Faculty of Engineering

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1 Faculty f Engneerng DEPARTMENT f ELECTRICAL AND ELECTRONIC ENGINEERING EEE 223 Crcut Thery I Instructrs: M. K. Uygurğlu E. Erdl Fnal EXAMINATION June 20, 2003 Duratn : 120 mnutes Number f Prblems: 6 Gd Luck STUDENT S NUMBER NAME SURNAME Prblem Pnts TOTAL 100

2 1. Cnsder e crcut n Fg. 1. a) Wuld yu use e ndal r mesh analyss t fnd e pwer absrbed by 20 V surce? Explan yur chce. b) Use e med yu prefer n (a) t fnd e pwer V ma SN1 100 Ω Ω Ω 200 Ω SN2 Fgure 1 a) There are 4 meshes, 2 current surces, 4 nnreference ndes and 2 ltage surces n e crcut. Ths means at, ere are 4(# f meshes)2(current surces) = 2 unknwn mesh currents and 4(nnreference ndes)2(ltage surces) = 2 unknwn nde ltages. We need t wrte 2 equtns fr each analyss med. Snce tw f e surces are dependent, e cntrl arables shuld be wrtten n terms f mesh currents r nde ltages. If we prefer ndal analyss and chse e bttm nde as reference nde en e cntrl arables wll be nde ltages and e calculatn f unknwn quanttes wll be ery easy. KCL at SN1: = = = (1) 1 2 KCL arund SN2: = = = (2) Multplcatn f Eq.(2) by 7 and en summatn f Eq(1) and (2) yelds: M. K. Uygurğlu, E. Erdl June 20, 2003

3 12.5 = = = 72 V 12.5 = 44 V and KCL at nde 2 20 : = 0 = 24 ma. 100 Therefre e pwer suppled by e 20 V surce s: P 20V = = (20) 480 mw suppled (480 mw absrbs) M. K. Uygurğlu, E. Erdl June 20, 2003

4 2. Use e prncple f superpstn t fnd e current n e crcut n Fg Ω 10 Ω 45 V 40 Ω 15 Ω 30 Ω 8 A Fgure 2 45 V ltage surce s acte and e er surces alues are set t zer. 5 Ω 10 Ω 45 V ' 40 Ω 15 Ω 30 Ω R e R e ' = //(10 30) = 40Ω 45 9 = = A 40 8 M. K. Uygurğlu, E. Erdl June 20, 2003

5 ltage surce s acte and e er surces alues are set t zer. 5 Ω 10 Ω 45 V '' 40 Ω 15 Ω 30 Ω 320 Re = //(5 15) = = = A R 320 Usng current dsn prncple '' e = = A R e 8A current surce s acte and e er surces alues are set t zer. KVL arund : 0 ''' = (1) KVL arund : ''' 40 80= (2) Multply Eq. (1) by 2 and en sum up Eq(1) and (2) ''' 80 = 480 ''' = 6A = = 7A ' '' ''' 5 Ω 10 Ω ''' 40 Ω 15 Ω 30 Ω 8 A 8 A M. K. Uygurğlu, E. Erdl June 20, 2003

6 3. The arable resstr ( R ) n e crcut n Fg.3 s adjusted untl t absrbs maxmum pwer frm e crcut. a) Fnd e alue f R. b) Fnd e maxmum pwer. 4 Ω 6 Ω R 2 A R 3 A V R Fgure 3 When R = R t wll absrb maxmum pwer. Maxmum pwer s P R max V = 4R 2 In rder t fnd R all ndependent surces are klled. 4 Ω 6 Ω R = 10 Ω Fr V : KCL at V V V = 2 4 V V = 8...(1) : V 4 Ω V 6 Ω KCL at V : V V V 10 = V 3V 2V 20= 36 2 A 3V 5V = 16...(2) Multply Eq. (1) by 5 and en sum up Eq(1) and (2) 2V = 24 V R P = 12 V = 10Ω R max 144 = = 3.6W 40 3 A M. K. Uygurğlu, E. Erdl June 20, 2003

7 4. Fnd 0 n terms f 1 and 2 e crcut n Fg.4. R 2 R 3 R R 1 0 V 1 x R 5 0 V 2 0 _ Fgure 4 KCL at e nertng termnal f OPAMP 1: x R2 = 0 x = R1 R2 R 1 KCL at e nertng termnal f OPAMP 2: R R R R R = 0 = = R R R R R R R R 2 x x M. K. Uygurğlu, E. Erdl June 20, 2003

8 5. In e crcut f Fg. 5, fnd t () fr t > 0. Assume at e swtch has been clsed fr a lng tme and pen at t = 0. t = 0 2 Ω 6 Ω 50 V 1 F 3 Fgure 5 At t = 0 KVL arund e lp: = 0 8 = 40 = 5A en 10 2 = 0 (0) = 20V Fr t > 0 : 2 Ω 6 Ω 50 V 2 Ω t t () = ( ) ( (0) ( )) e τ 1 2 τ = RC = 2 = s 3 3 In rder t fnd ( ), we wll assume at e crcut s under 1 F 3 dc cndtns. It s bus at ( ) = Therefre 3t 3t 2 2 t ( ) = 10 (20 10) e = 10(1 e ) V M. K. Uygurğlu, E. Erdl June 20, 2003

9 6. Fr e crcut n Fg. 6, fnd: t = 0 a) (0 ) and (0 ), b) d(0 ) d(0 ) and, dt dt c) ( ) and ( ). 12 V 6 Ω 4 Ω Snce e capactr ltage and nductr current cannt change nstantaneusly, (0 ) = (0 ) and 2 H 0.4 F (0 ) = (0 ). Therefre at t = 0 Fgure 6 6 Ω 4 Ω 12 (0 ) = (0 ) = = 2 A 6 (0 ) = (0 ) = 12V 12 V At t = 0 d d = 0 = 10 dt dt 2 and ( ) d d = 0.4 = 2.5 dt dt d(0 ) 1 = ( (0 ) 10 (0 )) = 4 A/ s dt 2 d(0 ) = 2.5 (0 ) = 5 V / 2 dt 6 Ω 4 Ω 2 H 0.4 F M. K. Uygurğlu, E. Erdl June 20, 2003

10 at t = ( ) = ( ) = 0 6 Ω 4 Ω 2 H M. K. Uygurğlu, E. Erdl June 20, 2003