Model answers for the 0 Electricity Revision booklet: SAMPLE ASSESSMENT SCHEDULE Physics 973 (.6): Demonstrate understanding of electricity and electromagnetism Assessment Criteria Achievement Achievement with Merit Achievement with Excellence Demonstrate understanding involves writing statements that show an awareness of how simple facets of phenomena, concepts or principles relate to a described situation. Demonstrate in-depth understanding involves writing statements that give reasons why phenomena, concepts or principles relate to a described situation. For mathematical solutions, the information may not be directly usable or immediately obvious. Demonstrate comprehensive understanding involves writing statements that demonstrate understanding of connections between concepts. Evidence Statement Not achieved Achievement Achievement with Merit Achievement with Excellence One NØ N No response; no relevant evidence ONE correct A3 ONE correct M5 ONE correct E7 ONE correct of E ONE correct of A N TWO correct A4 TWO correct M6 TWO correct E8 ONE correct of E ONE correct of M The electron gains kinetic energy. The electron gains potential energy. Describes that an electron that is free to move is attracted to the positive plate, but no mention of Describes the electron gains kinetic energy when free to move in an electric field. Describes the electron gains potential energy when forced to move against Demonstrates in-depth understanding by stating the electron loses electric potential energy and gains kinetic energy when moving in the electric field AND P a g e
the energy changes involved. the electric field Demonstrates in-depth understanding by stating the electron loses kinetic energy and gains electric potential energy when forced to move against the electric field. Solves: Solves: Solves: E V d E F q E V d 00 0.004 E = 5.0 0 4 Vm E p Fd E p 8.0 0 5 0.004 E p = 3. 0 7 J 3. 0 7 0.5 mv v 7. 0 9 v = 8.4 0 6 ms F Eq F 5.0 0 4.6 0 9 E k = 3. 0 7 J F 8.0 0 5 No force. Force increases or force decreases. Describes that the force is the same throughout. Describes the size of the force being much larger than the weight force of an electron. Explains the force is the same throughout because the electric field is uniform. P a g e
Not achieved Achievement Achievement with Merit Achievement with Excellence Two NØ N No response; no relevant evidence ONE correct A3 TWO correct M5 TWO correct E7 ONE omission N TWO correct A4 THREE correct M6 THREE correct E8 ALL correct 4 5 9 3 4 5 4 3 5 4 5.. 4 5 3 5 5. 7. Attempts calculation but uses incorrect value for resistance. I V R 9.0 7. I =.5A V 3 = 3.5 = 3.75 V V =.5 =.50 V V 5 = 9.0 (3.75 +.50) V 5 =.75 V V =..5 =.77 V The 5 resistor draws more power. The greater the resistance, the greater the voltage across it for the same current. Describes power depends on both voltage and current. Describes ONE correct calculation for power. This could be either 3 resistor = 3.75.5 = 4.69 W Explains by calculating power drawn by 3 resistor = 3.75.5 = 4.69 W Explains by calculation power drawn by 5 resistor Explains in detail power depends on both voltage and current. P = VI Voltage across 5 resistor is.75 V and the voltage across 3 resistor is 3.75 V. The current through 3 resistor is.5 A. Hence power drawn 3 P a g e
power drawn by 5 resistor = V.75.75 R 5 =.5 W Identifies 3Ω resistor as using more power due to it drawing more current having a higher voltage across it. = V R =.5 W.75.75 5 Explains that power depends on voltage and current and the 3Ω resistor draws more current AND has a higher voltage across it. by 3 resistor is 4.69 W. Power drawn by 5 resistor is.5 W. Explains in detail both the voltage across and the current through the 5 resistor is less than that of the 3 resistor. Hence the 3 resistor draws more power from the battery. Not achieved Achievement Achievement with Merit Achievement with Excellence Three NØ N No response; no relevant evidence TWO correct A3 ONE correct M5 ONE correct E7 ONE omission N THREE correct A4 TWO correct M6 TWO correct E8 BOTH correct The electrons feel a force. This causes