# Physical Sciences Grade 12 Paper 1 Free State June 2025 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Free State June Exam 2025, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-memo-p1-may-june-2025-free-state/16bdb802-4fec-4db9-88c6-6f5c91ca214b
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Free State June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 14

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## About this document

Physical Sciences · Grade 12 · Free State June Exam · 2025. Memorandum, 14 pages. Read online or download the PDF.

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### Page 1

education

Department ot
Education

FREE STATE PROVINCE

JUNE EXAMINATION
JUNIE EKSAMEN

GRADE/GRAAD 12

PHYSICAL SCIENCES: PHYSICS (P1)

FISIESE WETENSKAPPE: FISIKA (V1)

JUNE 2025

MARKS/PUNTE: 150

MARKING GUIDELINES
NASIENRIGLYNE

This marking guideline consists of 15 pages.
Hierdie nasienriglyne bestaan uit 15 bladsye.

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FS/VS/ JUNE 2025

Grade/Graad 12 June Exam/Junie Eksamen.

Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

11. Cvv (2)
12 Avv (2)
13 CyvYv (2)
14 Avv (2)
15 Avv (2)
16 BYv (2)
17 Bvv (2)
18 Dvv (2)
19 Cvv (2)
1.10 Bvv (2)
[20]

QUESTION/VRAAG 2
2.1.1 Kinetic frictional force, fk, is a force that opposes the motion of a moving

object relative toa surface.vv (2 or 0)

Kinetiese wrywingskrag, fx, is 'n krag wat die beweging van 'n bewegende

voorwerp relatief tot 'n opperviak teenstaan. “~~ (2 of 0) (2)
2.1.2 The two blocks are made up of two different materials. v (1)

Die twee blokke bestaan uit twee verskillende materiale. “ (1)
21.3 v v

N F

; Accept the following symbols / Aanvaar die volgende
sSimbole
' F | Force applied / Toegepaste krag / Fa/ v
56,68 N
T__| Fr/Force on the rope / Krag op die tou v
w__| Fg/ weight / gewig v
wv N- | Fn/ Normal / Normaal / Normal force / v
, Normaalkrag
| fk] f Friction / Wrywing / Frictional force /
Wrywingskrag / Kinetic frictional force / v
Kinetiese wrywingskrag

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Grade/Graad 12 June Exam/Junie Eksamen.
Marking Guidelines/Nasienriglyne

NB: Allocate ONE mark for each correct arrow and label.
Deduct one mark for any additional forces.

NB: Ken EEN punt toe vir elke korrekte pyl en byskrif.
Trek een punt af vir enige bykomende kragte.

FS/VS/ JUNE 2025

(5)

2.1.4 Marking criteria
e Correct formula: Fret = ma ~
e Correct calculation of both frictional forces Vv
e Correct calculation of Fav ¥ (either)
e §©Correct substitution of a into the Fret = mav (either)
e Equating both tension equations v
e Final answer of 1,16 - 1,2 kg ¥

Kriteria vir die nasien

¢ Korrekte formule: Foe: = ma
Korrekte berekening van beide wrywingskragte Vv ¥
Korrekte berekening van Fav ¥ (enigeen)
Korrekte vervanging van a in die Fnet = ma v¥
(enigeen)
Gelykstelling van beide spanningsvergelykings ¥
e Finale antwoord van 1,16 tot 1,20 kg ¥

Block B/ Blok B/ 4m

eee

Fret = ma

T+ Fg//+f=ma

T + pkN + Fg// = ma

T + uxmgcos30° + mgsin30° = ma

T- 0,4x4m(9,8)c0s30° - 4m(9,8)sin30° = 4m(2)

T - 33,179 m = 8m Any ¥
Block A/ Blok Alm
F+T+Fgi/+f=ma either Vv
‘Any ¥

F+T + ukN + Fo/ = ma

F+T+ u«mgcos30 + mgsin30° = ma

56.38 -T - 0,1xm(9,8)cos30° - m(9,8)sin30° = m(2)
56,83 - T -— 5,7487 m = 2m

T-33,179m 2 4(56,83 - T- 5,7487 m)

m= 1,20 kg ¥ (7)

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Grade/Graad 12 June Exam/Junie Eksamen.

