ET 143 (2009 vstel + memo)T

Toets Vraag 1 Memo a) [ 10 ] i) P = i(t)*v(t) = 5*4e -2t A - - - - 1 P = 20e -2t W - - - - - - - - - - - - 1 (marks subtracted for incorrect units) ii) w = int(P) = int(20e -2t )dt - - - 1 w = [-10e -2t ] 0 tot ∞ - - - - - 2 w = - (-10) = +10 J - - - - - - - 1 (marks subtracted for incorrect units) iii) Passive convention of figure shown (or energy +) Energy is absorbed - - - - - 2 iv) Energy is absorbed, and current and voltage are not derivatives of each other - - 2 b) [ 10 ] i) Two series resistors add 10+10 = 20 ohm - - - - - - - - - -1 Four parallel resistors add 20 || 20 || 20 || 20 = 5 ohm - - 1 Two series resistors add 5+15 = 20 ohm - - - - - - - - - - 1 Two parallel resistors add 20 || 20 = 10 ohm - - - - - - - - 1 Two series resistors add Req = 5 + 10 = 15 ohm - - - - - -1 (Marks subtracted for any of these steps left out) ii) Voltage division V = Vs*R/Rt - - - 1 V = 10*30 / (10 + 5) - - - - - - - - 1 V = 20 V - - - - - - - - - - - - - - - - 1 iii) I = V/R = 30 / 15 = 2 A - - - - - 2

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Vraag 4: Kapasitors & Induktors [20] Question 4: Capacitors & Inductors [20]

Vanaf die onderstaande sketse, From the sketches shown below,

is(t)+vL(t)−

200 mH

0 10 20 30 40 500

1

2

3

4

5

t

is(t)

[ms]

[A]

is(t)+

vC(t)−1000 µF

(a) bereken die stuksgewys kontinuefunksie vir die spanning oor dieinduktor, vL(t), en stip die funksiem.b.v. ’n netjiese vryhand skets vir dietydsverloop vanaf 0 tot 50 ms.

[6] (a) calculate the piece-wise linear func-tion for the voltage across the inductor,vL(t), and plot the function using a neatfreehand drawing for the time intervalfrom 0 to 50 ms.

[6]

The question is almost similar to Hambley Example 3.6.

The first step is to write is(t) as a piece-wise linear function.

is(t) =

250t 0≤ t ≤ 20 ms5 20≤ t ≤ 40 ms−500t +25 20≤ t ≤ 50 ms

The next step is to solve vL(t) (piece-wise) for each time interval.

For 0≤ t ≤ 20 ms:

vL(t) = LdiL(t)

dt(given)

= 0,2ddt

(250t)

= 50V 4

For 20≤ t ≤ 40 ms:

vL(t) = LdiL(t)

dt(given)

= 0,2ddt

(5)

= 0V 4

For 40≤ t ≤ 50 ms:

vL(t) = LdiL(t)

dt(given)

= 0,2ddt

(−500t +25)

=−100V 4

bladsy 1 van 5 page 1 of 5

Thus for 0≤ t ≤ 50 ms:

vL(t) =

50 0≤ t ≤ 20 ms0 20≤ t ≤ 40 ms−100 20≤ t ≤ 50 ms

Finally we plot vL(t) for all three time intervals.

See also Hambley Figure 3.17 (b).

10 20 3040 50

-100

-80

-60

-40

-20

0

20

40

60

80

100

t

vL(t)

[ms][V]

444

(b) bereken die stuksgewys kontinuefunksie van die energie gestoor in dieinduktor, wL(t), en stip die funksiem.b.v. ’n netjiese vryhand skets vir dietydsverloop vanaf 0 tot 50 ms.

