Ebn111 2014 st1 14 vraag 35 2 stel n

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EBN111 2014-ST1 14 Vraag 3.5 [2] Stel 'n matriksvergelyking d.m.v. inspeksie op in terme van die lusstrome. Question 3.5 [2] Use inspection to set up a matrix equation in terms of the mesh currents. + - + - 6g1848 2Ω 20Ω 7g1848 12Ω 8Ω 6Ω g1861 g2870 g1861 g2869
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EBN111 2014-ST1 15 Vraag 3.6 [4] Gegee: g2010=99 , g1848 g3003g3006 =0.7g1848 en I B = 5 μA. a) Bepaal V CE . b) Bepaal V S . (Vrae a) en b) is onafhanklik van mekaar.) Question 3.6 [4] Given: g2010=99 , g1848 g3003g3006 =0.7g1848 and I B = 5 μA. a) Determine V CE . b) Determine V S . (Questions a) and b) are independent of one another.) + - + - g1848 g3020 20g1863Ω 5g1863Ω 15g1848 1g1863Ω
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EBN111 2014-ST1 16 Vraag 3.7 [4] a) Bepaal k indien i 1 = 6 A en i 2 = 6 A. = - - 2 0 0 2 4 0 0 1 1 1 10 3 2 1 i i i k b) Bepaal V 2 . - = - - 2 2 4 1 4 5 2 1 V V Question 3.7 [4] a) Determine k if i 1 = 6 A and i 2 = 6 A. = - - 2 0 0 2 4 0 0 1 1 1 10 3 2 1 i i i k b) Determine V 2 . - = - - 2 2 4 1 4 5 2 1 V V
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EBN111 2014-ST1 17 EBN111 SEMESTER TOETS/TEST 1 (13 MAR 2014) ANTWOORDBLAD / ANSWER SHEET Van / Surname: Studentenommer: Student number: Voorletters / Initials: Taal / Language Afrikaans English Q1 Q2 Q3 Tot: /45 VRAAG 1 – QUESTION 1 (7) 1.1 a) p(1s) = b) W = c) d) Q = 1.2 P 1 = P 2 = P 3 = VRAAG 2 – QUESTION 2 (16) 2.1 Onafhanklike lusse: Independent loops: Takke: Branches: Nodusse: Nodes: 2.2 I= v o = R= 2.3 v y = 2.4 I 1 = G 1 = 2.5 R eq = 2.6 a) R cb = b) R ab = VRAAG 3 – QUESTION 3 (22) 3.1 V 0 = 3.2 i X = 3.3 a = b = c = d = 3.4 e = f = g = h = 3.5 = 2 1 i i 3.6 a) V CE = b) V S = 3.7 a) k = b) V 2 =
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