Marking Guidelines/Nasienriglyne

FS/VS/ JUNE 2025

2.2.1 Each body in the universe attracts every other body with a force that is
directly proportional to the product of their masses and inversely proportional

to the square of the distance between their centres. vv
Elke liggaam in die heelal trek elke ander liggaam aan met 'n krag wat

direk eweredig is aan die produk van hul massas en omgekeerd eweredig aan

die kwadraat van die afstand tussen hul middelpunte. v

(2)

2.2.2
OPTION 1 / OPSIE 7
M
g=G5Y

12Vv = 6,69 x 107"
M = 881x107 kg ¥

M

—————-_ ¥
(700 x 103)

| OPTION 2/ OPSIE 2

r
100 x 12¥

= 6,69 x 10711 1008

QUESTION/VRAAG 3

3.1 Oms?v

M = 881x102? kg ¥

(700 x 103)?

(4)

[21]

(1)

3.2 An object which has been given an initial velocity and then it moves under
the influence of the gravitational force only.“~Y (2 or 0)
‘n Voorwerp wat 'n aanvanklike snelheid gegee is en dan beweeg dit slegs
onder die inviloed van gravitasiekrag.¥¥ (2 of 0)

3.3 O02sv

3.4 Marking criteria

e Penalise once if upward is taken as positive (option 2)

e Correct formula to calculate vi: vr = vi + gAt / or any formular

e Correct substitution into the formula leading to the answer

e Final answer of 1,96 m-st

Kriteria vir die nasien
e Penaliseer een keer as opwaarts as positief beskou word (opsie 2)
e Korrekte formule om v; te bereken: v; = vi + gAt / of enige formule
e Korrekte vervanging in die formule wat tot die antwoord lei
e Finale antwoord van 1,96 ms

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(1)

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3.5

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Grade/Graad 12 June Exam/Junie Eksamen.
Marking Guidelines/Nasienriglyne

OPTION 1 / OPSIE 1
DOWNWARDS POSITIVE
AFWAARTS POSITIEF

vr = vi + gAtY

0 = vit (9,8)(0,2) ¥

vi = -1,96 m-s*
velocity/snelheid = 1,96 m-s! ¥

OPTION 2/ OPSIE 2
UPWARDS POSITIVE
OPWAARTS POSITIEF

vi = vi + gAt¥

0 = vi + (-9,8)(0,2) ¥

vi = 1,96 m-s"
Velocity/snelheid= 1,96 m-s? ¥

Max / MAKS (2)

Marking criteria

(3)

e Correct formula to calculate vi: Ay = viAt + YagAt? / or any Ay formular
e Correct substitution into the formula leading to the answer

e Addition of 30 m
e Final answer 1,96 ms
Kriteria vir die nasien

e Korrekte formule om vi te bereken: Ay = viAt + “gAt? / of enige

Ay-formule

e Korrekte vervanging in die formule wat tot die antwoord lei

e Byvoeging van 30m

e Finale antwoord 1,96 m-s?

POSITIVE MARKING FROM QUESTION 3.3
POSITIEWE NASIEN UIT VRAAG 3.3

OPTION 1 / OPSIE 1

Ay = viAt + “gAt?v¥
yr = viAt + “gAt? + yi

= —30,196
Height / Hoogte = 30,196 mv

Whole motion / Hele beweging

= -1,96(0,2)+ ¥2(9,8)(0,2)? V+ (-30)¥

OPTION 2/ OPSIE 2

v,e+ UF

Ay = ( )ary
_ (eet 0

~ 2

)o2v

= —0,196 m——)

OPTION 3 / OPSIE 3

vp = vj + 2gAyv
0? = (—1,96)? + 2(9,8)AyY
Ay = -0,196m

| OPTION 4/ OPSIE 4

Ay = viAt + “gat?¥
0 = -1,96(0,2) + ¥2(9,8)(0,2)2
Ay=- 0,196 m

a)

a
Height! Hoogte = -0:1,96 + (-30} v

= -30,196
= 30,196 m ¥

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(4)