[6] (b) calculate the piece-wise linear functionof the energy stored in the inductor,wL(t), and plot the function using aneat freehand drawing for the time in-terval from 0 to 50 ms

[6]

The question is almost similar to Hambley Example 3.6.


is(t) =

250t 0≤ t ≤ 20 ms5 20≤ t ≤ 40 ms−500t +25 20≤ t ≤ 50 ms

The second step is to find the general expressing for wL(t) as a function of iL(t).

wL(t) =∫ t

−∞pL(t)dt

=∫ t

−∞iL(t)vL(t)dt

=∫ t

−∞iL(t) ·L

diL(t)dt

dt

= L∫ iL(t)

iL(−∞)=0iL(t)diL

= 12LiL2(t) (similar to 1

2mv2 or 12Jω

2)


The next step is to solve wL(t) (piece-wise) for each time interval.

For 0≤ t ≤ 20 ms:

wL(t) = 12LiL2(t)

= 12 ·0,2 (250t)2

= 6250 t2 4

For 20≤ t ≤ 40 ms:

wL(t) = 12LiL2(t)

= 12 ·0,2 (5)2

= 2,5 J 4

For 40≤ t ≤ 50 ms:

wL(t) = 12LiL2(t)

= 12 ·0,2 (−500t +25)2

= 25000 t2−2500 t +62,5 4

Thus for 0≤ t ≤ 50 ms:

vL(t) =

6250 t2 0≤ t ≤ 20 ms2,5 20≤ t ≤ 40 ms25000 t2−2500 t +62,5 20≤ t ≤ 50 ms

Finally we plot wL(t) for all three time intervals.

See also Hambley Figure 3.17 (d).

0 10 20 30 400

0.5

1

1.5

2

2.5

3

t

wL(t)

[ms]

[J]

444

(c) bereken die stuksgewys kontinuefunksie vir die spanning oor diekapasitor, vC(t), en stip die funksiem.b.v. ’n netjiese vryhand skets virdie tydsverloop vanaf 0 tot 50 ms, metvC(0) = 50 V.

[8] (c) calculate the piece-wise linear functionfor the voltage across the capacitor,vC(t), and plot the function using a neatfreehand drawing for the time intervalfrom 0 to 50 ms, with vC(0) = 50 V.

[8]


This question is very similar to the example done by Mr. Nelius Bekker in one of the Thuesday lunchhour lectures.


is(t) =

250t 0≤ t ≤ 20 ms5 20≤ t ≤ 40 ms−500t +25 20≤ t ≤ 50 ms

The second step is to find the general expressing for vC(t) as a function of iC(t).

iC(t) = CdvC(t)

dt(given)

∴∫ t

t0dvC(t) =

1C

∫ t

t0iC(t)dt

⇒ vC(t) =1C

∫ t

t0iC(t)dt + vC(t0)

The next step is to solve vC(t) (piece-wise) for each time interval.

To minimise the writing, we set:

t0 = 0mst1 = 20mst2 = 40mst3 = 50ms

For t0 ≤ t ≤ t1:

vC(t) =1C

∫ t

t0iC(t)dt + vC(t0)

=1

1000×10−6

∫ t

0(250t)dt +50

= 125000 t2 +50 4

Calculate vC(t1), the initial value for the next time interval:

vC(t1) = 100,V 4

For t1 ≤ t ≤ t2:

vC(t) =1C

∫ t

t1iC(t)dt + vC(t1)

=1

1000×10−6

∫ t

20×10−3(5)dt +100

= [5000 t]t20×10−3 +100

= 5000 t 4

Calculate vC(t2), the initial value for the next time interval::

vC(t2) = 200,V 4


For t2 ≤ t ≤ t3:

vC(t) =1C

∫ t

t2iC(t)dt + vC(t2)

=1

1000×10−6

∫ t

40×10−3(−500t +25)dt +200

=[−250000 t2 +25000 t

]t40×10−3 +200

=−250000 t2 +25000 t−400 4

Calculate vC(t3), the final value of the voltage (see plot):

vC(t3) = 225,V

Thus for t0 ≤ t ≤ t3:

vL(t) =

125000 t2 +50 t0 ≤ t ≤ t1500 t t1 ≤ t ≤ t2−250000 t2 +25000 t−400 t2 ≤ t ≤ t3

Finally we plot vC(t) for all three time intervals.

