ES 50 Pracice Final Exam. Given ha v 8 V, a Deermine he values of v o : 0 Ω, v o. V 0 Firs, v o 8. V 0 + 0 Nex, 8 40 40 0 40 0 400 400 ib i 0 40 + 40 + 40 40 40 + + ( ) 480 + 5 + 40 + 8 400 400( 0) 000 480 Then 480 + 5 000 0 Ω 0 480 + 5 8 5 c 40 + + 40. Given ha 0 in his circui, consider hese wo observaions: When Ω hen v 4 V i A. When 6 Ω hen v 6 V i A. Fill in he blanks in he following saemens: a. The maximum value of i is 4 A. b. The maximum value of v is 8 V. c. The maximum value of p iv occurs when Ω. d. The maximum value of p iv is 8 W. We can replace he par of he circui o he lef of he erminals by is Thevenin equivalen circui: Using volage division v + v oc v oc using Ohm s law i + voc By inspecion, v voc will be maximum when. The + + maximum value of v will be v oc. Similarly, i + v oc will be.
maximum when 0. The maximum value of i will be v oc isc. The maximum power ransfer heorem ells use ha p iv will be maximum when. Then voc voc p iv voc. + + + Le s subsiue he given daa ino he equaion i. + voc voc When Ω we ge 4+ voc. When 6 Ω we ge 6+ v + 6 + voc 8 So 6+ 4+ Ω voc 4+ 8 V. Also isc 4 A. Now he blanks can be easily filled-in. v oc oc.. The inpu o his circui is he volage, v s. The oupu is he volage v o. The volage v b is used o adjus he relaionship beween he inpu oupu. Deermine values of 4 v b ha cause he circui inpu oupu have he relaionship specified by he graph v b.6 V 4 6.5 kω. ecognize he volage divider, volage follower noninvering amplifier o wrie 0 0 4 4 4 4 v o vs + + vb vs + + v 0 0 + 5 0 0 0 0 0 75 0 0 0 (Alernaely, his equaion can be obained by wriing wo node equaions: one a he noninvering node of he lef op amp he oher a he invering node of he righ op amp.) 5 The equaion of he sraigh line is vo vs + 5 Comparing coefficiens gives b 4 5 5 75 0 4 6.5 0 6.5 kω 75 0
4 6.5 0 5 + 5 vb + vb.08vb vb.6 V 0 0 0 0.08 4. Consider his inducor. The curren volage are given by 5 4.6 0 0. i() a+ b 0. 0.5 c 0.5.5 0 < < 0. v() 5 0. < < 0.5 0 > 0.5 where a, b c are real consans. (The curren is given in Amps, he volage in Vols he ime in seconds.) Deermine he values of he consans: a 0 A/s, b 5.6 A c 0.6 A A 0. s For 0. 0.5 i ( 0.) 5( 0.) 4.6.6 A i() 5dτ 6 0τ 6 0 0. ( 0 ) 6 0 5.6 A 5..... 0. A 0.5 s For 0.5 i ( 0.5) 0( 0.5) 5.6 0.6 A 0dτ 0.6 0.6.5 () 0.5 i Checks: A 0. s i ( 0.) 0( 0.) 5.6.6 A For 0. 0.5 d d.5.5 0 5.6.5 0 5 V d d v () i () ( ) 0.5 0.6 (.6) i( 0.5) i( 0.) 5dτ 0( 0.5 0.) A.5 0.