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Marking Guidelines/Nasienriglyne

3.6 Marking criteria

© Correct formula to calculate vi: Ay = viAt + %gAt? / or any Ay formula

« Correct substitution into the formula leading to the answer

« Final answer of 2,68 s

Kriteria vir die nasien

¢ Korrekte formule om v; te bereken: Ay = viAt + “gAt? / of enige
Ay-formule
Korrekte vervanging in die formule wat tot die antwoord lei
Finale antwoord van 2,68 s

POSITIVE MARKING FROM QUESTION 3.4 AND QUESTION 3.5
POSITIEWE NASIEN UIT VRAAG 3.4 EN VRAAG 3.5

OPTION 1 / OPSIE 1 OPTION 2/ OPSIE 2

Whole motion / Hele beweging From maximum height / Vanaf

1 maksimum hoogte
Ay = v,At + 5gAt?v

30 = -1,96At + 5(9,8)At?¥ Ay = vjAt+ 5gAt?v
At = 2,68 sv 30,196 = OAt + 5(9,8) At? v
At = 2,48
OPTION 3/ OPSIE 3 At = 2,48 + 0,2
At = 2,48 sv

From point of launch downwards
Vanaf die punt van lansering

Ay = v,At + gay
30 = 1,96At + 3(9,8)At?v

At = 2,28
At = 2,28+0,4
At = 2,68 s¥ (3)

3.7 Marking criteria

e Correct formula to calculate vr: vr = vi + gAt / Ay = viAt + YgAt?
or any Ay formula

e Correct substitution into the formula leading to the answer

e Final answer of 2,68 s

Kriteria vir die nasien

e Korrekte formule om v; te bereken: vr = vi + gAt / Ay = viAt + ZgAt?
of enige Ay-formule

e Korrekte vervanging in die formule wat tot die antwoord lei

e Finale antwoord van 2,68 s

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= 2 v
Copyrigh = (0)? +2(9,8)(18,196)

FS/VS/ JUNE 2025
Grade/Graad 12 June Exam/Junie Eksamen.

Marking Guidelines/Nasienriglyne

POSITIVE MARKING FROM QUESTION 3.4 AND QUESTION 3.6
POSITIEWE NASIEN UIT VRAAG 3.4 EN VRAAG 3.6

OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2 OPTION 3/ OPSIE 3
Whole motion Maximum height Point of launch
Hele beweging Maksimum hoogte Punt van lansering
Up = V, + gGAtY
vp = 4 + gAtY = 0+ 9,8(2,48)¥ | Y= + gAty
—1,96 + 9,8(2,68)¥ = 24,304m:-s71v = 1,96+ 9,8(2,28)¥
= 24,304 m:s7!v OR = 24,304m:s71v
‘ ; OR up = vp + 2gdy : , OR
oy Fi my Spy = 0? + 2(9,8)(30,196)| 7 = % + 294y
= (-1,96)? + 2(9,8)(30) | = 2433m-57 = (1,96)? + 2(9,8)(30)
= 24,33 m:s7? OR = 24,33 m:s}
OR vy + Y; OR
A = (4%) ay = 2 Jae A = ("*1) at
io a a vy +0 rN 2
vy — 1,96 30,196 = (=) (2,48) vy, +0
30 = (>) (2,68) 30,196 = ( ) 248)
2 vp = 24,35 m-s7} 2
vy = 24,35 m-s7t Vp = 24,355 m:s74
Range: 24,30 m-s" to / tot 24,36 m-s"

3.8 Marking criteria

¢ Correct formula to calculate vi: v7 = v? + 2gAy
e

Correct substitution of vi and Ay into the formula leading to the answer
e

Final answer of 18,88 m-s*
Kriteria vir die nasien

¢ Korrekte formule om v; te bereken: v7 = vj + 2gAy

e Korrekte vervanging van vi en Ay in die formule wat tot die antwoord lei
Finale antwoord van 18,88 m-s*

