3[The second mark is lost if axes are not labelled, unless 2 A and 12 V arepresent, with the units, to make sense of the axes.][8] 140.A[1] 141.D[1] 142.B[1] 143.A[1] 144.See c = 3 108 (m s1)1converts MHz to Hz1 = 3.13 m1 Example of calculation = (3 108 m s1) / 95.8 106 Hz = 3.13 m[3]145.QWC spelling of technical terms must be correct and the answer must beorganised in a logical sequenceIdentifies two rays of light (1)Two rays have same frequency/come from same source/are coherent (1)Path difference (between the two reflected rays) (1)They superpose (when they meet) /constructive and destructiveinterference occur (1)If they meet in phase/n / path difference, constructive interference/bright fringe (1)If they meet in antiphase / (n+ ) / path difference, destructiveinterference/dark fringe (1)Max 5[5] 146.(a)n = sin 48 / sin 301n = 1.5 (common answer will be 1.49)1(n = 0.67 scores 1 mark for idea of ratio of sin of angles) (b)(i)QWC spelling of technical terms must be correct and the answermust be organised in a logical sequenceAs x increases, y increasesORAt a certain angle / critical angle, y = 90 / the light travels along theboundary (do not allow reflects at 90)1For angles greater than the critical angle (in glass)1total internal reflection occurs1(do not accept TIR) (ii)Use of sin c = 1/n1c = 421ecf n from (a) unless n= 0.67 which scores 0 here[7] 147.(a)(i)v = f (words or symbols not numbers)1length of string = /2 OR wavelength = 2 lengthOR node to node = /21(ii)d2/4 OR (D3/2)2 OR (D3/2)^2(this mark is lost if there is a * length / A3)(ignore powers of ten)1 (iii)E4 * density OR E4 * 7800(ignore powers of ten)ORvolume of 1 metre length density1 (iv)5.12(spreadsheet answers must be correct to same number dec places sodo not accept 5.116 or 5.11)(correct answer on spreadsheet scores mark irrespective of whatswritten on next page)1 (v)See T = v2 OR T =v (not just quoting given equation)1T = 82 (N)1(do not penalise dec places twice, 82.1 could score both marks ifmore than 3 dec places given in (iv))(correct answer on spreadsheet scores both marks) (b)Plot a graph of v T, v2 T, f T, or f 2 T1Graph should be a straight line through the origin1Statement of what gradient equals (consistent with what has beenplotted)1(For this experiment is a constant. A graph using a variable canscore max 1 mark for the correct gradient)[10] 148.(a)Wavelength of microwaves < wavelength radiowavesOR statement that wavelength of radiowaves is larger1(need some comparison and do not credit frequency)Less / no diffraction OR beam spreads out less1 (b)(i)Frequency (of reflected wave) would be higher1 (ii)There is a link between frequency (change) and speed(ing)1(Car is speeding) when frequency (change) exceeds some limit1(answers may be given in terms of wavelength)(c)(i)Use of intensity area1 0.081energy/sec = 6 J or J s1 or W1 Example of calculationEnergy per second = 500 W m2 0.5 m 0.3 m 0.08 = 6 J (ii)Use of E = Pt with any relevant time e.g. 8 hours, 480 min or 28800s1t = 28800 s1Number of flashes = 1700Ecf answer to (c)(i) [Take their answer to (c)(i) and multiply by 288 tocheck their answer for full marks]1 Example of calculationNumber of flashes = (6 8 3600)/100 =1728[11] 149.D[1] 150.B[1] 151.B[1] 152.A[1] 153.C[1]154.(a)Unpolarised light oscillates/vibrates in many planes/ directions whilepolarised oscillates/vibrates in one plane/direction only OR labelled diagram1 (b)Filters at 90 to the (polarised) reflected light.1sunglasses cut out the reflected light/polarise light/glare1But not the light from the fish OR light from fish is unpolarised.1 (c)Sound is a longitudinal wave OR sound is not a transverse waveOR oscillations in one direction already OR only transverse wavescan be polarised.1[5] 155.