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FIRST ORDER RL RC CIRCUITS

FIRST ORDER RL RC CIRCUITS. Perpared by: Ertuğrul Eriş Reference book: Electric Circuits, Nielsson, Riedel Pearson, Prentence Hall,2007. Updated 1: October 2011. COURSE ASSESMENT MATRIX. FOUR FIRST ORDER RC/RL CIRCUITS. Compare a and b Compare c and d. SOLUTIONS. Mathematical

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FIRST ORDER RL RC CIRCUITS

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  1. FIRST ORDER RL RC CIRCUITS Perpared by: Ertuğrul Eriş Reference book: Electric Circuits, Nielsson, Riedel Pearson, Prentence Hall,2007 Updated 1: October 2011

  2. COURSE ASSESMENT MATRIX Ertuğrul Eriş

  3. FOUR FIRST ORDER RC/RL CIRCUITS Compare a and b Compare c and d Ertuğrul Eriş

  4. SOLUTIONS • Mathematical • Homogenous solution (Homojen çözüm ): independent sources deactivated • parameters not calculated, will be calculated on general solution • particular solution (Özel çözüm ) source type function, • estimated function parameters are calculated. • General solution (tam çözüm) =homogenous solution+particular • Parameters in Homogenous solutions are calculated by using initial conditions of inductor currents and capacitor voltages • Circuit based solutions • Natural response (Öz çözüm ): independent sources deactivated, energy sources are initial conditions of inductors and capacitors • Forced response (Zorlanmış çözüm ): independent sources are activated, initial conditions are zero. • This solution will be in the form of mathematical general solution • General Solution (tam çözüm) =natural response + forced response • Comparison • Each solution in circuit based correspond to two different circuit solutions. • Circuit based «Forced response» include part of mathematical homogenous solution and whole particular solution; • Circuit based «Natural response» include part of mathematical homogenous solution • Different solution can be observed for a square wave input Ertuğrul Eriş

  5. OBJECTIVES • will be able to solve first order linear electrical circuits in time-domain by using differential equations. • In other words • Analysis/Modeling First order circuits • Emotional requirements • Extra concentration, attention, Patient • Otherwise Circuit analysis can not be understood • Mathematical tools to be used • Integration, Differential Equation Ertuğrul Eriş

  6. MATHEMATICAL SOLUTION OF A DC-SOURCE SERIAL RL CIRCUIT : INTEGRATION General form of a first order differantial equation Why ınductor current is chosen as unknown? General solution What is solution of a dif equation? Interpretation of the solution Transient solution Steadty solution Ertuğrul Eriş

  7. MATHEMATICAL SOLUTION OF A DC-SOURCE SERIAL RL CIRCUIT: DIFFERENTIAL EQUATION General form of a first order differantial equation Homogenous solution + Particular solution = General solution Common form for all first order linear circuits • Compare particular solution and the DC source • What hapens after a few seconds? Ertuğrul Eriş

  8. NATURAL RESPONSE FOR RL AND RC CIRCUITS • Natural response (Öz çözüm ): independent sources deactivated, energy sources are initial conditions of inductors and capacitors • Difference from homogenous solution? Ertuğrul Eriş

  9. INDUCTOR INITIAL VALUE COULD BE SET TO ANY GIVEN VALUE Compare big and small circuits, source transform!! When the switch is on for a long time what would be inductor’s current? When the switch is of what will you get? Ertuğrul Eriş

  10. NATURAL RESPONSE OF A RL CIRCUIT When t=0 Instantaneous change in v, that is v is unknown Current is known i0 Power known t=0+ and onwards Energy is know from t=0 and onwards Ertuğrul Eriş

  11. TIME CONSTANT (ZAMAN SABİTİ ) τ = time constant= L/R Dimension of time constant? Ertuğrul Eriş

  12. STEP RESPONSE (FORCED SOLUTION) OF A SERIAL RL CIRCUIT General solution: Forced Solution: Natural solution (response): Tam çözüm (general solution): homojen çözüm (homogenous solution) + özel çözüm (particular solution) Tam çözüm (general solution): Öz çözüm (natural response): sources deactivated, initial con.exist+ zorlanmış çözüm(forced response): sources active+initial con. =0. Ertuğrul Eriş

  13. STEP RESPONSE (FORCED SOLUTION) OF A SERIAL RL CIRCUIT Tam çözüm (general solution): homojen çözüm (homogenous solution) + özel çözüm (particular solution) Tam çözüm (general solution): Öz çözüm (natural response): sources deactivated, initial con.exist+ zorlanmış çözüm(forced response): sources active+initial con. =0. Ertuğrul Eriş