POSITIVE MARKING FROM QUESTION 3.4 AND QUESTION 3.7
POSITIEWE NASIEN UIT VRAAG 3.3 EN VRAAG 3.7

OPTION 1 / OPSIE 1

OPTION 2/ OPSIE 2
Whole motion / Hele

Point of launch / Punt van
beweging

lansering
vp = v7 + 2gAyY

vp = v7 + 2gAyY
= (—1,96)? + 2(9,8)(18)”

= (1,96)? + 2(9,8)(18)¥
Ve = 18.88 m+ s71¥ vp = 18,88 m- stv
OPTION 3/ OPSIE 3 OPTION 4/ OPSIE 4
Maximum height / From the ground / Vanaf die
Maksimum hoogte grond

v7 =v? + 2gAyY vp = v7 + 2gAyv

(24,304)? = v? + 2(9,8)(12)¥
vp = 18,885 m- stv vp = 18,85 m:s71¥ peseniotom

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Grade/Graad 12 June Exam/Junie Eksamen.

Marking Guidelines/Nasienriglyne

(3)
Range 18,85 m-s™ to / na 18,93 m-s*

3.8 Marking criteria
« Indicating the value of 30,196 m and 0,2s
« Indicating 30 m andOsor0,4s
* Correct shape of the graph (parabolic)
e Correct labels of both x and y axes.

Kriteria vir die nasien

e Dui die waarde van 30,196 m en 0,2 s aan
e Dui 30men0s of 0,4 s aan
e Korrekte vorm van die grafiek (parabolies)
e Korrekte byskrifte van beide x- en y-as.

POSITIVE MARKING FROM QUESTION 3.4
POSITIEWE NASIEN UIT VRAAG 3.4

0,2 0,4 2,68
a Time (s)
EE Tvd (s)
5 2
83
aq
30,196 (4)
7” [24]
QUESTION/VRAAG 4
4.1.1. The net force acting on an object is equal to the rate of change of
momentum of the object in the direction of the net force. vv (2 or 0)
Die netto krag wat op 'n voorwerp inwerk, is gelyk aan die tempo van
verandering van momentum van die voorwerp in die rigting van die
netto krag.V¥~Y (2 of 0) (2)
4.1.2 True v
Change in momentum is zero / Car is moving at constant velocity. v (3)
Waar ¥

Verandering in momentum is nul / Motor beweeg teen konstante snelheid. VV

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Marking Guidelines/Nasienriglyne

AASB
OPTION 1/ OPSIE 1 OPTION 2/ OPSIE 2
East positive / Ooswaarts positief East positive / Ooswaarts positief

Lp(before / voor) = =p(after / na) AP carimotor = - APbakkie~
Picar/motor + Pibakkie = Pfcar/motor + Diet Y | 6 867 — (-12 495)” = 12 753 - pw”
-12 495 + 12 753 W= 6867 + prrakkieW” 6 609 = — prbakkie
Ptoakkie = - 6 609 Pfakkie = — 6 609
= 6 609 kgm:s", west / wes ¥ = 6 609 kgm:s", west / wes V

(4)

4.1.4 Marking criteria
e Correct formula to calculate Fret: Free = 2

e Correct substitution into the formula leading to the answer
e Final answer of 1,94x105 N west
Kriteria vir die nasien
4p

e Korrekte formule om Fret te bereken: Free = he

e Korrekte vervanging in die formule wat tot die antwoord lei
e Finale antwoord van 1,94x10° N wes

POSITIVE MARKING FROM QUESTION 4.1.3
POSITIEWE NASIEN UIT VRAAG 4.1.3

OPTION 1/ OPSIE 1-Bakkie | OPTION 2/ OPSIE 2- Car
| Motor
_ dp A
Fnet = 7% fs
=e 6867 ga a8)
aa ry ane
= —1,936 x 10° q
, = 1,936 x 10°
= 5 v ‘
1,94 x 10° west/wes = 1,94105 west/wes¥

(3)
4.1.5 1,94x105 east / oosv (1)
4.2.1 (U+k)A=(U+K)C
mgh + Yamv? = mgh + Yamv? | Any/Enige V

12x 9,8 x5 + OY =0 + Vaxt2vev
v=6,26 mslt¥ (4)

