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VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING DEPARTMENT: PROCESS CONTROL AND COMPUTER SYSTEMS BACCALAUREUS TECHNOLOGIAE: ENGINEERING ELECTRICAL SUBJECT : CIRCUIT ANALYSIS IV EICAM4A ASSESSMENT : UNIT FIRST ASSESSMENT DATE : 3 AUGUST 207 DURATION : 4H00 5H30 EXAMINER : R FITCHAT MODERATOR : L COETSEE REQUIREMENTS: Pocket calculator may be ued INSTRUCTIONS:. Neat work requred 2. Number your anwer clearly and correctly Full mark = 50 Total = 50 QUESTION PAPER CONSISTS OF: typed page DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED

Crcut Analy IV EICAM4A Unt Frt Aement 3 Augut 207 Page Queton For the crcut n Fgure, the wtch S cloed at tme t = 0. Calculate (t) for all tme t. (0) Queton 2 For the crcut hown n Fgure 2, fnd an expreon for (t) for all tme t, f the ource current, (t), gen by: (t) for t 0 for t 0 (2) V (t) amp 0 t F H S t=0 Fgure Fgure 2 Queton 3 Refer to the crcut n Fgure 3 and determne the zero tate tep repone of the oltage (t), acro the capactor. (2) = u(t) 3 H F Fgure 3 Queton 4 For the crcut hown n Fgure 4, the wtch S wa open for a long tme, and t then cloed at tme t = 0. Determne the current (t) n the ½ H nductor and the oltage (t) acro the ¼ F capactor, for all tme t. (6) 2 V S H 2 t=0 F 4 Fgure 4 ---ooo000ooo--- Total: 50 Appendx Trgonometrc dentte: Network model Acot + Bnt = A2 B 2 co (t tan - (B/A)) co = n( + /2) = B 2 A2 n (t + tan - (A/B)) n = co( /2) Euler rule: e j = co + jn co = ½ (e j + e j ) n = ½ j (e j e j L ) Laplace tranform: L[f(t)] = F() = (0) 0 f(t)e - tdt L(0) f(t) F() f(t) F() df(t)/dt F() f(0) te -at u(t) /(+a) 2 t 0 f(t)dt F()/ nt u(t) /(2 + 2 ) (t) cot u(t) /( 2 + 2 ) u(t) / e -at nt u(t) /[(+a) 2 + 2 ] e -at u(t) /(+a) e -at cot u(t) (+a)/[(+a) 2 + 2 ] tu(t) / 2 2Ae -at co(t+)u(t) A A a j a j C (0) C(0) C L

Faculty: Department: Dploma: Subject: Internal code: VAAL UNIVERSITY OF TECHNOLOGY Engneerng Coer Sheet for Memorandum (Unt Frt Aement 3 Augut 207) Proce Control and Computer Sytem Baccalaureu Technologae: Engneerng: Electrcal Crcut Analy IV EICAM4A Nne dgt code: 080820706 Hour: Examner: Moderator: ½ R Ftchat L Coetee Total Mark: 50 Full mark: 50 Sgnature of Examner: Sgnature of Moderator: Date: Date:

Stroombaananale IV EICAM4A Eenhed Eerte Ealuae 3 Augutu 206 Memorandum Blady. t < 0: (t) = V [2] t 0: = d/dt + d/dt + 2 = [5] = Ae 2t + ½ [] (0) = A = ½ [] [0] (t) = ½(e 2t + ) [] t < 0 t 0 V 2. t < 0: (t) = ½ A [] t 0: Node N: = +... () t < 0 t 0 N Loop L: d/dt = 0 = + d/dt... (2) L (2) n (): = + + d/dt d/dt d/dt + 2 = [6] [2] (t) = ( /2) + Ae 2t [2] but (0) = ½ A = [] (t) = ½ + e 2t amp, for t 0 [2] 3. t < 0: (t) = 0 [½], (t) = 0 [½] zero tate t 0: 3 d/dt = 0 en = + d/dt 3 d/dt 3( + d/dt) d/dt( + d/dt) = 0 3 3d/dt d/dt d 2 /dt 2 = 0 d 2 /dt 2 + 4d/dt + 4 = [6] = 2 en n = 2 krtek gedemp (t) = Ate -2t + Be -2t + ¼ [2] (0) = 0 B + ¼ = 0 B = ¼ d/dt = Ae -2t 2Ate -2t 2Be -2t d(0)/dt = A 2B V 3 d(0)dt = (0) (0) = 0 A 2B = 0 A 2( /4) = 0 A = /2 [2] (t) = ½te -2t ¼e -2t + ¼ [3] [6] ½ 0 4. t < 0: (t) = V [] en (t) = A [] t 0: Loop contanng ½H and ¼F (loop L): L + = 0 ½d/dt + = 0 = ½d/dt... () 2 V Node N: = C + R = ¼d/dt + / = ¼d/dt +... (2) () n (2): {or (2) n ()} = ¼d/dt( ½d/dt) + ( ½d/dt) { = ½d/dt(¼d/dt+} (/8)d 2 /dt 2 + ½d/dt + = 0 {(/8)d 2 /dt 2 +½d/dt+=0} d 2 /dt 2 + 4d/dt + 8 = 0 [6] {d 2 /dt 2 +4d/dt+8=0 [6] } = 2 and n = 8 underdamped wth d = 2 r/ (t) = Ae -2t co2t+be -2t n2t [2] {(t)=ae -2t co2t+be -2t n2t [2] } t (0) = A = {(0)= A=} d/dt = 2Ae 2t co2t 2Ae 2t n2t 2Be 2t n2t + 2Be 2t co2t d(0)/dt = 2A + 2B {d/dt= 2Ae 2t co2t 2Ae 2t n2t 2Be 2t n2t+2be 2t co2td(0)/dt= 2A+2B} but (0) = ½d(0)/dt (from ()) = ½d(0)/dt d(0)/dt = 2 {but (0)=¼d(0)/dt+(0) (from (2)) =¼d(0)/dt+ d(0)/dt=0} 2() + 2B = 2 B = 0 (t) = e 2t co2t for t 0 [4] { 2() + 2B = 0 B = (t) = e 2t co2t + e 2t n2t [4] } and d/dt = 2e 2t co2t 2e 2t n2t {and d/dt = 2e 2t co2t 2e 2t n2t 2e 2t n2t+2e 2t co2t = 4e 2t n2t} (t) = ½d/dt = ½( 2e 2t co2t 2e 2t n2t) = e 2t co2t + e 2t n2t [2] {(t)=¼d/dt+=¼[ 4e 2t n2t] + e 2t co2t + e 2t n2t = e 2t co2t [2] } F V L F 2 4 N d/dt t < 0 t 0 L C R