  14. INDUCTOR CURRENT FORCED SOLUTION Forced solution: sources acvtive, initial con. are (0) Ertuğrul Eriş

  15. INDUCTOR VOLTAGE FORCED SOLUTION Forced solution: sources avtive, initial con. are (0) Ertuğrul Eriş

  16. EXAMPLE-1 iL(t)= 20 e-5t A t≥0; i0= -4 e-5t A ; v0= -160 e-5t V; p 10Ω= 2560 e-10tt≥0+; W 10Ω=256 J.t≥0 Ertuğrul Eriş

  17. EXAMPLE-2 iL(0)= - 12.5 A; WL initial=625 mJ; τ = 4 ms; iL(t)= - 12.5e-250t A; Ertuğrul Eriş

  18. EXAMPLE-3 V0 = - 8e-10t V , t≥ 0 Ertuğrul Eriş

  19. EXAMPLE Switch is in (a) initial condition Switch is in (b) initial condition paralel RL circuit general solution i(t) = 12+ (-8-12) e –t/0.1 A, t≥0. v(t) = 40 e –t10V, t≥0. Ertuğrul Eriş

  20. MATHEMATICAL SOLUTION OF A DC-SOURCE SERIAL RC CIRCUIT : INTEGRATION General form of a first order differantial equation Why capacitor voltage is chosen as unknown? What is solution of a dif equation? General solution Interpretation of the solution Transient solution Steadty solution Ertuğrul Eriş

  21. MATHEMATICAL SOLUTION OF A DC-SOURCE SERIAL RC CIRCUIT: DIFFERENTIAL EQUATION General form of a first order differantial equation Homogenous solution + Particular solution = General solution Common form for all first order linear circuits • Compare particular solution and the DC source • What hapens after a few seconds? Ertuğrul Eriş

  22. CONDUCTOR INITIAL VALUE COULD BE SET TO ANY GIVEN VALUE When the switch is in position (a) for a long time what would be Vc? Ertuğrul Eriş

  23. NATURAL RESPONSE OF A RC CIRCUIT Sometimes holding capacitor terminals may cause electrical shock why? When t=0 Instantaneous change in i, that is v is unknown Voltage is known i0 Power known t=0+ and onwards Energy is know from t=0 and onwards Ertuğrul Eriş

  24. DISCHARGE OF A CAPACITOR: DEFIBRILLATOR C=200μF R=10Ω Vc(0)=5000V Stored energy: W=(1/2)Cv2=2500 J Energy absorbed by human body: t=100µs→ W(t)=2161J t=200µs→ W(t)=2454J 25.2 km/saat hızda giden(7m/s) ve ağırlığı1000kg olan bir arabanın, bir duvara çarptığında 10 ms de tahrip olduğunu düşünürsek, Bu arabanın enerjisi (1/2)mv2=2500J dür. Bu enerjinin çarpma sonucu (hızın (0) düşmesi) sonunda ortaya çıkan kuvvet F=ma=m (dv/dt)→ F=m (Δv/Δt) ; M=100kg Δ V=7m/s Δ t=10ms F=100*(7/10*10-3 )=7*105 Nw→70 ton Ertuğrul Eriş

  25. WHAT IT MEANS 2500J-ENERGY IN A SHORT TIME INTERVAL Assume that a car with a weight of ve 1000kg is moving at a speed of 25.2 kmh (7m/s) and strike a wall and destroyed within 10 ms. Kinetic energy of the car is =(1/2)mv2=2500J . Thge force of this strike is F=ma=m (dv/dt)→ F=m (Δv/Δt) ; m=100kg Δ V=7m/s Δ t=10ms F=100*(7/10*10-3 )=7*104 kg→70 ton. Ertuğrul Eriş

  26. STEP RESPONSE OF RC CIRCUIT General solution sources active, initial condition different from (0) If the initial condition was (0) then general solution becomes forced solution. Ertuğrul Eriş

  27. GENERAL SOLUTION Forced Response Vc (0)=0, Is=Is Natural Response Vc (0)=Vo , Is=0 Vc (0)=V0 , Is=Is General solution= =Forced Response + Natural response Ertuğrul Eriş

  28. EXAMPLE vc(t)= 100 e-25t V; v0 (t)= 60 e-25t V ; i0= e-25t mA; p 60kΩ= 60 e-50t mW t≥0+ ; W 60kΩ=1,2 mJ Ertuğrul Eriş

  29. EXAMPLE V(t) = 20 e -t V, t ≥ 0, i(t) = 80 e –t μA t ≥ 0+ ; v1(t) = (16 e –t – 20) V , t ≥ 0 ;v2(t) = (4 e –t + 20) V , t ≥ 0 W1 = 40 μJ , W2 = 5760 μJ Ertuğrul Eriş