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Marking Guidelines/Nasienriglyne

4.2.2 ME at/ by B = Total ME of the pendulum at A — energy gained by crate
Totale ME van die pendulum by A — energie verkry deur krat
= mgh + Yamv2 - Ymv2
= 12x 9,8 x 5 + OY - %x4(2)? ¥
= 588 —-8
= 580 J
580 = mgh + “mv? ¥

= 12x 9,8 x 1 + %x12v?

v=8,78 ms*

p=mvv ra
= (12(8,78) ¥
= 105,36 kgm-st¥ (6)

[22]

QUESTION/VRAAG 5

5.1.1 A force for which the work done (in moving an object between two points)
depends on the path taken. Vv“
'‘n Krag waarvoor die arbeid wat verrig word (om 'n voorwerp tussen twee
punte te beweeg) afhanklik is van die pad wat gevolg word. ~~” (2)

5.1.2 Negative/Negatief Vv (1)

5.1.3. Spanningskrag~ en wrywingskrag. ~“ (2)

5.1.4 Fv

NY Accept the following symbols / Aanvaar die
volgende simbole
Ww EF, /mg/weight/gewig v
fk/Friction/wrywingfrictional
f force/wrywingskrag/kinetic frictional v
force/kinetiese wrywingskrag
N FN / Normal / Normaal / Normal force / v
Normaalkrag
F T/Force applied/Tension/FT/Force on v
the rope / Krag op die tou (4)

wv

5.1.5 The work done on an object by a net force is equal to the change in the
object's kinetic energy. vv
Die arbeid wat deur 'n netto krag op 'n voorwerp verrig word, is gelyk aan die
verandering in die voorwerp se kinetiese energie. VV (2)

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Grade/Graad 12 June Exam/Junie Eksamen.

5.2

Marking Guidelines/Nasienriglyne

OPTION 1/ OPSIE 1

Whet = AK
We + Wi+ Weg = AK Anyi Enigesn”
FAxcos@ + fAxcos0+ masindaxcos® = = AK “

F(24,29)cos180° + 20,79(24,29)cos180° + 60x9 8(35 ) 24, 29cos0° =

24,29,
F = 187,39 NY

OPTION 2/ OPSIE 2

Whe = AK + AU
Wer + Wi+ = AK + AU Any / Enigeenv
FAxcos@ + fAxcos@ = 2m(vj — v?) + mg(hy — hi)

sf 7 7
F(24,29)cos1 80" + 20,79(24 29)cos180° =0+ 60 x 9,8(0 — 8,6)
F = 187,39 Nv

(6)

rp.”
~ At
_ mgh— 0

At

Any/enige ¥

_ 400x9,8%5,6 /,

4
= 5488WV (3)
[20]

QUESTION/VRAAG 6

6.1

It is the (apparent) change in frequency (or pitch) of the sound (detected

by a listener) ¥ because the sound source and the listener have different
velocities relative to the medium of sound propagation. Dit is die (oénskynlike)
verandering in frekwensie (of toonhoogte) van die klank (waargeneem deur 'n
luisteraar) “ omdat die klankbron en die luisteraar verskillende snelhede het
relatief tot die medium van klankvoortplanting. “

OR/OF

An (apparent) change in (observed/detected) frequency (pitch),

(wavelength) “because of the relative motion between a source and

an observer (listener). “

‘n (oénskynlike) Verandering in (waargenome) frekwensie (toonhoogte),
(golflengte) “as gevolg van die relatiewe beweging tussen 'n bron en

'‘n waarnemer (luisteraar). ~ (2)

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Marking Guidelines/Nasienriglyne

