Physical Sciences Grade 12 P1 KZN June 2023 Paper & Memo

REPUBLIC OF SOUTH AFRICA GRADE 12 NATIONAL SENIOR CERTIFICATE SBS RRR RR RRS RZ ETERS Sees esasasaese PHYSICAL SC 1 (PHYSICS) ®. CT ” cy o a . “ . " w . © Sese2e Seeeeeseeseeeeest MARKS: 150 TIME: 3 Hours This question paper consists of 15 pages and 3 data sheets. Please turn over Copyright reserved

pnyeepinleaded from Stanmorephysics.com June 2023 Common Test NSC INSTRUCTIONS AND INFORMATION 4, Write your NAME in the appropriate spaces on the ANSWER BOOK. Z, This question paper consists of EIGHT questions. Answer ALL the questions in the ANSWER BOOK. 3. Start EACH question on a NEW page in the ANSWER BOOK. 4. Number the answers correctly according to the numbering system used in this question paper. 5. Leave ONE line between two subquestions, for example between QUESTION 2.1 and QUESTION 2.2. 6. You may use a non-programmable calculator. rs You may use appropriate mathematical instruments. 8. Show ALL formulae and substitutions in ALL calculations. 9. Round off your final numerical answers to a minimum of TWO decimal places. 10. Give brief motivations, discussions et cetera where required. 11. You are advised to use the attached DATA SHEETS. 12. Write neatly and legibly. Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Physical Sciences P1 3 June Common Test NSC QUESTION 1: MULTIPLE-CHOICE QUESTIONS Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question number (1.1-1.10) in the ANSWER BOOK, for example 1.11 D. 141 An object is moving to the left at constant velocity when a force F acts on it. The magnitude of the frictional force acting on the object is... zero. smaller than F.° equal to F. 00 80 > larger than F. (2) 1.2 A horizontal force F is applied to the block X of mass M, which is in contact with another block Y of mass m on a smooth surface. Xx Y m ee wer a Y . The magnitude of the force exerted on block Y by block X is... A mF M 5 a M+F c m _mF_ : D M+m (2) Copyright reserved Please turn over

pryRawalaaded from Stanmorephysics.com NSC June Common Test 13 Four planets |, Il, Ill and IV are shown below. Their masses are M, 4M, 4M and 16M respectively. OOO Radius= R Radius= R Radius= 2R Radius=2 R Which of the above planets will have the same gravitational accelerations? A }1, Illand iv B J Il, Iland tv. c |!andil Ill and IV p |!and til Il and IV (2) r Copyright reserved Please turn ove

Prydoadteaded from Stanmorephysics.com June Common Test NSC 1.4 A person dives from a high platform into a pool as shown below. Ignore the effect of air friction. c D 4B Pewars At which ONE of the positions A, B, C or D will the magnitude of his momentum be a maximum? A _ Position A B_ Position B C_ Position C D Position D 1.5 If the momentum of an object is doubled, then the kinetic energy of the object m Bic A halved. B doubled. c three times greater. D four times greater . (2) Copyright reserved Please turn over

Phydda Giphsettd from Stanmorephysics.com June Common Test 1.6 Consider the statements below: |. Work is done on an object only when a force displaces the object in the direction of the force. iT chanical energy is conserved only in the absence of non- c ative forces. Ul. ork done on an object by a net force is equal to the kinetic energy object. Which of the above statements is/are TRUE? Only I. A Bland Il only. C_ilonly. D1, land Wl. (2) yg The hooter of a car emits sound waves of constant frequency as the car moves away from a stationary listener. Which of the following properties of the sound waves heard by the listener will NOT change? A Velocity B Frequency C Wavelength and frequency D Frequency and loudness (2) Copyright reserved Please turn over

Physevsemedad from Stanmorephysics.com June Common Test NS 1.8 Three charged spheres, P, Q and R are arranged as shown below. Q is POSITIVELY charged. Spheres Q and R are a distance d from each other and carry charges of +q and —q respectively. The magnitude of the charge on sphere P is NINE times greater than that on sphere Q. — Q R l€ lk ——! I x I d I Which ONE of the following combinations of factors will cause the net electrostatic force on charge Q to be ZERO? Charge on sphere P Distance X A Positive 45d B Positive 3d Cc Negative 45d D Negative 3d (2) 1.9 A cell is connected to a resistor and an open switch. Five points are labelled D, E, F, G and H respectively. Ee | D oe I oe 1! s \__»_____[R_}-—e—_+- 2 F S H A VOLTMETER has a ZERO reading if it is connected across points... A ED B FH C FG D GH (2) Copyright reserved Please turn over