the electrons to move. Describes that the electrons feel a magnetic force and so move to one end of the wire leaving the other end positive. Explains the separation of charge causes an electric field to be formed. The electrons will also experience a force due the formation of the electric field. Explains in detail that the force due to the electric field is opposite to the force due the magnetic field as experienced by the electrons. AND Describes that the electrons feel a magnetic force and states correct direction of the force (to the left). Charge separation continues until the magnetic force is equal and opposite to the electric force. B BvL V 0.80 65 Calculates: V 0.80 0.65 4 P a g e
Current is anticlockwise. V = 6.4 V (or consequential error accepted for M in 3 ) Calculates the current correctly. Shows states that the current is anticlockwise: V = IR Calculates: I 6.4 4.5 I =.39 A AND Shows/states correct direction. 5 P a g e
QUESTION 4: ELECTROSTATIC SWING (NCEA 0, Q) Q Evidence Achievement Merit Excellence ONE The left hand plate, A. Field lines go from positive to negative. ONE part correct. BOTH parts correct. Field lines show the direction a positive test charge would move. The field lines are equally spaced. Correct answer. E = V d V = Ed Correct working except for ONE error. Correct answer. V = 3.33 0 6 0. V = 399 600 = 4.0 0 5 V = 400 kv (d) V = DE q DE = Vq DE = 4 0 5.5 0-0 Correct calculation of energy change. F = 4.995 0 4 Correct working with ONE error. a = 0.0998 Correct working and answer. DE = 6 0-5 J E K = mv = 6 0-5 J v = 6 0-5.5 0 - = 4.8 0-3 = 0.069 m s - 6 P a g e
(e) When the ball touches the negative plate, it will gain electrons until it has an overall negative charge. It then experiences a force in the opposite direction to the field ( is attracted to the positive plate is repelled from the negative plate). ONE correct idea. Eg moves towards positive plate TWO correct ideas. Moves towards + and attraction/force/ repulsion Full explanation linking the charging process and the force due to the field. M+ Electron movement and repetition When the ball touches the positive plate, it loses electrons until it has an overall positive charge. It then experiences a force in the same direction as the field ( is attracted to the negative plate is repelled from the positive plate). QUESTION 5: STATIC ELECTRICITY (NCEA 00, Q) Q Evidence Achievement Achievement with Merit Achievement with Excellence ONE Upward arrow (. E V d 0.0 E 3.0 0 6.667 3 03 ÊVÊm Correct working and answer without the unit. Correct answer including correct alternate unit N C. Correct answer Alternate unit is NC 7 P a g e
(d)(i) (ii) At the negative plate, the electron has electric potential energy. As it goes towards the positive plate electric potential energy is changed to kinetic energy. The electron accelerates towards the positive plate. Idea of EITHER the electron possessing electric potential energy at the negative plate. Electron gaining kinetic energy as it approaches the positive plate. Electron accelerating towards positive plate. Potential to kinetic + accelerating down/towards positive plate. (e) As it reaches the positive plate, E p = E k E p Eqd E p 6.667 0 3.6 0 9 3.0 0 3 E p 3.0 0 8 Recognition that E p = E k Finds F =.07 0 5 Correct except for one error. Eg finds a =.9 0 5 uses d = 3 Correct answer for speed of electron. 3.0 0 8 9.0 0 3 v v.67 0 6 ÊmÊs 8 P a g e
QUESTION 6: SPRAY PAINTING (NCEA 005, Q) 9 P a g e
QUESTION 7: MASS SPECTROMETER (NCEA 009, Q) Q Evidence Achievement Merit Excellence ONE Arrow towards top of page. Correct answer. Correct charge. Correct answer. The force remains constant because the electric field strength is constant (F = Eq).. Correct answer (d) If the voltage increases, the force on the ion increases. Greater force means greater acceleration, which means greater maximum velocity. Greater velocity. Except if based on V=Bvl, Achievement plus partial explanation. Correct answer and full and concise explanation. If the voltage increases, the ion gains more kinetic energy, and therefore has a greater velocity. (e) F Eq Correct answer. E F q 3. 0 5.6 0 9.0 04 ÊNÊC (orêvêm ) (f) KE mv E Eqd.0 0 4.6 0 9 0.04 E.8 0 6 ÊJ Correct KE. Correct working except for one error. Correct working and answer. mv.8 0 6 5.3 0 6 v.8 0 6 v.8 0 6 5.3 0 6 v 69Ê434ÊmÊs Ğ v 6.9 0 4 ÊmÊs 0 P a g e