(a)ANY THREESound waves are longitudinal waves (1)Air molecules vibrate (1)Parallel to the direction of travel of the wave (1)In a series of compressions and rarefactions (1)3 (b)Frequency is the number of cycles/oscillations/waves per second/per unit timeOR number of cycles/oscillations/waves passing a point per second.1 (c)Use of v = f1V = 3000 m s11 Example of answerv = 1500 m s1 2 Hzv = 3000 m s1 (d)Animals detect infrasound / lower frequencies than humans / vibrationsthrough the ground1Infrasound travels faster than the tidal wave1[8] 156.(a)Coherent: Waves of constant phase relationship1Standing wave: no (net) transfer of energy OR pattern of nodes and antinodesOR points of maximum displacement and zero displacement1(b)QOWCWork must be clear and organised in a logical sequenceCalculation to show a path of 24 cm or 42 cm OR paths of 2 and 3.5 1Path difference is 1 OR divide path difference by 121Waves at in antiphase /180out of phase/ radians out of phase1destructive interference1 Example of answerOne path length = 18 cm + 6 cm = 24 cmOther path length = 30 cm + 12 cm = 42 cmPath difference = 42cm 24 cm = 18 cmNumber of wavelengths = 18/12 = 1.5 (c)Food moves through hot and cold spots1Over time period all parts of food receive similar amount of energy.1[8] 157.(a)Diagram:Smaller wavelength before gap1Less diffraction and same wavelength1 (b)Two sets of concentric circles1equal spacing1Identification of a line of points of destructive interferenceIdentification of a line of points of constructive interference1 (c)(i)Attempt to use inverse relationship (e.g. 1.2 0.60 =constant)1Separation =1.8 mm1 Example of answer1.2 = constant / 0.6Constant = 0.72Spacing = 0.72/ 0.4 = 1.8 mm(ii)(Initially bands) will get close together1Eventually gap too large for overlap to occur, no fringes seenOR reference to fringes produced providing overlap still occurs1[10] 158.D[1] 159.B[1] 160.C[1] 161.B[1] 162.D[1] 163.A[1] 164.(a)Diffraction is the spreading out of the wave1As it passes through an aperture/around an obstacle1 (b)(i)Electrons can behave as waves OR electrons have wave like propertiesOR electrons act like wave particles1(ii) spacing/gap between atoms OR the size of the atoms ORspacing/gap in the graphite1[4]165.(a)Doppler1 (b)Ambulance moving towards,higher frequency/pitch(1)Wavelength shorter/waves bunch together(1)Ambulance moving awaylower frequency/pitch(1)wavelength increased/waves spread out(1)(wavelength marks may be awarded on a diagramMax 3 (c)Reference to a higher/lower frequency/wavelength/pitch scores 1Change in frequency is greater OR even higher/lower frequency ORrange of frequencies greater scores 22[6] 166.(a)Ray drawn along edge of prism (labelled X) (ignore areflected ray)1 (b)(i)n = 3 108 1.96 1081n = 1.53(no unit, ue if one given)1(ii)Use of sin (critical angle) = 1/ n OR use of sin i/sin r = v1/v2 = n1c = 411 (c)Red light: refraction towards normal at first face but lessthan refraction for blue light1Refracts into air at second face with angle in air > angle in glass1[7] 167.(a)The answer must be clear, organised in a logical sequenceand uses specialist vocabularyInterference (pattern) produced / superposition occurs/standing wave formed1Maxima related to constructive interference/antinode and/orminima related to destructive interference/node1Maxima/antinode formed where the waves are in phase / pathdifference n1Minima/node formed where the waves are in antiphase / pathdifference = ( n+)[out of phase is not sufficient]1(b)(i)Distance between adjacent maxima = /21Wavelength = 0.1 m1 (ii)Use of v = f with their from (b)(i)1Speed = 330 m s1 ecf their 1Example of answerv = 3300 0.1v = 330 m s1 (c)(i) (ii)(mark (i) and (ii) as one section(minima never zero) because there is not completecancellation/overall displacement is not zero/ not totaldestructive interference1Because the waves have different amplitudes/amplitudedecreases with distanceORenergy loss due to reflection or spreading out1ORreflection off other surfacesAs the microphone moves towards the plate, the pathdifference decreases1Amplitudes (of waves) get similar1[12] 168.B[1] 169.D[1] 170.B[1]171.B[1] 