VAAL UNIVERSITY OF TECHNOLOGY FACULTY OF ENGINEERING DEPARTMENT: PROCESS CONTROL AND COMPUTER SYSTEMS BACCALAUREUS TECHNOLOGIAE: ENGINEERING ELECTRICAL SUBJECT : CIRCUIT ANALYSIS IV EICAM4A ASSESSMENT : UNIT 2 FIRST ASSESSMENT DATE : 28 SEPTEMBER 207 DURATION : 0H00 H30 EXAMINER : R FITCHAT MODERATOR : L COETSEE REQUIREMENTS: Pocket calculator may be ued INSTRUCTIONS:. Neat work requred 2. Number your anwer clearly and correctly Full mark = 50 Total = 50 QUESTION PAPER CONSISTS OF: typed page DO NOT TURN THE PAGE BEFORE PERMISSION IS GRANTED

Crcut Analy IV EICAM4A Unt 2 Frt Aement 28 September 207 Page Queton Refer to the crcut n Fgure. a) Fnd the teady tate current repone (t) n the tme doman, olng the dfferental equaton for, f the upply oltage, gen by the real nuod (t) = co2t olt. Ge your anwer n the form, Aco(t + ). (0) b) Fnd the teady tate current repone (t) n the tme doman, olng the dfferental equaton for, f the upply oltage the complex nuod = e j2t olt. Ge your anwer n the form Ae j(t + ). (6) c) Ue frequency doman analy to determne the phaor current I, f the upply oltage the phaor V = 0 olt. Recontruct the tme form of (t), Aco(t + ), agan from the phaor current I. (6) Queton 2 Refer to the crcut n Fgure 2. a) Determne the reonance frequency of the crcut from the perpecte that the mpedance een by the ource, mut become real. (0) b) Calculate the mpedance of the crcut at reonance. (2) Queton 3 Refer to the crcut n Fgure 3. The upply oltage to the crcut gen by (t) = co2t u(t) olt. The ntal current through the nductor (0) = 0 amp and the ntal oltage acro the capactor (0) = 0 olt. Draw an equalent Laplace network model for the crcut and ue th model to fnd an expreon for (t). (6) co2t e j2t 0 ---ooo000ooo--- Total: 50 = co2t u(t) F F H 2 (0) = 0 H Fgure F (0) = 0 H Fgure 2 Fgure 3 Appendx Trgonometrc dentte: Network model Acot + Bnt = A2 B 2 co (t tan - (B/A)) co = n( + /2) = B 2 n (t + tan - A2 (A/B)) n = co( /2) Euler rule: e j = co + jn co = ½ (e j + e j ) n = ½ j (e j e j ) L Laplace tranform: L[f(t)] = F() = - t (0) 0 f(t)e dt f(t) F() f(t) F() L(0) df(t)/dt F() f(0) te -at u(t) /(+a) 2 t 0 f(t)dt F()/ nt u(t) /(2 + 2 ) (t) cot u(t) /( 2 + 2 ) u(t) / e -at nt u(t) /[(+a) 2 + 2 ] e -at u(t) /(+a) e -at cot u(t) (+a)/[(+a) 2 + 2 ] tu(t) / 2 2Ae -at co(t+)u(t) A A a j a j C (0) C(0) C L

