# Physical Sciences Grade 11 P1 North West November 2018 Memo > Grade 11 Physical Sciences Paper 1 memorandum from the North West November 2018, 16 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-grade-11-nov-2018-memo-afr-eng/e51c9a42-01e5-4262-8c0d-610862f1583c - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 11 - Document type: Memorandum - Year: 2018 - Exam period: North West November - Paper: 1 - Curriculum: CAPS - Pages: 16 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Physical Sciences](https://www.hlayiso.com/library/grade-11/physical-sciences) › [Physical Sciences Grade 11 P1 North West November 2018 Memo](https://www.hlayiso.com/library/physical-sciences-p1-grade-11-nov-2018-memo-afr-eng/e51c9a42-01e5-4262-8c0d-610862f1583c) ## About this document Physical Sciences · Grade 11 · North West November · 2018. Memorandum, 16 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2018 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 16 pages. Hierdie nasienriglyne bestaan uit 16 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 A  (2) 1.2 C  (2) 1.3 C  (2) 1.4 D  (2) 1.5 B  (2) 1.6 D  (2) 1.7 B  (2) 1.8 A  (2) 1.9 B  (2) 1.10 C  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Resultant (net) vector/Resultante (netto) vektor  (1) 2.2.1 F y = Fsin θ = 50sin 30°  OR/OF 50cos 60º = 25 N  (2) 2.2.2 POSITIVE MARKING FROM QUESTION 2.2.1 POSITIEWE NASIEN VANAF VRAAG 2.2.1 F x = 50cos 30°  = 43,3 N  R x = 80 – 43,3  = 36,7 N Substitution marks awarded within the question even if calculations for Fx and Rx are wrong F net 2 = R x 2 + F y 2 Substitusiepunte toegeken in die vraag selfs indien = 36,72 + 252  berekeninge vir Fx en Rx verkeerd bereken word. = 44,41 N  (5) 2.2.3 POSITIVE MARKING FROM QUESTION 2.2.1 AND 2.2.2 POSITIEWE NASIEN VANAF VRAAG 2.2.1 EN 2.2.2 OPTION 1/OPSIE 1 36,7  tan θ = Fnet = 44,41 N Ry = 25 N 25 θ = 55,74°  θ ϕ Rx = 36,7 N OPTION 2/OPSIE 2 25 cos θ =  44,41 θ = 55,74°  OPTION 3/OPSIE 3 36,7 sin θ =  Accept direction as /Aanvaar rigting as 44,41 θ = 55,74°  ϕ = 90º - θ = 34,26º  OPTION 4/OPSIE 4 25 cos θ =  44,41 θ = 55,74°  (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 The force that opposes the motion of a moving object relative to a surface. Die krag wat die beweging van 'n bewegende voorwerp relatief tot 'n oppervlak teenwerk. [2 or/of 0] (2) 3.2 A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.  'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid bly/volhard tensy 'n nie-nul resulterende/netto krag daarop inwerk. [Penalise -1 if key words/phrase is omitted/ Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2) 3.3 F x = 90cos 50°  OR/OF 90sin 40º = 57,85 N  (2) 3.4 N = Fg - Fy  NOTE/NOTA: N = 45(9,8)  - 90sin 50°  Weight and the vertical component can be N = 372,06 N  calculated separately, award one mark each even (4) if the formula for N is incorrect Gewig en vertikale komponent kan apart bereken word, een punt elk selfs indien die formule vir N verkeerd is. 3.5 POSITIVE MARKING FROM QUESTION 3.3 and 3.4 POSITIEWE NASIEN VANAF VRAAG 3.3 en 3.4 fk = µkN  57,85  = µ k (372,06)  µ k = 0,16  (4) 3.6 No  The coefficient is dependent on the (nature of) the surfaces / type of material in contact.  Nee. Die koëffisiënt is afhanklik van die (tipe) oppervlakke / soort materiaal in kontak. (2) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object.  Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Die versnelling is direk eweredig aan die netto krag en omgekeerd eweredig aan die massa van die voorwerp. [Penalise -1 if key words/phrase is omitted/ Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2) 4.2 Accept any set of coordinates from the graph, e.g.: Aanvaar enige kombinasie van koördinate vanaf die grafiek, bv.: 2,5 – 0  Gradient/Helling = =2 1,25 – 0  OR/OF 2,1 – 1,7  Gradient/Helling = =2 1,05 – 0,85  (3) 4.3 OPTION 1/OPSIE 1 1 1 Gradient/Helling = = =2 ma Fnet 1 F net = = 0,5 N  Accept/Aanvaar F net = 0,5 N  2 OPTION 2/OPSIE 2 F net = ma Accept any coordinates from graph = (1)(1/ 2 )  Aanvaar enige koördinate vanaf grafiek = 0,5 N  (2) 4.4 Acceleration is inversely proportional to the mass of an object (if the net force is kept constant)  Accept: The inverse of acceleration is directly proportional to the mass of the object (if the net force is kept constant) 1 OR a α m Versnelling is omgekeerd eweredig aan die massa van die voorwerp (indien die netto krag konstant bly) Aanvaar: Die omgekeerde van die versnelling is direk eweredig aan die massa van die voorwerp (indien die netto krag konstant bly) 1 OF a α m (2) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 N OR/OF N f f Fgll Fg⊥ Fg (3) Accepted Labels/Aanvaarbare Byskrifte Mark/Punt w weight/F G /F g OR Both components for one mark  gewig/gravitasiekrag/swaartekrag OF Beide komponente vir een punt N Normal force/F N  Normaalkrag/F N f Friction/F f  Wrywingskrag/F f Any additional force: deduct 1 mark (maximum ⅔) Enige addisionele krag: trek 1 punt af (maksimum ⅔) Omission of arrow heads: deduct 1 mark (maximum ⅔) Pylpunte uitgelaat: trek 1 punt af (maksimum ⅔) Lines must touch object otherwise (maximum ⅔ ) Lyne moet voorwerp raak anders (maksimum ⅔) Do not penalise if vectors are not to scale Moenie penaliseer indien vektore nie op skaal is nie 5.2 F net = ma  Any one/Enige een mgsin θ = ma 25(9,8)sin 15°  = 25a  NOTE/NOTA: a = 2,54 m·s-2  Award one mark for the parallel OR/OF component if calculated separately 25(9,8)cos 75º  = 25a  Ken een punt toe indien die parallel -2 a = 2,54 m·s  komponent apart bereken is (4) 5.3 Up the slope/Teen die helling op  (1) 5.4 F net = ma  Any one/Enige een F g// + (-f) = ma 25(9,8)sin 15° - f  = 25(-1,2)  f = 93,41 N  OR/OF 25(9,8)cos 75° - f  = 25(-1,2)  f = 93,41 N  Note/Let wel: Accept if calculation is done with direction up the slope as positive Aanvaar indien berekening gedoen is met rigting teen die helling op as positief (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 Each particle in the universe attracts every other particle with a gravitational force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.  Elke deeltjie in die heelal trek elke ander deeltjie aan met 'n krag wat direk eweredig is aan die produk van hulle massas en omgekeerd eweredig is aan die kwadraat van die afstand tussen hulle middelpunte. [Penalise -1 if key words/phrase is omitted/ Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2) 6.2 Gm1m 2 F=  r2 �6,67 x 10-11 ��6,39 x 1023 �(m) 3 338  =  3 2 (3 390 x 10 ) m = 900 kg  OR/OF Gm g=  r2 �6,67 x 10-11 ��6,39 x 1023 � g=  3 2 (3 390 x 10 ) g = 3,71 m·s-2 F g = mg 3 338 = m(3,71)  m = 900 kg  (899,73 kg) (4) 6.3 POSITIVE MARKING FROM QUESTION 6.2 POSITIEWE NASIEN VANAF VRAAG 6.2 w = mg = 900(9,8)  = 8 820 N  (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 Refraction/Refraksie  (1) 7.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 n i sin θ i = n r sin θ r  sin θi n=  1sin θ i = 1,33sin 40º  sin θr θ i = 58,75° sin θi 1,33 = sin 40°  sin θ i = 1,33sin 40º θ i = 58,75° Therefore the angle between ray and surface/Daarom is die hoek tussen invallende straal en oppervlak θ = 90°- 58,75°  = 31,25° (4) 7.3 n i sinθ i = n r sinθ r 1,33sin 40°  = n r sin 35º  n r = 1,49  (3) 7.4 58,75º air/lug 40º water 40º glass/glas 35º Allocation of marks/Toekenning van punte: (5) Light ray bends towards normal in water  Ligstraal breek na die normaal in water Light ray bends further towards normal in glass  Ligstraal breek nog meer na die normaal in glas Angle of incidence 58,75º shown (OR 31,25º)  Invalshoek 58,75º aangedui (OF 31,25º) Angles in water (40º)  Hoeke in water (40º) Angle in glass (35º)  Hoeke in glas (35º) If normal lines are not indicated, penalise with one mark Indien normaal lyne nie aangedui is nie, penaliseer met een punt If arrows are omitted, penalise -1 (maximum 4/ 5 ) Indien pylpunte weggelaat word, penaliseer -1 (maks 4/ 5 ) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 7.5 c n=  v 3 × 10 8 1,5 =  v v = 2 x 108 m·s-1  (3) 7.6 No/Nee  (1) [17] QUESTION 8/VRAAG 8 8.1 Diffraction is the ability of a wave to spread out in wave fronts  as the wave passes through a small aperture or around a sharp edge.  