172.B[1] 173.D[1] 174.Point A is half a wavelength from X (1)At Y arrow drawn vertically downwards (1)B marked at one of three positions of max displacement (1)[3] 175.QWC Work must be clear and organised in a logical manner usingtechnical wording where appropriateAny threeReflected light is polarised (1)Polarised light vibrates/oscillates in one plane/direction (1)Polaroid filter only allows vibrations/oscillations in one direction/plane topass through (1)When planes are parallel puddle appears light OR when perpendicular puddleappears dark (1)max 3(for 2nd and 3rd mark only one reference to vibrations/oscillations is needed)(candidates who make no reference to puddle and answer in terms of twofilters can score 2nd and 3rd marks only)[3] 176.(a)(i)Node correctly placed (1)1 (ii)Arrow at Y moving up (1)Arrow at Z moving down (1)2 (b)Identifies a factor of 3 (1)Fundamental frequency = 0.5 Hz (1)2[5]177.(a)Waves must have same frequency or wavelength (1)Waves must have same amplitude (1)Waves must be 180, wavelength, half a cycle, radians apart or inantiphase (1)3 (b)Noise of a vibrating object has a constant pitch/frequency (1)Speech/sound varies in pitch and/or amplitude (1)The idea of the difficult of matching a changing signal (1)3[6] 178.(a)Use of distance = speed time (1)Recognising distance travelled is twice the measurement or halves thetime given (1)Distance = 4.1 m (1)3Example of calculationDistance = (330 m s1 25 103 s) 2Distance = 4.125 m (b)One pulse must return before the next one is sentORSo that time interval between transmitted and received pulses canbe measuredORNo overlap between pulsesORNo interference between pulses1[4] 179.A[1] 180.C[1] 181.B[1]182.C[1] 183.D[1] 184.(a)D1 (b)WavelengthUse of v = f l (1)Use of f = 1/T (1)Answer T = [0.002 s] (1)[give full credit for candidates who do this in 1 stage T = /v]Example of answerv = flf = 330 / 0.66T = 1/f = 0.66 / 330T = 0.002 s3[4] 185.Direction of travel of light is water air (1)Angle of incidence is greater than the critical angle (1)2[2] 186.(a)Transverse waves oscillate in any direction perpendicular to wave direction (1)Longitudinal waves oscillate in one direction only OR parallel to wavedirection. (1)Polarisation reduces wave intensity by limiting oscillations and wavedirection to only one plane OR limiting oscillations to one direction only. (1)(accept vibrations and answers in terms of an example such as a ropepassing through slits)3 (b)Light source, 2 pieces of polaroid and detector e.g. eye, screen, LED ORlaser, 1 polaroid and detector (1)Rotate one polaroid (1)Intensity of light varies (1)3[6]187.Frequency unaltered (1)Wavelength decreases (1)Speed decreases (1)3[3] 188.The answer must be clear and the answer must be organised in a logicalsequence (QWCIt was known that X penetrated (1)It was not known that X rays were harmful (1)Doctors died because of too much exposure (1)Lack of shielding (1)New treatments may have unknown side effects (1)Treatments need to be tested / time allowed for side effects to appear (1)Max 4[4] 189.(a)[1.0 m] (1)1 (b)Ratio of (5 or 6 / 3 ) 60 (1)Answer [f = 100 Hz] (1)2[3] 190.Use of sin i / sin r = m (1)Use of either 80 or 1.33 (1)[r = 48] (1)3Example of answersin 80 / sin r = 1.33[r = 48]Both rays refracted towards the normalViolet refracted more than red2[5] 191.(a)Diffraction is the change in direction of wave or shape orwavefront (1)when the wave passes an obstacle or gap (1)2 (b)The energy of the wave is concentrated into a photon (1)One photon gives all its energy to one electron (1)2(c)Energy of photon increases as frequency increases OR reference toE = hf (1)Electrons require a certain amount of energy to break free and thiscorresponds to a minimum frequency (1)2[6] 192.