VAAL UNIVERSITY OF TECHNOLOGY Coer Sheet for Memorandum (Unt 2 Frt Aement 28 September 207) Faculty: Engneerng Department: Proce Control and Computer Sytem Dploma: Baccalaureu Technologae: Engneerng: Electrcal Subject: Crcut Analy IV Internal code: EICAM4A Nne dgt code: 080820706 Hour: ½ Examner: R Ftchat Moderator: L Coetee Total Mark: 50 Full mark: 50 Sgnature of Examner: Sgnature of Moderator: Date: Date:

[22] Stroombaananale IV EICAM4A Eenhed2 Eerte Ealuae 28 September 207 Memorandum Blady. a) ( )/ = + d/dt... () and = d/dt... (2) d/dt = + d 2 /dt 2 d 2 /dt 2 + d/dt + = d 2 /dt 2 + d/dt + = co2t [5] we nt that = Aco2t + Bn2t, therefore d/dt = 2An2t + 2Bco2t and d 2 /dt 2 = 4Aco2t 4Bn2t 4Aco2t 4 Bn2t 2An2t + 2Bco2t + Aco2t + Bnt = co2t [ 3A + 2B]cot + [ 3B 2A]nt = cot 3A + 2B = en 3B 2A = 0 A = (3/3) = 0.23076923 and B = (2/3) = 0.5384654 (t) = (3/3)co2t + (2/3)n2t [4] = 0.277350098co(2t 2.55359) ampere [] (0) b) For (t) = e j2t d 2 /dt 2 + d/dt + = e j2t j(2t + ) [] we agan nt that (t) mut hae the form, (t) = Ae j(2t + ) therefore d/dt = j2ae j(2t + ) and d 2 /dt 2 = j 2 4Ae j(2t + ) = 4Ae 4Ae j(2t + ) + j2ae j(2t + ) + Ae j(2t + ) = e j2t ( 4 + j2 + )Ae j(2t + ) = e j2t ( 3 + j2)ae j e j2t = e j2t ( 3 + j2)ae j = Ae j = /( 3 + j2) = 0.277350098 2.553590 Ae j = 0.277350098e j2.553590 A = 0.277350098 [2] and = 2.553590 r [2] (t) = 0.277350098 e j(2t 2.55359) amp [] (6) c) Z = + 0.5 90//290) = + [(0.5 90 290)/(0.5 90+290)] = + (0/.590) = + (2/3) 90=.2085043 33.6900676 [2] I t = V /Z = 0/.2085043 33.6900676 = 0.8320502933.6900676 A [2] I = [0.5 90/(0.5 90 + 290)] I t = [0.5 90/(.590] 0.8320502933.6900676 = 0.333333333 80 0.8320502933.6900676 = 0.277350097 46.309932 A [] (t) = 0.277350097co(2t 2.55359005 r ) [] (6) {or V = [(2/3) 90/(+(2/3) 90)]0=0.5547002 56.309932 I = 0.5547002 56.309932/290=0.277350 46.309932} 2. a) Z = + j + 2//(/j) [2] = + j + [2(/j)]/[2 + (/j)] = + j + [2/( + j2)] = + j + [2( j2)/( + 4 2 )] = + j + [2/( + 4 2 )] j[4/( + 4 2 )] j = { + [2/( + 4 2 )]} + j{ [4/( + 4 2 ]} [4] { = [( + 4 2 + 2)/( + 4 2 )] + j{[( + 4 2 ) 4]/( + 4 2 )} 2 /j = [(3 + 4 2 )/( + 4 2 )] + j[(4 2 3)/( + 4 2 )] } For reonance: Im{Z} = 0 [4/( + 4 2 ] = 0 4/( + 4 2 ) = + 4 2 = 4 4 2 = 3 = 3/2 (0.866925 r/a) [4] (0) [2] b) Z 0 = { + [2/( + 4 2 )]} = 3/2 = + 2/( + 3) = + 0.5 =.5 (2) [6] 3. [/( 2 + 4) ]/ = + /(/) /( 2 + 4) = + + But = /( 2 + 4) = + + 2 ( 2 + + ) = /( 2 + 4) = /( 2 + + )( 2 + 4) [4] co2t e j2t = 2 r/ = /[( 2.0944)( 2.0944)( 2.5708)( 2.5708)] [2] = 0.60290.76665/( 2.0944) + 0.6029 0.76665/( 2.0944) + 0.38675 2.55359/( 2.57) + 0.386752.55359/( 2.57) [4] = 0.60290.76665/( + 0.5 j0.866025) + 0.6029 0.76665/( + 0.5 + j0.866025) + 0.38675 2.55359/( + 0 j2) + 0.386752.55359/( + 0 + j2) (t) = 20.6029e -0.5t co(0.866025t + 0.76665) + 20.38675e 0 co(2t 2.55359) = 0.320258e -0.5t co(0.866025t + 0.76665) + 0.27735co(2t 2.55359) [2] V 0 V 2 4 [] [] I t [] 0.5 90 V I 290 / []

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