QUESTION 8: THE PARTICLE ACCELERAT (NCEA 007, Q) Q Evidence Achievement Achievement with Merit Achievement with Excellence ONE Left to right. Correct answer. Electrical/potential to kinetic. Correct answer. Electrical to kinetic Potential to kinetic E k m v f m v i E k.67 0 7 ((8.8 0 5 ) (6. 0 5 ) ) Attempts to use or states E Eqd E k 3.5 0 6 ÊJ Calculates a kinetic energy. Calculates the gain in energy correctly uses E Eqd Correct working and answer. E E k qd 3.5 0 6.6 0 9 0.0.0 05 ÊVÊm uses F = ma Or Finds a v f v ad gives a 9.75 0 F ma and F Eq give E.67 0 7 9.75 0.6 0 9 0765 00 000 (d) N C Correct unit. (e) V Ed 00 000 0.0 000ÊV Correct answer. (f) (g) Towards the top of the page. F Bvq F 3.5 0 3 8.8 0 5.6 0 9 F 4.9 0 6 UnroundedÊisÊ4.98 0 6 Correct answer. Upward. Correct answer Accept correct substitution into formula. P a g e
sig figs. Correct sf. For any attempt to find F. QUESTION 9: CHARGED PARTICLES (NCEA 008, Q) Q Evidence Achievement Achievement with Merit Achievement with Excellence ONE Downward line. Evenly spaced parallel lines with curved end(s). + Curves towards negative plate. - The electric force is at right angles to the direction in which the positive particle is moving. This causes the particle to describe a parabolic path. Force is down. Repelled from +. Attracted to. Achievement plus has link to forward motion or constant downwards force and parabolic path (d) Magnetic field into the page. Correct answer. (e) The electric force depends only on the electric field strength and the size of the charge. Hence is not affected by the velocity of the particle. The magnetic force F = Bqv increases as the velocity of the particle increases, as the magnetic force is directly proportional to the velocity, provided the magnetic field strength is a constant. Electric force is not affected by the velocity Magnetic force increases as velocity increases. Electric force is not affected by the velocity, but the magnetic force increases as the velocity increases. Merit, plus F depends only on E and q; eg F=E q AND F = Bqv depends on v. (f) Correct formula used to find E, but did not convert cm to m.(e=44 ) Correct value for E (4400) F using cm Correct answer. 7.0 0 8 F = 7.0 0 8 N Correct sig figs. P a g e
Any attempt to find F correct to sf (g) Correct except for charge, eg 3.5 05 I 0 3.5 0 4 Correct answer. 5.6 0 5 A QUESTION 0: CATHODE RAY TUBE (NCEA 006, Q) Upper plate is positive. Top plate positive + + Correct direction Evenly spaced parallel lines. Curved ends. Direction and: Evenly spaced parallel lines. Curved ends. Direction with evenly spaced parallel lines and Curved ends. E V d E = 45 8.0 0 3 E = 45 8.0 0 3 (d) One unit correctly derived. any correct E = V Ê(V) dê(m) E = N C E = J C m - (e) F = Eq 9 0 6 F = 565.6 0 9 N F = 9.0 0 6 N 3 P a g e
(f) The electron experiences an electric force and is moving in the same direction as the electric force, hence it is losing electrical potential energy but gaining kinetic energy as it accelerates. the electron experiences work / attraction / repulsion / force / acceleration. the electron moves in the electric field / to the positive plate. any correct linked ideas All 3 correct linked ideas. Velocity or Kinetic energy increases. QUESTION : CAMP TCH (NCEA 006, Q) Q Evidence Achievement Achievement with Merit Achievement with Excellence V E q E = Vq.5 J (Can be answered from definition, so does not need to show working.) Correct..5 I V R I.5 5 I = 0.3 A Correct 0.3 Resistance is the slowing down of electrons as they flow through a conductor A correct concept. Slowing electron/ 4 P a g e