(a)(i)Use of speed = distance over time (1)Distance = 4 cm (1)Answer = [2.7 105 s] (1)Example of answert = 4 cm 1500 m s1t = 2.7 105 s3 (ii)Use of f = 1/T (1)Answer = [5000 Hz] (1)2 (iii)Time for pulse to return greater than pulse interval (1)All reflections need to reach transducer before next pulse sent. (1)Will result in an inaccurate image. (1) (Max 2)Need to decrease the frequency of the ultrasound. (1) (Max 3)Max 3 (iv)X-rays damage cells/tissue/foetus/baby but ultrasound doesnot (need reference to both X-rays and ultrasound) (1)1 (b)The answer must be clear, use an appropriate style and be organisedin a logical sequence (QWC)Doppler shift is the change in frequency of a wave when the source or thereceiver is moving (1)Requirement for a continuous set of waves (1)Two transducers required (one to transmit and one to receive) (1)Change in frequency is directly related to the speed of the blood (1)4[13] 193.(a)(i)Demonstrating the stationary waveMove microphone between speaker and wall OR perpendicular to wallOR left to right OR towards the wall [could be shown by labelledarrow added to diagram] (1)Oscilloscope/trace shows sequence of maxima and minima (1)2(ii)How nodes and antinodes are producedSuperposition/combination/interference/overlapping/crossingof emitted/incident/initial and reflected waves (1)Antinodes: waves (always) in phase OR reference to coincidenceof two compressions/rarefactions/peaks/troughs /maxima/minima,hence constructive interference/reinforcement (1)Nodes: waves (always) in antiphase/exactly out of phase ORcompressions coincide with rarefactions etc, hence destructiveinterference / cancellation (1)3 (iii)Measuring the speed of soundMeasure separation between (adjacent) nodes / antinodes anddouble to get /this is [not between peaks and troughs] (1)Frequency known from/produced by signal generator ORmeasured on CRO / by digital frequency meter (1)Detail on measurement of wavelength OR frequencye.g. measure several [if a number is specified then 3] nodespacings and divide by the number [not one several times]OR measure several (3) periods on CRO and divide by the numberOR adjust cro so only one full wave on screen (1)Use v (allow c) = f4 (b)(i)Application to concert hallLittle or no sound /amplitudeOR you may be sat at a node (1) (ii)Sensible reasonExamples:Reflected wave not as strong as incident waveOR walls are covered to reduce reflectionsOR waves arrive from elsewhere [reflections/different speakers]OR such positions depend on wavelength / frequency (1)2[11] 194.(a)(i)Condition for reflectionAngle of incidence greater than critical angle [accept i > c] (1)1(ii)Description of path of rayAny two from:Ray refracted at A and CDescription of direction changes at A and CTotal internal reflection at B (1)(1)2 (b)(i)Things wrong with the diagramAngle of refraction cant be 0 / refracted too much (1)No refraction on emergence from prism (1)[Allow 1 mark for correct reference to partial reflection]2 (ii)Corrected diagramemergent ray roughly parallel to the rest of the emergent rays (1)direction of refraction first surface correct (1)direction of refraction second surface correct (1)3[8] 195.(a)(i)Add standing waves to diagramsMark for each correct diagram (1)(1)2 (ii)Mark place with largest amplitude of oscillationantinode marked [allow clear indication near centre of wave otherthan an X, allow correct antinode shown on diagrams B or C] (1)1 (iii)Name of place marked(Displacement) Antinode [allow ecf from (a) (ii)] (1)1 (b)(i)Calculation of wavelengthCorrect answer [5.6 m]Example of calculation:= 2 2.8 m= 5.6 m (1)1(ii)Calculation of frequencyRecall of v = f (1)Correct answer [59 Hz] [ecf] (1)Example of calculation:v = f f = 330 m s1 / 5.6 m= 58.9 Hz2 (c)(i)Explanation of difference in soundas the room has a standing wave for this frequency / wavelength /it is the fundamental frequency(allow relevant references to resonance) (1)1 (ii)Suggest another frequency with explanationAppropriate frequency [a multiple of 59 Hz] [ecf] (1)Wavelength 1/2, 1/3 etc (stated or used) (1)2 (d)Explain change in frequencieswavelengths (of standing waves) bigger / f = v/2l (1)hence frequencies smaller/lower (1)2[12] 196.