_ {vive a v
62 f= GEyhy oo f= (Zh
330
bv = ——. v
615 330-0, x 500
v, =61,71m:s1 Vv (4)
6.3 POSITIVE MARKING FROM QUESTION 6.2
POSITIEWE NASIEN UIT VRAAG 6.2
al v
Vv =
At
30,85 (3)
61,71 = —— ¥
At
At =0,499sv
6.4.1 Away from / Weg van ¥
A lower frequency is detected.” The spectral lines are shifted to the red
end of the spectrum. ”
Laer frekwensie word waargeneem. “ Die spektraallyne word na die
rooikant van die spektrum verskuif. “ (3)
6.4.1 Universe is expanding / Die heelal word groter/brei uit ¥ (1)
[13]
QUESTION/VRAAG 7
7.1 The electric field at a point is the electrostatic force experienced per unit
positive charge placed at that point. v “ (2)
Die elektriese veld by 'n punt is die elektrostatiese krag wat ervaar word per
eenheid positiewe lading wat op daardie punt geplaas word. Vv
72
Criteria for sketch / Kriteria vir skets
ea Correct direction M
wy eae Korrekte rigting
Correct shape (radial) | ¥
Korrekte vorm
(2)
7.3. The magnitude of the electrostatic force exerted by one point charge on

another point charge is directly proportional to the product of the

magnitude of the charges” and inversely proportional to the square of

the distance between them. “

Die grootte van die elektrostatiese krag uitgeoefen deur een puntlading

op 'n ander puntlading is direk eweredig aan die produk van die groottes

van die ladings en ~ omgekeerd eweredig aan die kwadraat van

die afstand tussen hulle. ~ (2)

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74

7.6

Grade/Graad 12 June Exam/Junie Eksamen.

Marking Guidelines/Nasienriglyne

Fiav

Fa2¥

_ 9X 10° x (100 x 1076)(100 x 107%) |
- (3 x 10-2)?
=1x105NY

POSITIVE MARKING FROM QUESTION 7.5
POSITIEWE NASIEN UIT VRAAG 7.5

_ 9x10? x (100 x 107°)Q;
7 (3 x 10-2)2
= Q3 x 10°N

(Fnee)® = (Fiz)? + (Fs2)?
(1,41 x 105)? = (1 x105)? + (Q, x 10°)2¥
Q3 = 9,88 x 107°C v

QUESTION/VRAAG 8

8.1

8.2.1

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3.11 J Yof energy is transferred per one coulomb of charge. ¥
3,11 J Yvan energie word oorgedra per een coulomb lading. ¥

OPTION 1 / OPSIE 1 OPTION 2/ OPSIE 2

Any / _
1,6 = 0,2R,v igeenv - neat 1,6Y
R, = 82 _ 2, 4VL ,
V = IR E

=0,2(4 +8)¥
=24VY

V = IR, Vext = vata Any / Enigeen ¥

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(3)

(4)
[15]

(2)

(4)

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8.2.2

8.2.4

Grade/Graad 12 June Exam/Junie Eksamen.
Marking Guidelines/Nasienriglyne

FS/VS/ JUNE 2025

OPTION 1/ OPSIE 1| OPTION 2/ OPSIE 2 | OPTION 3/ OPSIE 3
V=IR 1 1 1 Ratios Resistance: R1:R2
2,4 =1(2)¥ Ry Ry - Ry Weerstandverhouding 6 : 1v
1=1,2A 1 1 re Current:1 : 6
= 3+a48 = 7(0,2) v
trot = 1,2 + 2¥ a =1,4 AY
=1,4 Av ry
V, = IpRy
2,4 = In(1,71) ¥
Ip = 1,4A¥ (3)
8.2.3 POSITIVE MARKING FROM QUESTIONS 8.2.1 AND 8.2.2
POSITIEWE NASIEN UIT VRAE 8.2.1 EN 8.2.2
OPTION 1/ OPSIE1__ | OPTION 2/ OPSIE 2
e=I(R+r) Viost = € — Vext
3,11¥= 1,4(1,71+r)¥ = 311-24”
r =0,510V = 0,71
Viese = Ir
0,71 = 14rv
r= 05104 (3)
OPTION 1/ JOPTION2/ =] OPTION 3/
OPSIE 1 _ OPSIE 2 OPSIE 3
P=Viv 'P=PRY pe Be
= 0,714(1,4) v = (1,4)?(0,51) v (714)?
= 0,9966 WY = 0.9996 WY = Ope
| = 0,9996 WY
(3)
[15]

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GRAND TOTAL: / GROOTTOTAAL: 150

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