0 10 20 30 400

20

40

60

80

100

120

140

160

180

200

220

240

t

vC(t)

[ms]

[V]

444


FAKULTEIT INGENIEURSWESEFACULTY OF ENGINEERING

Elektrotegniek 143 Electro-Techniques 143Hoof KlastoetsMain Class Test

4 September 20094 September 2009

Tydsduur:Duration: 1.5h Totaal:

Total: 80Mede-Eksaminator Mede-EksaminatorCo-Examiner Co-Examiner

Prof. H.C. Reader Dr. H.A. Engelbrecht

Mede-Eksaminator Mede-EksaminatorCo-Examiner Co-Examiner

Mnr. P.J. Randewijk Mr. P.G. Wiid

Instruksies Instructions1. Beantwoord Vraag 1 & 2 in een antwoordboek

en Vraag 3 & 4 in ’n aparte antwoordboek.1. Answer Question 1 & 2 in one answering book

and Question 3 & 4 in a separate answering book.

2. Toon alle berekening stappe duidelik. 2. All calculation steps must be clearly shown.

3. Die finale antwoorde moet in ink geskryf word. 3. The final answers must be written in ink.

4. Punte sal afgetrek word indien antwoorde se een-hede nie getoon word nie.

4. Marks will be deducted if no units for answersare shown.

5. Sakrekenaars soos voorgeskryf vir die eerstetwee jaar van B.Ing. mag gebruik word.

5. Pocket calculators as prescribed for the first twoyears of B.Eng. may be used.

6. As daar, na u mening, enige parameters ontbreekof verkeerd is, maak redelike aannames, verklaarduidelik u aannames en gaan voort.

6. If, in your opinion, there are any missing orincorrect parameters, make reasonable assump-tions, stating them clearly and continue.

7. Indien daar ’n verskil tussen die “Afrikaanse” en“Engelse” gedeeltes is, stel duidelik watter taalse inligting u gebruik vir u antwoord.

7. If there is a difference between the “Afrikaans”and “English” section, state clearly which lan-guage’s information you are using.

8. Geen selfone word toegelaat nie. 8. No cellphones are allowed.


Vraag 1: Alg. Stroombaan Teorie [20] Question 1: General Circuit Theory [20]

(a) Vir die onderstaande stroombaan ele-ment, met v(t) = 5 V en i(t) = 4e−2t A,

(a) For the circuit element shown below,with v(t) = 5 V and i(t) = 4e−2t A,

+ v(t)−i(t)

i. bepaal die drywing geassosieermet die stroombaan element.

[2] i. determine the power associatedwith the circuit element.

[2]

ii. bepaal die energie oorgedra vanaft = 0 tot t = ∞.

[4] ii. determine the energy transferredfrom t = 0 to t = ∞.

[4]

iii. word die energie geabsorbeer ofgelewer deur die element?

[2] iii. is the energy absorbed or deliveredby the element?

[2]

iv. kom die element as ’n induktor,kapasitor of weerstand voor, enhoekom?

[2] iv. does this element resemble an in-ductor, capacitor of resistor, andwhy?

[2]

(b) Vir die stroombaan hieronder getoon, (b) For the circuit shown below,

−+30V

5Ω

i +v−

20Ω

15Ω

20Ω

10Ω

10Ω20Ω 20Ω

a

b

i. bepaal die ekwivalente weerstand,Rab, soos deur die bron gesien.

[5] i. determine the equivalent resis-tance, Rab, as seen by the source.

[5]

ii. bepaal v. [3] ii. determine v. [3]iii. bepaal i. [2] iii. determine i. [2]

Vraag 2: Thevenin & Norton [20] Question 2: Thevenin & Norton [20]

Vir die onderstaande stroombaan, For the circuit shown below,

−+72V 6Ω 2A

RLa b

+ −VRL


(a) bepaal die Thevenin ekwivalente weer-stand Rt soos gesien vanuit klemme aen b.