when the ends of the conductor are connected to a supply of electrical energy a measure of how much a component opposes the flow of electrons through itself current flow Opposing electron/ current flow ratio of V/I ratio of V/I V / Cs - (d) I Q t Q = It Q = 0.3 3 60 (consequential on ) Calculated without converting minutes to seconds 0.3 x 3 = 0.9 Correct answer with working. Q = 0.3 3 60 Q = 54 Q = 54 C 0.3 x 60 QUESTION : CLAIRE S CAR LIGHTS (NCEA 0, Q3) THREE When the bulb has V across it, the power output is 5 W. Correct answer. (Must link power of 5W to V) P = VI Correct answer. I = P V I = 5 = 0.4 A QUESTION 3: CAMP RADIO (NCEA 006, Q).5 3 = 4.5 V Correct answer. 4.5 R = 4 + ( 4.00 + 4.00 ) Adds resistors in Correct 5 P a g e
= 6.0 parallel ( 4.00 + 4.00 ).00 4 + 0.5 = 4.5 calculation. 4.0+.00=6.0 Sf = 3 3 significant figures Total current = I V R I = 4.5 6 Calculated total current. I = 0.8 Voltage across radio V = 0.8 4.0 V radio = 3.94 Working and answer correct. V lamp = 4.5 3.94 = 0.56 I = 0.8 A Voltage across radio I 4 ½ V lamp = 4.4 = 0.56 V = IR V = 0.8 4.0 V radio = 3.94 V Voltage across parallel resistance V parallel =.8 = 0.56 Voltage across lamp = 4.5 3.94 = 0.56 V 4.5 6 = 0.56 (consequential on (e),(f) ) QUESTION 4: DC ELECTRICITY (NCEA 00, Q) 6 P a g e
TW O Correct substitution. Eg /6 + /5 Correct calculation of effective resistance in series = 6.0 Correct except for one error. Eg finds.73. Correct answer. Correct answer. Consequential from. Correct answer to voltage across 3 resistor. 6.7 Correct answer to voltage across 5 resistor. 5.73 Correct answer..5 (d) The brightness of a lamp depends on its power output. Power depends on the current through and the voltage across the lamp. (P = VI or P = I R) The 3 lamp will be the brightest because its power output is the greatest. (P = 6.8.09 = 3. W). The current through the branch with the 4.0 resistor is only (.09.4) = 0.95 A. Hence the power output of that lamps will be 0.95 4.0 = 3.6 W Recognition that brightness of a lamp depends on its power output. Power depends on the current through and the voltage across a component. Shows the calculation for power for any one lamp in the circuit. Recognition that brightness of a lamp depends on its power output. AND Power depends on the current through and the voltage across a component. Shows the calculation for power for any one lamp in the circuit. Recognition that brightness of a lamp depends on its power output. AND Power depends on the current through and the voltage across a component. AND Shows the calculation for power for the two lamps in the circuit. QUESTION 5: THE MP3 PLAYER (NCEA 007, Q) 7 P a g e
TWO R V I 4.5 5 0 80Ê Correct except for 3 unit conversion. Correct answer. V.0 4.5 7.5ÊV Correct working. I t V R 7.5 4 0.035ÊA current through R 0.035 0.05 0.00ÊA R V I 4.5 0.0 450Ê Or Calculates total current. 0.035 Calculates current through R. 0.035-0.05=0.0 Correct method but makes a computational error All working correct. Needs 4.5/0.0=450 or equiv Or Evidence of solving R in parallel combination Circuit R = 0.035 34.4 Calculates Rt 34.4 R of parallel branch = 34.4 4 = 8.4 And R 8.4 80 gives R = 450 (d) Total resistance increases. (to 664) Total current decreases. (was 0.03 now 0.08) Current through 4 Ω resistor decreases. Voltage across 4 Ω resistor decreases (V= IR). Voltage decreases (across 4) Voltage increases across R (450 Ω) Total R increases Current decreases Two ideas Voltage decreases because current decreases Full and clear explanation clearly linking ideas. (Can have maths but needs written explanation) (was 7.5 now 3.85) (e) Both resistors are in series, therefore carry same current. 450 Ω resistor has higher resistance therefore higher voltage. (V = IR) Therefore higher power output, (P=VI), therefore more heat output in the same time P I R so same current means bigger resistor (450/R) gives more power and more heat. 450 Ω/R resistor produces more heat. Current through both 4 and R the same Biggest V gives biggest power 4 produces less Now a series circuit links power to heat Two linked ideas ie same current- higher V (gives more heat) same current so higher R (gives more heat) R as larger Full and clear explanation clearly linking ideas. Should mention that heat relates to power or (energy and volts)could be explicitly stated or by stating P=IV or P I R 8 P a g e