(a)Angles:Normal correctly added to raindrop (by eye)An angle of incidence correctly labelled between normal andincident ray and an angle of refraction correctly labelledbetween normal and refracted ray2 (b)Angle of refraction:Use of = sin i / sin rCorrect answer [20][allow 2021 to allow for rounding errors]eg.sin r = sin 27/ 1.3r = 202 (c)(i)Critical angle:The angle beyond which total internal reflection (of the light)occurs [allow T.I.R] / r = 901(ii)Critical angle calculation:Use of = 1 / sin CCorrect answer [50.3] [allow 50 51]Eg.Sin C = 1/1.3C = 50.32 (d)Diagram:i = 35 [allow 33 37]Ray of light shown refracting away from normal on leavingraindropSome internal reflection of ray also shown with i = r [by eye]Reflected ray shown refracting away from the normal as it leavesthe front of the raindrop / angle of refraction correctlycalculated at back surface4 (e)Refractive index:(Red light has) lower refractive index (than violet light)1[12] 197.(a)(i)How we know the speed is constantCrest spacing constant / circular crestsOr wavelength constant / equal wavelength (1)[Accept wavefront for crests][Dont accept wave]1 (ii)Calculation of speedl is 10 mm (1)[Allow 9 to 11]Use of v = fl (1)0.40 m s1 (1)[Allow 0.36 to 0.44Allow last two marks for correct calculation from wrong wavelength]3(40Hz)(10 103 m)= 0.40 m s1 (b)Line X1st constructive interference line below PQ, labelled X (1)[Accept straight lineIgnore other lines provided correct one is clearly labelled X]1(c)(i)Superposition along PQConstructive interference / reinforcement / waves of largeramplitude / larger crests and troughs (1)Crests from S1 and S2 coincide / waves are in phase / zero phasedifference / zero path difference (1)Amplitude is the sum of the individual amplitudes (OR twice theamplitude of the separate waves) (1)3 (ii)TableA constructive (1)B destructive (1)2[10] 198.(a)Amplitude(i)Amplitude remains constant (1)1(ii)Amplitude decreases then increases (1)Amplitude is zero at node (OR half way between X and Y) (1)2 (b)Phase difference(i)Phase difference increases / is proportional to distance XP (1)1(ii)Up to node phase difference is zero / in phase (1)Beyond the node phase difference is / 180 / half a cycle / inantiphase (1)[Do not allow completely out of phase]2[6] 199.(a)(i)Name processRefraction (1)1 (ii)Explanation of refraction taking placechange in speed / density / wavelength (1)1 (b)(i)Draw ray from butterfly to fishrefraction shown (1)refraction correct (1)2(ii)Explain what is meant by critical angleIdentify the angle as that in the denser medium (1)Indicate that this is max angle for refraction OR total internalreflection occurs beyond this (1)[angles may be described in terms of relevant media]2 (iii)Explain two paths for rays from fish A to fish Bdirect path because no change of medium/refractive index/density (1)(total internal) reflection along other path /angle of incidence > critical angle (1)direct ray correctly drawn with arrow (1)total internal reflection path correctly drawn with arrow (1)[lack of ruler not penalised directly] [arrow penalised once only]4[10] 200.(a)Ultrasound:High frequency sound / sound above human hearing range / soundabove 20 kHz / sound too high for humans to hear (1)1 (b)(i)Pulses used:to prevent interference between transmitted and reflected signals /allow time for reflection before next pulse transmitted / to allow forwave to travel to be determined (1) (ii)High pulse rate:Greater accuracy in detection of preys motion / position / continuousmonitoring / more frequent monitoring (1)2 (c)Size of object:Use of = v/f (1)Correct answer (0.0049 m or 4.9 mm) (1)[accept 0.0048 m or 0.005 m]example: = 340 m s1/ 70000 Hz= 0.0049 m = 4.9 mm (accept 5 mm)2(d)Time interval:Use of time = distance / speed (1)Correct answer (2.9 103 s) [allow 3 103 s][allow 1 mark if answer is half the correct value ie. Distance = 0.5mused] (1)example:time = 1 m / 340 m s1= 2.9 103 s2 (e)Effect on frequency:Frequency decreases (1)Greater effect the faster the moth moves / the faster themoth moves the smaller the frequency (1)2[9] 201.