[3] (a) determine the Thevenin equivalent re-sistance Rt as seen from terminals aand b

[3]

(b) bepaal die Thevenin ekwivalente span-ning vt soos gesien vanuit klemme a enb.

[5] (b) determine the Thevenin equivalentvoltage vt as seen from terminals a andb.

[5]

(c) bepaal die kortsluitstroom isc wat vloeivanaf klem a na klem b wanneer dieklemme gekortsluit word.

[4] (c) determine the short circuit current iscflowing from terminal a to b when theterminals are short-circuited.

[4]

(d) gebruik superposisie om die spanningVRL te bereken. Aanvaar die weerstandRL = 6 Ω.

[8] (d) use superposition to determine the volt-age VRL . Assume the resistor RL = 6 Ω.

[8]

Vraag 3: Knooppunt– en Maasanalise [20] Question 3: Nodal – and Meshanalysis

[20]

Vir die onderstaande stroombaan, For the circuit shown below,

−+10V

5Ω

10Ω

2Ω

2A

4Ω

v1v2 v3

i1 i2

i3

(a) gebruik die knooppunt span-ningsanalise metode om spannings v2en v3 te bereken.

[12] (a) use the nodal voltage analysis methodto calculate the voltages v2 and v3.

[12]

(b) maak gebruik van die maas-stroomanalise metode en skryf dietwee relevante vergelykings neerom die onbekende maas-strome tebereken.

[8] (b) use the mesh-current analysis methodto write down two relevant equations tosolve the unknown mesh currents.

[8]

N.B. Moenie die vergelykings vir (b)oplos nie.

N.B. Do not solve the equations for(b).


Vraag 4: Kapasitors & Induktors [20] Question 4: Capacitors & Inductors [20]

Vanaf die onderstaande sketse, From the sketches shown below,

is(t)+vL(t)−

200 mH

0 10 20 30 40 500

1

2

3

4

5

t

is(t)

[ms]

[A]

is(t)+

vC(t)−1000 µF

(a) bereken die stuksgewys kontinuefunksie vir die spanning oor dieinduktor, vL(t), en stip die funksiem.b.v. ’n netjiese vryhand skets vir dietydsverloop vanaf 0 tot 50 ms.

[6] (a) calculate the piece-wise linear func-tion for the voltage across the inductor,vL(t), and plot the function using a neatfreehand drawing for the time intervalfrom 0 to 50 ms.

[6]

(b) bereken die stuksgewys kontinuefunksie van die energie gestoor in dieinduktor, wL(t), en stip die funksiem.b.v. ’n netjiese vryhand skets vir dietydsverloop vanaf 0 tot 50 ms.

[6] (b) calculate the piece-wise linear functionof the energy stored in the inductor,wL(t), and plot the function using aneat freehand drawing for the time in-terval from 0 to 50 ms

[6]

(c) bereken die stuksgewys kontinuefunksie vir die spanning oor diekapasitor, vC(t), en stip die funksiem.b.v. ’n netjiese vryhand skets virdie tydsverloop vanaf 0 tot 50 ms, metvC(0) = 50 V.

[8] (c) calculate the piece-wise linear functionfor the voltage across the capacitor,vC(t), and plot the function using a neatfreehand drawing for the time intervalfrom 0 to 50 ms, with vC(0) = 50 V.

[8]

Formules Formulas

∫e−ktdt =−e−kt

k

∫cos(kt)dt =

1k

sin(kt)dt

R =ρ lA

q = Cv

C =εAd

ε0 = 8,854187×10−12 F/m

ε = ε0εr

iC(t) = CdvC(t)

dt

vL(t) = LdiL(t)

dt


ET 143 (2009 vstel + memo)T

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