PhysiftSeierersalted from Stanmorephysics .com June Common Test 1.10 In the circuits below, the emf of the batteries is the same and all the light bulbs are identical. The resistance of the connecting wires and internal resistance of the batteries are negligible. The power in bulb X is P. The power in bulb Y is... A MP B ye Cc 2P D 4P (2) [20] Copyright reserved Please turn over

PrysBomntoaded from Stanmorephysics.com June Common Test QUESTION 2 2.1 In the diagram below, a 2 kg block is connected to a 1 kg block by means of a light inextensible string. The blocks are pulled up an inclined plane, which of 28° with the horizontal. The sides of the blocks which touch ne have the same area. 30 N 28° The kinetic frictional force between the 1 kg block and the inclined plane is 4 N while the kinetic frictional force between the 2 kg block and the inclined plane is 8N. 2.1.1 State Newton's Second Law of Motion in words. (2) 2.1.2 Draw a labelled free-body diagram showing ALL the forces acting on the 1 kg block as it moves up the inclined plan. (5) 2.1.3 Calculate the magnitude of the tension in the string connecting the blocks by applying Newton’s second law separately to each of the blocks. (5) The ‘two blocks are interchanged so that the SAME 30 N force is now acting on the 2 kg block along the inclined plane, as shown in the diagram below. 30 N 2.1.4 How will the acceleration of the system ft Choose from: . INCREASES, DECREASES or REMAINS THE SAME. (1) 2.1.5 How will the tension in the string connecting the two blocks change? Choose from: INCREASES, DECREASES or REMAINS THE SAME. (1) Copyright reserved Please turn over

Phpeansloaded from Stanmorephysics.com June Common Test NSC 2.2 The mass of the Sun is 330 000 times greater than that of = Earth. The distance between the centres of the Sun and the Earth is 1,38 x 10° m. 2.2.1 State Newton's law of Universal Gravitation in words. (2) 2.2.2 Galculate the gravitational force that the Sun exerts on the Earth. (4) lan 2.2.3 {H6W will the gravitational force that the Earth exerts on the Sun compare jte-the answer calculated in QUESTION 2.2? Choose from: GREATER /FHAN, LESS THAN or EQUAL TO. Name the law that supports your answer. (2) [22] QUESTION 3 A stone is thrown vertically upwards from the top of a tower (point A). The stone then, on its way down, passes point A after 3 seconds. Ignore the effects of air friction. 41,6m ee eee eee eae Se 3.1 Define a projectile (2) 3.2 Calculate the magnitude of the velocity with which the stone was thrown. (3) 3.3 Calculate the maximum height above point A reached by the stone (3) 3.4 Onits way down the stone takes 0,1 s to pass a window of ytd m, as shown in the diagram above. The top of the window is at a height of 41 Sim from the ground. Calculate the height H of the tower. Fa (7) 3.5 USING THE GROUND AS REFERENCE, sketch a position-time graph for the entire motion of the stone. Indicate the height of the tower. (3) [18] Copyright reserved Please turn over

prieneuigaded from Stanmorephysics.com Se atin NSC QUESTION 4 4.1. Accricket ball of mass 0,15 kg is thrown vertically downwards with an initial speed of 8 m:s"' from a height of 2,5m. — 4.1.1 8,the Law of conservation of mechanical energy in words. ENERGY PRINCIPLES, determine the velocity with which the ball s the ground. 4.1.2 The ball bounces vertically off the ground and reaches a height H. The ball was in contact with the ground for 0,1 seconds. The ground exerts a force of 25,55 N on the ball. 4.1.3. Define Impulse 4.1.4 Calculate the height H 4.2 Students are performing an experiment in the laboratory. A 0,9 kg trolley is travelling with a constant velocity of 2,4 m-s*t on a long frictionless track when a block of mass m is dropped vertically onto it. The trolley with the block on it moves with velocity v as shown in the diagram. The following position vs time graph for the trolley is drawn. 10,2 f=---*-------2ee--c enn nn nnn nnn monn o5s Position (m) 4B fa-secesses-s-- No bene wn need Copyright reserved Please turn over (2) (4) (2) (6)