(f) Correct answer. 9 P a g e
QUESTION 6: ELECTRIC CIRCUITS (NCEA 009, Q) TWO I V R.0A Correct answer. R P = 4Ê R Tot = 0Ê I V R 0.ÊA current through Ê is. 0.4ÊA 3 One correct step. Eg 0 ohms or 4 from parallel. Correct answer except for one error. Eg. A Correct answer. V IR 0.4 4.8ÊV or V 4 4.8ÊV 0 If the resistance increases, the total resistance increases and the total current decreases. One correct idea. Must answer question correctly. Two correct ideas. Correct answer and explanation. This means the voltage across the two series resistors decreases. QUESTION 7: ELECTRIC CIRCUITS (NCEA 008, Q) TWO J Two grades here Correct number. Correct unit. This is a show question: 4.5 3.4 5. States R 3.4 5. This is a show question: Correct working. = 4.5 +.06 = 6.56 Correct answer. 0 P a g e
(d) Effective resistance of 3.4 and 5. =.06 and V=IR=.83.06 = 3.8 V Calculates voltage correctly. 3.8 V Voltage across 4.5 resistor is: V = IR = 8.35 V Voltage across the 3.4 resistor is 8.35 = 3.8 V (e) The voltage across the 5. resistor will also be 3.8 V, as it is in parallel with the 3.4 resistor. Mentions that the voltage is 3.8 V. Achievement, plus states that this is because it is in parallel with the 3.4 resistor. QUESTION 8: Mike s MotorBike (NCEA 005, Q) P a g e
QUESTION 9: ELECTROMAGNETISM (NCEA 006, Q3) 3 The moving electrons would experience a force and move towards A, leaving end B positive. moving charge experiencing a force. Electrons move up the rod. End A negative. End B positive. Moving electron experiences an upward force and either A negative or/and B positive. 3 Since the circuit is now complete, the induced voltage would cause an anticlockwise current in the circuit (or would cause electrons to flow in a clockwise direction). Current produced. Voltage produced. Electron flow. Voltage causes a current/ electron flow. Anticlockwise current. Clockwise electron flow. Voltage causes an anticlockwise current in the circuit. Voltage causes a clockwise electron flow. P a g e
3 V = BvL V = 0.8 4.0 0 0 Induced voltage calculated (ignore std form). Correct Voltage Correct current V = 0.3 V 3 0.3 0.6 I V R 6 I = 0.3 I = 0.6 A 3(d) Arrow going from N to S. Correct answer. 3(e) Up Correct answer. Up 3(f) When a conducting wire carrying current is placed perpendicularly in a magnetic field, the electrons moving in the wire experience a force causing the wire itself to move in a direction that is perpendicular to both the direction of the magnetic field, and the current. The charge is cutting across the field. Charge moving in the magnetic field. Current in the rod. Rod perpendicular to the magnetic field. Electrons travelling across the magnetic field. Current carrying rod across the magnetic field. The magnetic fields add / subtract. Explanation of the magnetic flux difference. Magnetic field around the rod. 3(g) F = BIL Correct answer. F = 0.90 3. 0.0 F = 0.88 N = 0.9 N 0.88 0.9 3 P a g e
Two sig figs. sig. figures. =.9 QUESTION 0: ELECTROMAGNETIC SWING (NCEA 0, Q) TWO Electrons flow from the negative terminal in the direction X Y These electrons are cutting across a magnetic field that is towards the bottom of the page. Each electron experiences a force in the direction A This causes the wire to experience a force and to swing in direction A can explain in terms of conventional current. The loop swings in direction A. Direction of charge flow and loop movement correct. Current direction and wire perpendicular/cros sing/cutting (not in) field. Full explanation. Loop movement + Current flow + perpendicular / cutting / across (not in). V = IR ONE correct calculation. All correct except for ONE error. Correct answer. I = V R I = 6.0.8 = 3.33 A F = BIl l = F BI = 0.5.0 3.33 l = 0.038 m Stops at 0.038 m. Must have = 3.8 cm. Electron direction C (or left). Correct answer. (d) V = Bvl Correct except for one error. Correct answer. v = V Bl v = 0-3 0.0375 v = 0.467 = 0.5 m s - QUESTION : MAGNETIC FIELDS (NCEA 009, Q3) THREE To the right. Correct answer. 4 P a g e