(a)Diffraction diagram:Waves spread out when passing through a gap / past an obstacle(1) stays constant (1)2 (b)Diagrams:Diagram showing 2 waves in phase (1)Adding to give larger amplitude (1)2 (c)Information from diffraction pattern:Atomic spacing (similar to )Regular / ordered structureSymmetrical structureDNA is a double helix structure (2)Max 2 (d)Electron behaviour:(Behave) as waves (1)1[7] 202.(a)(i)Diagram:i and r correctly labelled on diagram (1)i = 25 +/ 2 (1)r = 38 +/ 2 (1)[allow 1 mark if angles measured correctly from interfaceie. i = 65 +/ 2, r = 52 +/ 2] (1)3(ii)Refractive index:Use of ga = sin i / sin r [allow ecf] (1)Use of ag = 1/ga (1)example:ga = sin 25 / sin 38 = 0.686ag = 1/ ga = 1.462 (b)Ray diagram:Ray added to diagram showing light reflecting at interface withangles equal (by eye) (1)1 (c)Observation:Incident angle > critical angle (1)T.I.R occurs (1)2 (d)largest angle:sin C = 1/1.46 (allow ecf) (1)C = sin1 (1/1.46) = 43 (1)2[10] 203.(a)Experiment[Marks may be earned on diagram or in text]Named light source plus polaroid (OR polariser ORpolarising filter) / Laser / Named light source andsuitable reflector (e.g. bench) (1)2nd Polaroid plus means to detect the transmitted light (1)(i.e. eye OR screen OR LDR OR light detector ORinstruction to e.g. look through polaroids)Rotate one Polaroid [Only award if expt would work] (1)Detected intensity varies / No light when polaroids are at 90 (1)Maxima and minima 90 apart / changes from dark to light every 90 (1)[Use of microwaves, slits or blockers: 0/5Use of filters or diffraction gratings: lose first two marksUse of sunglasses to observe: lose mark 2]5(b)Why sound cant be polarisedThey are longitudinal / They are not transverse / Only transversewaves can be polarised / Longitudinal waves cannot be polarised /Because the (*) is parallel to the (**) (1)(*) = vibration OR displacement OR oscillation OR motion of particles(**) = direction of travel OR direction of propagation OR motion ofthe wave OR direction of energy transfer1[6] 204.(a)(i)Tablef2.4(110)1.22200.8330All wavelengths correct (2)[One or two wavelengths correct gets 1]Both frequencies correct (1)[Accept extra zero following wavelength figure, e.g. 2.40.Accept units written into table, e.g. 2.4 m, 220 Hz]3 (ii)Why nodesString cannot move / no displacement / zero amplitude /no oscillation / phase change of on reflection / two wavescancel out / two waves are exactly out of phase (1)(OR have phase difference of OR half a cycle) /destructive interference1 (b)Why waves with more nodes represent higher energiesMore nodes means shorter wavelength (1)Momentum will be larger (1)[OR Allow 1 mark for More nodes means higher frequency and E = hf]2[6]205.(a)Which transitionUse of ()E = hc/ OR ()E = hf and f = c/ (1)Use of 1.6 1019 (1)Correct answer [1.9 eV] (1)C to B / 1.5 to 3.4 (1)[Accept reverse calculations to find wavelengths]e.g.(6.63 1034 J s)(3.00 108 m s1)/(656 109 m)(1.6 1019 J eV1)= 1.9 eV4 (b)Explanation of absorption lineQOWC (1)Light of this wavelength is absorbed by hydrogen (1)In the outer part of the Sun (OR Suns atmosphere) (1)Absorbed radiation is reemitted in all directions (1)Transition from B to C (OR 3.4 to 1.5) (1)Max 4 (c)Why galaxy recedingWavelength increased (OR stretched) / red shift /frequency decreased1[9] 206.(a)Describe propagation of longitudinal wavesParticles oscillate / compressions/rarefactions produced (1)oscillation/vibration/displacement parallel to direction of propagation (1)2 (b)Calculation of wave speedRecall of v = f (1)Correct answer [7.2 km s1] (1)Example of calculation:v = f v = 9 Hz 0.8 km= 7.2 km s1 [7200 m s1]2(c)Determine if elephants can detect waves more quicklyRecall of v = s / t (1)Correct answer for t in minutes or hours [about 6 minutes] or