Physosvidaseted from Stanmorephysics.com June Common Test 4.2.1 | USE THE GRAPH to determine the magnitude of the velocity of the trolley after the block is dropped on the trolley. (3) 4.2.2 State the principle of conservation of linear momentum in words. (2) 4.2.3 Caléulate the mass m of the block. (4) a [23] QUESTION 5 ra 5.1 Acart of mass 2 kg is travelling with an initial speed of 1,5 m-s' when it is pulled by a horizontal force F of unknown magnitude over a distance of 2,5 m. A frictional force of 26 N acts on the cart as it moves. The cart reaches a speed of 4m:s". 2 kg F [7 B « > 2.5m 5.1.1 State the work-energy theorem in words. (2) 5.1.2 | Using ENERGY PRINCIPLES, calculate the magnitude of the force F (4) 5.2 A windmill on a farm is used to pump stationary water, from point A, in a well. The water flows past point B, 35 m above point A, at a speed of 2,1 m-s". 5.2.1. Define a non-conservative force. (2) 5.2.2 The pump in the windmill is rated at 500 W. Calculate the mass of water flowing past point B every minute. (5) [13] Copyright reserved Please turn over

Phyitawedaectad from Stanmorephysics.com June Common Test N QUESTION 6 6.1 The siren of a car emits sound waves of frequency 650 Hz. The car is driven at constant speed along a straight horizontal road . A detector placed on the side of the road ters changes in the pressure of the air caused by the sound waves. AIR PRESSURE VS TIME > an time (x 10+ s) Air pressure ' Assume that the speed of sound in air is 340 m:s". 6.1.1. State the Doppler effect in words. (2) 6.1.2 Is the car moving TOWARDS or AWAY FROM the detector? Motivate your answer by means of a calculation. (3) 6.1.3. Calculate the speed of the car. (4) 6.1.4. State TWO applications of the Doppler effect in medicine. (2) 6.2. The spectral lines observed for a distant star show that the star is moving away from the Earth. Explain, by referring to frequency, how one can deduce that the star is moving away from the Earth. aaa (2) foo [13] i Copyright reserved Please turn over

PryRawaeded from Stanmorephysics.com June Common Test NSC QUESTION 7 7.1 Two identical metal spheres, P and Q are suspended from a wooden bar by light inextensible strings as shown in the diagram below. The distance between the centres ie spheres is d metres. The charge on the spheres are —-4 pC iierececay, ar and +8 7.1.1 Sphere Q experiences an electrostatic force. In which direction will sphere Q move? Choose from: TO THE LEFT or TO THE RIGHT. The spheres are now brought into contact with each other and are then separated by distance d metres again. The spheres exert a force of 0,8N on each other. 7.1.2 Calculate the net charge on each sphere 7.1.3. Draw the electric field pattern due to the two spheres. 7.1.4 Calculate the number of electrons transferred during contact. 7.1.5 Calculate the distance ,d, between the two spheres. 7.2 Charges of +4 uC and —3 pC are placed a distance 0,2 m apart on a straight line, as shown below. Point P is located 0,1 m to the right of the -3 C charge. 7.2.1. Define electric field at a point. 7.2.2 Calculate the net electric field at point P A sphere carrying a charge of +6 uC is now placed at point P. It experiences an initial acceleration of magnitude 5 x 10? m-s7. 7.2.3 Calculate the mass of this sphere. Copyright reserved Please turn over (1) (2) (3) (2) (4) (2) (5) (3) [22]

prrewelaeded from Stanmorephysics.com June Common Test NSC QUESTION 8 8.1 The circuit below is used in an experiment to determine the resistance of resistor X. | | | 12V The 12 V battery has negligible internal resistance. When switch S is closed, the reading on the ammeter is 0,5 A. 8.1.1 State Ohm's law in words. (2) 8.1.2 Calculate the resistance of resistor X. (4) 8.1.3. During the experiment the switch is closed for very short periods of time. Give a reason for this. (1) 8.2 Study the circuit below. The battery has an emf of 12 V with negligible internal resistance. 129 8a A 169 49 <= 12V Switch S is closed. 8.2.1 Write down the potential difference across the 4 Q resistor. (1) 8.2.2 Calculate the reading on the ammeter. (4) 8.2.3. Calculate the energy dissipated in the 12 Q resistor i inutes. (4) 8.2.4 The 12 Q resistor is now removed from the circuit. How will the reading on the ammeter be affected?Choose form INCREASES, DECREASES or REMAINS THE SAME. Explain the answer. (3) [19] TOTAL: 150 Copyright reserved Please turn over