5. This circui is a seady sae when he swich opens a ime 0. a The capacior volage is v() A Be for 0 +. Deermine he values of he consans A, B, a: A 4 V, B 8 V a 0 s. Soluion: Before 0, wih he swich closed he circui a he seady sae, he capacior acs like an open circui so we have Using superposiion 60 60 60 0 v( 0 ) 6+ 6 6+ 6 V 0 + 60 60 60 + 60 0 4 ( ) ( ) The capacior volage is coninuous so v v 0+ 0 V. Afer 0 he swich is open. Deermine he Thevenin equivalen circui for he par of he circui conneced o he capacior: 6 The ime consan is τ C 0 0 5 0 0. s so 0. τ s The capacior volage is given by τ ( oc ) oc () 60 v oc 6 4 V 60 + 0 0 60 0 kω 0 0 v v0 + v e + v 4 e + 4 4 + 8 e V for 0
6. This circui is a seady sae before he swich closes a ime 0. Afer he swich closes, he inducor curren is given by 5 () i 0.6 0. e A for 0 Deermine he values of, L: 0 Ω, 0 Ω L 4 H Soluion: The seady sae curren before he swich closes is equal o 50 i 0 0.6 0.e 0.4 A. The inducor will ac like a shor circui when his circui is a seady sae so 0.4 i( 0) 0 + + Ω Afer he swich has been open for a long ime, he circui will again be a seady sae. The seady sae inducor curren will be 5 i 0.6 0.e 0.6 A The inducor will ac like a shor circui when his circui is a seady sae so 0.6 i( ) 0 Ω Then 0 Ω. Afer he swich is closed, he Thevenin resisance of he par of he circui conneced o he inducor is. Then 0 5 L 4 H τ L L L 7. The volage curren for his circui are given by v() 0 cos (0 + 5 ) V i().49 cos (0 + 6 ) A Deermine he values of he resisance,, capaciance, C: 9 Ω C 5 mf.
Soluion: In he frequency domain we have: V 0 5 0 j Z ( 5 6 ).4 48 8.98 j9.97 Ω 0C I.49 6.49 Equaing real imaginary pars gives 9 Ω C 5 mf. 0 9.97 8. This circui is a seady sae. The volage source volages are given by v () cos ( 90 ) V v () 5 cos ( + 90 ) V The currens are given by i () 744 cos ( 8 ) ma, i () 540.5 cos ( + 00 ) ma i() A cos ( 64 ) ma Deermine he values of A,,, L C: A 460 ma, 0 Ω, 0 Ω, L 6 H C 50 mf. Soluion: epresen he circui in he frequency domain using impedances phasors: ( j ) ( j ) ( 0.49 0.094) j ( 0.657 0.5) I I+ I 0.744 8 + 0.5405 00 0.49 0.657 + 0.094 + 0.5 + + 0.44 j 0.5 0.460 64 i() 460 cos ( 64 ) ma eplacing series impedances by equivalen impedances gives
Z Z + jω L j ω C From KVL Nex V 0I 90 0 0.460 64 ZI+ 0I V 0 Z I 0.744 8 ( j ) j 0 0.44 0.5 0.744 8 4.4 j0.75.6 67.6 0.744 8 0.744 8 5.6 50.4 0 + j Ω V 0I 5 90 0 0.460 64 ZI + V 0I 0 Z I 0.5405 00 ( j ) j5 0 0.44 0.5 0.5405 00 4.4+ j 6.5 7.66 54.7 0.5405 00 0.5405 00 4.4 55. 0 j0 Ω 0 + j + jω L + j L 0 Ω L 6 H 0 j0 j j 0 Ω C 0.05 F ω C C 0
9. The inpu his circui is he curren () is cos 5+ 5 A. In he frequency domain, his circui is represened by he node equaion where V a d + j0.5 j0.5 V 5 a j0.5 0.5 + je b 0 V V b are he phasor node volages d e are real numbers. Deermine he values of d e. d 0.5 Ω e.5 Ω Soluion: epresen he circui in he frequency domain using impedances phasors: Apply KCL a node a o ge Va Va Vb 5 + ( 0.5 + j0.5) Va ( j0.5) V 8 j Apply KCL a node b o ge Va Vb Vb Vb + j0.5( V V ) ( 0.5 j4) V j 4 j0.5 a b b a b ( j ) 0 j0.5v + 0.5.5 V Organize hese equaions ino marix form o ge 0.5 + j0.5 j0.5 Va 5 j0.5 0.5 j.5 b 0 V Compare his equaion o he given node equaion o see ha d 0.5 Ω e.5 Ω. b