I V R 8 0.666 A F BIl (4 0 4 ) 0.66 8 Correct current. Correct answer except for one error. Eg no unit. Correct answer. F. 0 3 N No. The two wires carry current in opposite direction. The force on the two wires is in the opposite direction. No force. Except if because yachts are stationary. Currents are equal and opposite. No. Currents opposite, forces equal and opposite. The forces are equal and so they cancel. (d) Yes there is a voltage induced because the two wires are cutting across a magnetic field. This causes an induced voltage. Correct answer. Must convey movement at 90 deg to /across field. (e) F Bvq Correct answer. F 4 0 4 3.0.6 0 9 F.9 0 N QUESTION : THE MODEL RAILWAY (NCEA 007, Q3) THREE Battery causes electrons to flow in axle. These electrons are moving perpendicular to a magnetic field. The electrons experience a force perpendicular to the axle and the field. The electrons are trapped in the axle so the whole axle experiences a force. one idea Charge moving through a field experiences a force Current flowing makes magnetic field. Force on charges. moving in magnetic field results in force on the axles. The two magnetic fields interact and produce force (on axle). In / (arrow indicating left to right) Correct answer. 5 P a g e
F BIl F 0.05 I F Bl 0.06 0.5 35 0 3 I.97ÊA batteryêcurrent 5.94ÊA V IR V 5.94 0.55 V 3.3ÊV Correct equation and calculation of current..97 or rounded 5.94 Correct process for calculating voltage but with one error. Eg does not double =.6 / allow incorrect length conversion but not 35. Correct working and answer. Accept any rounding eg 3., 3.7 etc. (d) V Bvl V 0.5 0.9 35 0 3 V.5ÊmV Only penalise the same incorrect length conversion once from c and d Correct answer except for one error eg for unit conversion of either length or to mv incorrect length combining both axles. Correct answer. (e) As the carriage rolls, the axles (and the electrons) cut across the magnetic field, the electrons in the wire get pushed to one end of the wire. One correct idea. Force /push on electrons Charge moving through mag field Full and clear explanation clearly linking ideas. electrons then move / shift towards one end. This causes a build-up of negative charge at one end of the axle. (f) The axle has an induced voltage across it, but the connecting wire is also cutting across the magnetic field. It also has an induced voltage. The two voltages oppose each other, so the induced current is zero. No current flows, Induced voltage in axle Induced voltage in wire Idea of two induced voltages. lamp does not operate. Contradictory statements will not negate achievement. Full and clear explanation clearly linking ideas. Two opposite induced Voltages cancel. No change in flux as entire circuit / loop in field means no light / current. QUESTION 3: THE ELECTRIC MOT (NCEA 00, Q3) 6 P a g e
Correct answer to current..67 Correct use of F= BIL with incorrect value of current. Correct answer for force on a single turn. One mistake in calculation, eg missing cm conversion. Correct answer. Wire AD is parallel to the magnetic field. The wire does not cut the field. Or equivalent. Correct answer. Increase strength of magnetic field. Increase current / voltage / batteries. Increase length of coil or have more turns of wire. Not increase the length. Any TWO correct answers. QUESTION 4: Generator (NCEA 008, Q3) THREE V BvL 0.75 0.0 0.46 V 0.044ÊV Correct except for one mistake. Either incorrect unit conversion (4.4) or missing x (0.0) Correct answer. 0.044 Stronger magnetic field, longer length of wire in the field, increasing the speed with which the wire is made to move in the magnetic field. Correct answer. QUESTION 5: INDUCTION (NCEA 004, Q3) 7 P a g e
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L E+M Matching Answers:. ZZ. C 3. W 4. WW 5. TT 6. CC 7. Q 8. QQ 9. G 0. J. UU. S 3. F 4. XX 5. O 6. DD 7. BB 8. PP 9. AAA 0. FF. Z. RR 3. X 4. I 5. MM 6. SS 7. A 8. B 9. M 30. H 3. VV 3. I I 33. U 34. YY 35. Y 36. FF 37. NN 38. N 39. LL 40. P 4. V 4. CCC 43. KK 44. D 45. AA 46. OO 47. T 48. K 49. L 50. E 5. R 5. GG 53. BBB 54. EE 55. HH 9 P a g e
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