Physiiswenteded from Stanmorephysics.com June Common Test ISC DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) at GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 VRAESTEL 1 (FISIKA) TABLE 1--PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL | VALUE/WAARDE jihles saminda G 6,67 x 10 Nm2kg? po iia c 3,0 x 10° m-s* ali ier h 6,63 x 10 J-s persotee aie k 9,0 x 10°N-m?-C? ta peat £ 16x10"C zeal Me 9,11 x 10°" kg a ser dacd M 5,98 x 10" kg fos vga Re 6,38 x 10° m Copyright reserved Please turn over

phrenalgaded from Stanmorephysics.com June Common Test NSC TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING Vp = Vi + aAt> Ax=v,At+iadt? or/of Ay =v,At+ ast? Zon, V, =v? +2aAx or/of v,2=v,2+2aay | Ax= (“Jat orlof Ay = (“eau lana 2 2 oh FORCE/KRAG Fret = Ma p=mv en =U,N f, =—,N FieeAt = nett = Ap wamg Ap= mv, - mv, M F=oue oof F=G ae g -c4 oof 9g =Gz WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W-=Faxcos6 U=mgh offof Ep =mgh ; / W,,-4K oof W,,=AE, K=-mv? offof_—sE, => 2 AK=K,-K, offof AE, =E,,-E,, W,, =AK+AU O1/0f Wye = AE, +AE, Pa Pa =Fva / Pagemid = FVgemia WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG v=f2 viv + mall © {at viv, viv, E=hf and/en W, =hf, and/en Eynar =SV2 E=W, +Exmay ONOfE=W, +K,., where/waar 1 orfof Kj, =>MVv?.,, Copyright reserved Please turn over

Physhosataaded from Stanmorepfaysics.com June Common Test ELECTROSTATICS/ELEKTROSTATIKA ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE R V emf (€ = (R+1) . emk (€)= 1(R +1) R, =Ry+Ry+ V4 4 q=I At — = + — R, R, R W=Vq pW At W=VIat P=VI W=FRat oi P=PR W= 2 R piv R Copyright reserved

Downloaded from Stanmorephysics.com EDUCATION REPUBLIC OF SOUTH AFRICA 150 This Marking Guideline consists of 13 pages.

Downloaded from Stanmorephysics.com QUESTION 1: MULTIPLE-CHOICE QUESTIONS 1.1 Cvv 1.2 Dv 1.3 Dv 1.4 AVY 1.5 Dv v (Accept B) 1.6 Cvv 1.7 AVY 1.8 Dvv 1.9 Cyvv 1.10 Dvv [20]

Downloaded from Stanmorephysics.com QUESTION 2 2.1 2.1.1. When anet force is applied to an object, it accelerates the object in the direction of the net force. The acceleration is directly proportional to the net force and inversely-proportional to the mass of the object.v (2) N/Fy Fa T /Fs w/F, Notes: Accepted Labels MARK w_| F,/Fw/weight/9,8 N/mg/gravitational force v f__| Filfk/friction v N_| FN/Normal v T_| FT/Tension force v Fa | 30 N/F v Any additional force: deduct 1 mark maximum (maximum 4/5) Lines must touch dot/square otherwise (maximum 4/5) Accept if components of gravitational force provided instead of Fg (5) 2.1.3 For the 1 kg block: Fret = ma F - (T+mg sin@ + fx) = ma 30 - {T+(1)(9,8)(sin28°) +4} = (1)(a)Y¥ 21, 399-T=a (1) Any one ¥ For the 2 kg block: Fret = ma T-(mg sin@ + fx) = ma T-{(2)(9,8)(sin28°) + 8} = (2)(a)¥ T - 17,201 = 2a ............ (2) ura Substitute for T from equation (1) fan (21,4 - a) - 17,2 = 3av a 4,198 = 3a a=1,4 ms? T=21,399-a 1,399 - 1,399 ONY (5)