Diffraksie is die vermoë van 'n golf om uit te sprei in golffronte soos wat die golf deur 'n klein opening of om 'n skerp rand/kant beweeg. (2) 8.2 Criteria for investigative question/Riglyne vir ondersoekende vraag The dependent and independent variables are stated correctly.  Die afhanklike en onafhanklike veranderlikes korrek genoem. Ask the relationship between the dependent and independent variables  in a question, not as a statement. The question may not be written in a way that the answer is yes or no. Vra die verband tussen die afhanklike en onafhanklike veranderlike as 'n vraag, nie 'n stelling nie. Die vraag mag nie op so 'n manier geformuleer word dat die antwoord ja of nee is nie. Dependent variable/Afhanklike veranderlike: degree of diffraction/mate van diffraksie Independent variable/Onafhanklike veranderlike: wavelength/golflengte Examples/Voorbeelde: What is the relationship between the wavelength of a light ray and the degree of diffraction? Wat is die verband tussen die golflengte van 'n ligstraal en die mate van diffraksie? OR/OF How does a change in wavelength affect the degree of diffraction? Hoe beïnvloed 'n verandering in golflengte die mate van diffraksie? (2) 8.3 Degree of diffraction is directly proportional to the wavelength.  Mate van diffraksie is direk eweredig aan die golflengte. OR/OF Degree of diffraction α λ.  Mate van diffraksie α λ. (2) 8.4 Red/Rooi  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 8.5 Line should indicate inverse proportionality  Lyn moet omgekeerde eweredigheid aandui (2) Degree of diffraction/ Mate van diffraksie Slit width (mm)/ Spleetwydte (mm) [9] QUESTION 9/VRAAG 9 9.1 The magnitude of the electrostatic force exerted by two point charges on each other is directly proportional to the product of the (magnitudes of the) charges  and inversely proportional to the square of the distance between them.  Die grootte van die elektrostatiese krag wat deur twee puntladings op mekaar uitgeoefen word, is direk eweredig aan die produk van die (groottes van die) ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. [Penalise -1 if key words/phrase is omitted/ Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] NOTE: If learners refers to masses, no marks awarded NOTA: Indien leerder na massa verwys, geen punte (2) 9.2 OR/OF 20º T 20º Fg T FE Fg FE Accepted Labels/Aanvaarbare Byskrifte Mark/Punt w weight/F G /F g  gewig/gravitasiekrag/swaartekrag T Tension/F T  Spanning/F T FE Electrostatic force  Elektrostatiese krag One angle indicated  Een hoek aangedui (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 9.3 OPTION 1/OPSIE 1 If F A and F B were used/Indien F A en F B gebruik word kQ1Q 2  F= NOTE/NOTA: r2 Due to information given in the question, (9 x 109 )(8 x 10-9 )(7 x 10-9 )  accept all possible options = 0,03 2  As gevolg van die inligting in die vraag = 5,60 x 10-4 N  gegee, aanvaar alle moontlike opsies OPTION 2/OPSIE 2 If F g and F E were used/Indien F g en F E gebruik word F g = mg = (0,2 x 10-3)(9,8)  = 1,96 x 10-3 N  F E = (1,96 x 10-3)tan 70º  = 7,13 x 10-4 N  OPTION 3/OPSIE 3 If F g and F E were used/Indien F g en F E gebruik word F g = mg = (0,2 x 10-3)(9,8)  = 1,96 x 10-3 N FE Fg = sin 20° sin 70°  FE (1,96 x 10-3 ) =  sin 20° sin 70° F E = 7,13 x 10-4 N  (4) 9.4 POSITIVE MARKING FROM QUESTION 9.3 POSITIEWE NASIEN VANAF VRAAG 9.