EE203-SUNYBuffalo-22-Chapter13-03

# EE203-SUNYBuffalo-22-Chapter13-03 - SMALL for Big Things...

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Unformatted text preview: SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Appendix A: Solution of Linear Simultaneous Equations EE 203 Circuit Analysis 2 Lecture 22 Chapter 13.3 Applications (continued) Kwang W. Oh, Ph.D., Assistant Professor SMALL (nanobioSensors and MicroActuators Learning Lab) Department of Electrical Engineering University at Buffalo, The State University of New York 215E Bonner Hall, SUNY-Buffalo, Buffalo, NY 14260-1920 Tel: (716) 645-3115 Ext. 1149, Fax: (716) 645-3656 E-mail: [email protected], http://www.SMALL.Buffalo.edu a11 x1 + a12 x2 = c1 a21 x1 + a22 x2 = c2 a11 a12 x1 c1 = a21 a22 x2 c2 Δ= EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 1/11 2 x 2 Matrix Matrix Matrix 2 1 x1 0 = 1 1 x2 − 1 a11 a12 21 Characteristic = a11a22 − a12a21 Determinant Δ = = 2 ×1 − 1×1 = 1 a21 a22 11 N1 = c1 c2 N2 = a11 c1 = a11c2 − c1a21 a21 c2 x1 = Two Linear 2 x + 1x = 0 1 2 Simultaneous 1x1 + 1x2 = −1 Equations a12 = c1a22 − a12c2 a22 N1 N , x2 = 2 Δ Δ Numerator Determinant Cramer ’s Method N1 = 01 = 0 × 1 − 1 × (− 1) = 1 −1 1 N2 = 2 1 −1 1 x1 = = 1, 1 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] 0 = 2 × (− 1) − 0 × 1 = −2 x2 = −2 = −2 1 Lecture 22 | Chapter 13 | 3/8 | 2/11 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Example: Solution of Linear Simultaneous Equations Matrix 42 + 8.4 s − 42 I1 336 s ⎧(42 + 8.4 s ) I1 − 42 I 2 = 336 s ⇒ = ⎨ 90 + 10 s I 2 0 − 42 ⎩ − 42 I1 + (90 + 10 s ) I 2 = 0 − 42 42 + 8.4 s Δ= = (42 + 8.4 s )(90 + 10 s ) − (−42)(−42) − 42 90 + 10 s Step Response of a Multiple Mesh Circuit The initial energy stored zero = 84( s 2 + 14 s + 24) = 84( s + 2)( s + 12) 336 s − 42 0 90 + 10 s N (336 s )(90 + 10 s ) − ( − 42)(0) 40( s + 9) I1 = 1 = = = 42 + 8.4 s − 42 Δ 84( s + 2)( s + 12) s ( s + 2)( s + 12) − 42 90 + 10 s 42 + 8.4 s 336 s I2 = − 42 0 (42 + 8.4 s )(0) − (336 s )(−42) 168 N2 = = = 42 + 8.4 s − 42 84( s + 2)( s + 12) Δ s ( s + 2)( s + 12) − 42 90 + 10 s EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 3/11 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 4/11 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Use of Thevenin’s Equivalent (2) Use of Thevenin’s Equivalent Q: Find the capacitor current, iC Initial condition: The stored energy energy prior to closing is zero Thevenin voltage: Open-circuit voltage voltage across terminals a, b VTh VTh ZTh Voltage Division, no current across 60 Ω If I0 is zero If V0 is zero EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 5/11 60 + (20 || 0.002 s) Voltage Division, no current across 60 Ω EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 6/11 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Use of Thevenin’s Equivalent (3) 480s s + 104 s s + 104 ⋅ ⋅ 80( s + 7500) 2 × 105 s + 104 s + 4 s + 10 s 480s = 80( s + 7500) s + 2 × 105 ( s + 104 ) 480s = 80 s 2 + 80 × 7500s + 80 × 2500s + 80 × 25 × 106 6s 6s =2 = s + 10000s + 25 × 106 ( s + 5000) 2 6s K1 K2 = + ( s + 5000) 2 ( s + 5000) 2 ( s + 5000) K1 : × ( s + 5000) 2 Use of Thevenin’s Equivalent (4) dvC 1 dvC = iC dt dt C 1 ∫ dvC = ∫ C iC dt t 1 vC = (6 − 30000 x)e −5000 x dx −6 ∫ − 0 5 ×10 iC = C t = 2 × 105{ 6 (−5000 x − 1) −5000 x e −5000 x − 30000 e } (−5000) 2 − 5000 0 = 2 × 105{−30000 s = −5000 − 5000 −5000t te } = 12 ×105 te −5000t (−5000) 2 6s s = −5000 = K1 + K 2 ( s + 5000) s = −5000 − 3000 = K1 [ d × ( s + 5000) 2 s = −5000 ds d [6s]s=−5000 = d [K1 ]s=−5000 + d [K 2 ( s + 5000)]s=−5000 ds ds ds 6 = K2 K2 : EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 7/11 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 8/11 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Circuit with Mutual Inductance (1) Circuit with Mutual Inductance (2) Inductor is a short circuit at t = ∞ Q: to derive the timedomain expression for i2 The make-before-break switch has been in position a for a long time inductor is is a short circuit at t = ∞) One independent voltage source Th The initial value of the current in the 2 H th inductor i1(0-) + i2(0-) = 5 A. The branch carrying i1 has no voltage source because L1- M = 0 Mesh equations T-equivalent circuit Appendix C.1 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 9/11 SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York Circuit with Mutual Inductance (3) Δ= 3 + 2s 2s = (3 + 2 s )(12 + 8s ) − (2s )(2s ) 2s 12 + 8s = 12s 2 + 48s + 36 = 12( s 2 + 4 s + 3) = 12( s + 1)( s + 3) 3 + 2s 10 2s 10 N2 10(3 + 2s ) − 10(2s ) = = Δ 12( s + 1)( s + 3) 12( s + 1)( s + 3) 2.5 K K = = 1+ 2 ( s + 1)( s + 3) s + 1 s + 3 2.5 × ( s + 1) s = −1 ⇒ = K1 = 1.25 −1+ 3 2.5 × ( s + 3) s = −3 ⇒ = K 2 = −1.25 − 3 +1 1.25 1.25 I2 = − ⇒ i2 = 1.25(e −t − e −3t )u (t ) s +1 s + 3 I2 = di2 = 1.25(−e −t + 3e −3t ) = 0 dt e −t = 3e −3t − t = ln 3 − 3t 2t = ln 3 1 t = ln 3 = 0.54931 2 i2 = 1.25(e −t − e −3t ) EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] t = 0.54931 = 0.48113 Lecture 22 | Chapter 13 | 3/8 | 11/11 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | [email protected] Lecture 22 | Chapter 13 | 3/8 | 10/11 ...
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## This note was uploaded on 10/19/2011 for the course EE 203 taught by Professor Staff during the Spring '08 term at SUNY Buffalo.

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