Downloaded from Stanmorephysics.com 2.1.4 REMAINS THE SAMEY (1) 2.1.5 DECREASESY (1) 2.2 2.2.1 Every in the universe attracts every other body with a force that is directly al to the product of their massesY and inversely proportional to the squaréofithe distance between their centres. ¥ (2) co 2.2.2 =m Vv (5,98%10* )[(330000)(5,98x10%)] ¥ (1,38x10°P vy F=4,13x10"N ¥ (4) F =(6,67*10"")x 2.2.3 Equal to.” Newton's third law is obeyed.v (2) [22] [EEE

Downloaded from Stanmorephysics.com QUESTION 3 3.1. Object movingY under the influence of the gravitational force onlyY (2) 3.2 OPTION 14 OPTION 2 Upward positive: Upward positive: \tv Ay = viAt + % aAt? ¥ O= Vi +(4918)1,.5¥ 0 =vi(3) + % (-9,8)(3)? ¥ « Vi = 44% mst ¥ «Vi = 14,7 mst v¥ Upward negative: Upward negative: (3) Vr=Vi + adtY Ay = viAt + % aAt? ¥ 0 = Vi +(9,8)1,5¥ 0 =vi(3) + % (9,8)(3)? ¥ - Vi =-14,7 ms « Vi =-14,7 ms"! «Viz 14,7 mst v «Viz 14,7 mst v¥ 3.3 3.4 POSITIVE MARKING FROM Q3.2 Upward positive: ve = vi2 + ZaAyY 0 = 14,72 + 2(-9,8) AyY Ay = 11.025mYv Upward negative: ve = vi2 + ZaAyY 0 = (-14,7)? + 2(9,8) AyY Ay = -11.025m -. max height reached = 11,025 mv (3) POSITIVE MARKING FROM Q3.3 Ay = viAt + % aAt? ¥ ve = vit + 2ady v =vi(0,1) + % (9.8)(0,1)? ¥ 19,51v = 14,72 +2(9.8)AyY vi = 19,51 m-st Ay = 840m (7) Height = 41,6 +v 8,4 = 50 mv

Downloaded from Stanmorephysics.com 3.5 POSITIVE MARKING FROM Q3.4 Upward positive: 50 position(m) 0 > time (s) Upward negative: A Oo > time (s) position(m) ch ro) v Criteria for graph Marks Correct shape. Final position lower than initial position. Graph ends on x axis and starts from height of tower (3) [18]

Downloaded from Stanmorephysics.com QUESTION 4 4.1 4.1.1 Total mechanical energy in an isolated system remains constant / is C (2) 4.1.2 Ew t p) = Em total (bottom) | Any one Y (Ep + Ek)top = (Ep + Ek)sottom (0,15 x 9,8 x 2,5) + % (0,15) x 8*V¥ = 0 + 40,15 vv V = 10,63 m-s'¥ OPTION 2 Whe = AEp + AEx ¥ O = [0-(0,15x9,8x2,5)] ¥ + [40,15 v2 - % (0,15) x 87] ¥ V = 10,63 m-s'¥ OPTION 3 Wnet = AEx Vv Wrg = %mve - %mvi Any one mgAx Cos@ = %mvr - % mvt 0,15(9,8)(2,5) Cos 0°Y = % 0,15 v2 - % (0,15) x 82 ¥ V = 10,63 m-s'v (4) 4.1.3 Product of the net force and the time” for which the force actsY (2) 4.1.4 Fground = Fnet + Fgravity 25,55 = Fnet_+ (0,15 x 9,8) ve = vi2 + ZaAyY Fret = 24,08 N 0 =5,422 + 2(-9,8) VAy Fret.At = Ap Ay = 1,5 mv 24,08 x 0,1Y = 0,15v -0,15(-10,63)” = 5,42 m:s™ (6) = =

Downloaded from Stanmorephysics.com 4.2 4.2.1 4.2.2 4.2.3 In an isolated system the total linear momentum is conserved. Vv ¥ (p total) before = (p total) after v Any one miv1 + Mav2 = (m1 + Ma)vr (0,9)(2,4) + 0 v= (0,9 + m)(1,8) ¥ m = 0,3 kg¥ QUESTION 5 5.1 5.2 5.1.1 5.1.2 5.2.1 5.2.2 The net work done on an object is equal to the change _in kinetic energy of the object.vv W. =AE, Fnet Ax cos © = % mv - % mvi2 Fnet x 2,5”_ =14(2 x 42) -% (2 x 1,52) ¥ Fnet =5,5N Fnet = F - fk 5,5 =F -26 F =31,5NY A force for which the work done in moving an object between two points is dependent on the path taken. vv P= Whe At P = AEp + AEk At 500Y = m(9,8) (35 — 0) ¥ + % m (2,12-0) ¥ 60 v Any one m =87kgy (86,9 kg) [EEE (3) (2) (4) [23] (2) (4) (2) (5) [13]