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Using F g and F E /Gebruik F g en F E Using F g and angle/Gebruik F g en F g = mg hoek = (0,2 x 10-3)(9,8)  Fg -3 T= = 1,96 x 10 N sin 70° T2 = (1,96 x 10-3)2 + (5,6 x 10-4)2 1,96 x 10-3   T =  sin 70° T = 2,04 x 10-3 N  -3 T = 2,09 x10 N  OPTION 3/OPSIE 3 NOTE/NOTA: Using F E and angle/Gebruik F E en Due to information given in the hoek FE question, accept all possible T= options cos 70° As gevolg van die inligting in die -4  vraag gegee, aanvaar alle 5,6 x 10 T= moontlike opsies cos 70°  T = 1,64 x10-3 N  (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1 + Criteria for marking/Nasienkriteria Shape of the field (minimum of 4 field lines)  Vorm van veld (minimum van 4 veldlyne) Direction of the field  Rigting van veld Lines don't touch charge/lines cross etc. (maximum ½) Lyne raak nie lading/lyne kruis ens. (maksimum ½) (2) 10.2.1 16 : 1  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 10.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 EP : ET kQ 16 : 1  Ep = 2  r 9 × 10 9 Q Because/Omdat 4 × 10 6 =  1 r2 Eα 2  9 x 109Q = (4 x 106)r2 …(1) r rP : rT kQ 1:4 ET = r2 r : r + 3 mm 5 9 × 10 9 Q 2,5 x 10 =  r = 1 mm  (0,001 m) (r + 0,003 ) 2 9 x 109Q = (2,5 x 105)(r + 0.003)2 …(2) Equation/Vergelyking (1) = (2) (4 x 106)r2 = (2,5 x 105)(r + 0,003)2 16r2 = r2 + 0,006r + 9 x10-6 r = 1 mm  (0,001 m) (4) 10.2.3 POSITIVE MARKING FROM QUESTION 10.2.2 POSITIEWE NASIEN VANAF VRAAG 10.2.2 kQ kQ Ep = OR/OF ET = r2 r2 9 × 10 9 Q 9 × 10 9 Q 4 × 10 6 =  2,5 × 10 5 =  (0,001) 2 (0,004 ) 2 Q = 4,44 x 10-10 C  Q = 4,44 x 10-10 C  (2) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 11/VRAAG 11 11.1 The magnitude of the induced emf across the ends of a conductor is directly proportional to the rate of change in the magnetic flux linkage with the conductor.  Die grootte van die geïnduseerde emk oor die punte van 'n geleier is direk eweredig aan die tempo van verandering van die magnetiese vloedkoppeling met die geleier. [2 or/of 0] (2) 11.2 ε= − N Δϕ  Δt − 400 Δϕ 7=  0,08 -3 ∆Φ = - 1,4 x 10 Wb  (-0,0014) (3) 11.3 POSITIVE MARKING FROM QUESTION 11.2 POSITIEWE NASIEN VANAF VRAAG 11.2 ∆Φ = AB(cos θ f – cos θ i )  -0,0014  = (0,03)2B(cos 45º - cos 0º)  B = 5,31 T  (4) 11.4 Increase/Toeneem  (1) 11.5 - 1 εα Δt  OR/OF Emf is inversely proportional to time. Emk is omgekeerd eweredig aan tyd. If the time decreases, the emf will increase. Indien die tyd verminder, sal die emk toeneem. (1) 11.6 North/Noord  (1) 11.7 From A to B/Van A na B  (1) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 12/VRAAG 12 12.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 1 1 1 R1R 2 = + Rp = R p R1 R 2  R1 + R 2  1 1 1 4R × 6R = +  4,8 =  4,8 4R 6R 4R + 6R R=2Ω R=2Ω  (3) 12.2 POSITIVE MARKING FROM QUESTION 12.1 POSITIEWE NASIEN VANAF VRAAG 12.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 V 4R = IR 4R V 4R = IR 4R = 1,8(4)(2)  = 1,8(4)(2)  = 14,4 V = 14,4 V V V I6 R = IT = R 6R RT 14,4 14,4 I 6R =  IT = 12 4,8 = 1,2 A =3A V 2R = IR  I 2R = 3 – 1,8  = 1,2(4)  = 1,2 A = 4,8 V  V 2R = IR  = 1,2(4)  = 4,8 V  OPTION 3/OPSIE 3 OPTION 4/OPSIE 4 R1 : R2 V 4R = IR 4R 4:6 = 1,8(4)(2)  I1 : I2 = 14,4 V 6:4 6 R : 2R : 3R x I = 1,8 A  1:2:3 10 IT = 3 A V R : V 2R : V 3R 1:2:3 I 2R = 3 – 1,8   = 1,2 A 2 V 2R = x 14,4  6 V 2R = IR  = 4,8 V  = 1,2(4)  = 4,8 V  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 12.3 POSITIVE MARKING FROM QUESTION 12.1 AND 12.2 POSITIEWE NASIEN VANAF VRAAG 12.1 EN 12.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 W = I2R ∆ t  W = VIΔt  V 2 ∆t 2 = 1,8 (8)(120)  = (14,4)(1,8)(120)  W=  R = 1036,8 J  = 3110,4 J  (14,4) 2 (120 ) W=  8 W = 3110,4 J  (3) 12.4 Decrease/Neem af  (1) - 12.5 The ammeter has such a low resistance  It short circuits the parallel part and all current flows through the ammeter.  OR The ammeter short circuits the resistors  No current flows through resistor 2R  Die ammeter het so 'n lae weerstand Dit kortsluit die parallelgedeelte en al die stroom vloei deur die ammeter. 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