Downloaded from Stanmorephysics.com QUESTION 6 6.1 6.1.1 loppler effect is the change in frequency (or pitch) of the sound detected listener, because the sound source and the listener have different ities relative to the medium of sound propagation.v ¥ pler effect is the change in frequency (or pitch) of the sound detected by a listener, when there is relative motion between the sound source and the listener. v ¥ 6.1.2 1,25 Pie 1,25 _ 1 f= 17,36 x10~* I f= 13,89 x1074 f = 720,05 Hz fl >fs.v Towards. ¥ POSITIVE MARKING FROM 6.1.2 _V#V, VEVs v 720,05 Co v S, Vear = 33,06 mst¥ if fs¥ Determine whether arteries are clogged.v Determine heartbeat of foetus. v 6.2 Frequencies of the spectral lines decreasev v. OR frequencies of light emitted by these stars are red shifted. (2) (3) (4) (2) (2) [13]

Downloaded from Stanmorephysics.com QUESTION 7 7.1 7.1.1 (1) 7.1.2 maul -4x10%+8x10% y —Ora = 2 Q,,, =2*10°C “.Q, =2*10°C - Any one v 1Qg =2*10°C (2) 7.1.3 Criteria for sketch: Marks Correct direction of field lines. v Shape of the electric field. v No field line crossing each other / Field lines start from the sphere / | ¥ No field lines inside the spheres. (3) 7.1.4 POSITIVE MARKING FROM 7.1.2 AQ=Q-Q, = 210% ~ (-4x105) a — al = 6x106C e n= 6x10* v 1,6«10"9 n= 3,75x108V (2)

Downloaded from Stanmorephysics.com 7.2 7.2.1 7.2.2 7.2.3 cl oe: (9x10°)(2 x10°)(2x10° e-? od g, = (9*10°X4x10%)v E,=4x10°N-C* _(9*10°)(3x10%) (0,1)° Es = 2,7 x 10°N-C* E, E,+E,=E, E,=4x10°-2,7x10° ¥ E, =2,3x10°N-C" Left ¥ Fij=mai vy -.QE=ma _ QE a - (6x10*)(2,3x10°) v (5x 107) m = 2,7 x 107 kg ¥ m POSITIVE MARKING FROM QUESTION 7.1.2 F= Force experienced per unit positive charge placed at that point. vv MARKING CRITERIA: e Electric field formula ¥ e Substitution into electric field due to 4uC charge. ¥ e Substitution into electric field due to -3yuC charge. ¥ e Subtracting the two electric fields. ¥ e Final answer, including direction. ¥ MARKING CRITERIA: e Equating electrostatic force to net force v e Substitution for Q, E anda. ¥ e Final ancwetat i (4) (2) (5) (3) [22]

Downloaded from Stanmorephysics.com QUESTION 8 8.1 @ potential difference across a conductor is directly proportional to the nt in the conductor at constant temperature. v ¥ (2) ION 1 Rv tot 12=(0.5)R,¥ “R= 240 wR.= (24-8) =16QV OPTION 2 V=IRY = (0,5)(8) ¥ =4V oV=(12-4)¥ =8V V=IR 8 = (0,5)(R,) aR = v R=160 (4) 8.1.3 Keep temperature of conductor constant ¥ (1) 8.2 8.2.1 12Vv (1) 8.2.2 OPTION 4 OPTION 2 VE LR a 12=1(4)v BR Rt R I=3A V.=ILR IR=1R 12=1,16 OR (3)(4)=1,,(16) Rp =3,2.9 |= 0,754 = 0,754 | = (3 +¥ 0,75) = 3,75 AY (4)

Downloaded from Stanmorephysics.com 8.2.3 OPTION 14 OPTION 2 _ Vat v =I(R,+R,) R oe = (20) g - Kv 1=0,6A eye _ v w=P RAtY © (0.6)'(12)Y (120) v 518,4 Jv (4) 8.2.4 REMAINS THE SAME-V. Potential difference across the 16Q and 4 Q resistors in parallel remain the samev Vv (3) [19] TOTAL: 150