  30. EXAMPLE V(0-)= 200V; τ=20ms; v(t)=200e-50t V t≥0; WC(0)=8 mJ; Ertuğrul Eriş

  31. EXAMPLE V0(t) = 8 e -25t + 4 e -10t V , t≥0 Ertuğrul Eriş

  32. EXAMPLE V0 = - 60 + [30 – (-60)] e -100 t V t≥0; i0 = - 2.25e -100 t mAt≥0+. Ertuğrul Eriş

  33. SERİ RL ve SERİ RC İÇİN GENEL ÇÖZÜM Ertuğrul Eriş

  34. GENERAL SOLUTION EXAMPLE VC= 90 + [ -30 – 90)] e -5 t V t≥0; i0 = 300 e -5t μAt≥0+. Ertuğrul Eriş

  35. FORCED SOLUTION EXAMPLE Kaynak dönüşümü: Vc(t)= 150-150 e-200t V İ(t)= 3 e-200t mA V(t) = 150-60 e-200t V Ertuğrul Eriş

  36. FORCED SOLUTION EXAMPLE İ(t) = 20- 15 e -1 2.5t A Ertuğrul Eriş

  37. RL RC CIRCUIT ANALYSIS VS MATH • RC/RL circuit analysis→ first order ordinary differential equation • Independent variable: time • Ordinary differential equation not partial • A: constant, real • Linearity valid • b independent sources, knowns • For RLC circuits second order differential equation is required. Ertuğrul Eriş

  38. MATHEMATICAL SOLUTION OF A FIRST ORDER DIFFERANTIAL EQUATION • Homogenous solution : • b(t) =0 • Homogenous solutıon is in the form of decreasing exponential : • Two parameters: C and s. • (s) will be calculated from the equation. • C will be calculated by using the initial condition, after the general solutıon found. • particular solution = source type function, solutıon • Source DC then particular solution is a constant. • Source AC then particular solution is a sinusoidal function of which amplitude, and phase should be calculated. Frequency remains the same as source frequency. • General solution = Homogenous + Particular Transient solution, why? Steady(kalıcı) solution, why? What is solution of a differantial equation? Ertuğrul Eriş

  39. SUMMARY Mathematical: • Homojen çözüm (homogenous solution) : b(t) =0 • Özel çözüm (particular solution) = b(t) Type Solution • General sol= homogenous solution+ particular solution. • Circuit based solution: • Öz çözüm (Natural response): initial conditions exist, independent sources deactivated • Zorlanmış çözüm (Forced response): initial conditions are (0), independent sources activated • General sol= natural sol.+ forced sol. Aynı denklemin farklı iki çözümü olur mu? Ertuğrul Eriş

  40. SEQUENTIAL SWITCHING Ertuğrul Eriş

  41. SEQUENTIAL SWITCHING t<0 t ≥ 35 ms iL(t) = 6 e -40t A, 0 ≤ t ≤ 35 ms iL(35ms) = 1.48 A. iL(t) = 1.48 e – 60(t-0.035) A, t ≥ 35 ms 0≤ t ≤35 ms Ertuğrul Eriş

  42. SEQUENTIAL SWITCHING t < 0 anahtar (a) 0 ≤ t ≤ 15 ms anahtar (b) 15 ms < t anahtar (c) v(t) = 400 + (0-400) e – 100t V 0 ≤ t ≤ 15 ms v(15ms) = 310.75 V. v(t) = 310.75 e – 200(t – 0.015) V 15ms ≤ t Ertuğrul Eriş

  43. SEQUENTIAL SWITCHING Switch 1 is on for a long time. v(t) = 80 e – 40t V 0 ≤ t ≤ 0.01 s v(10ms) = 53.63 V. v(t) = 53.63 e – 50(t – 0.01) V 10ms ≤ t Ertuğrul Eriş

  44. SEQUENTIAL SWITCHING i(t) = 3 - 3 e – 0.5t A 0 ≤ t ≤ 1 s i(1s) = 1.18 A. i(t) = -4.8 + 5.98 e –1.25(t – 1) A t ≥ 1 s Ertuğrul Eriş

  45. UNBOUNDED RESPONSE:ASTABLE CIRCUIT Rth= - 5 KΩ V0(t) = 10 e 40t V t ≥ 0. What would be the voltage after 1 sec.? Ertuğrul Eriş

  46. INTEGRATING CIRCUIT Ertuğrul Eriş

  47. PROGRAM DESIGN DEPT, PROGRAM G R A D U A T E S T U D E N T STUDENT P R OG R A M O U T C O M E S PROGRAM OUTCOMES P R OG R A M O U T C O M E S STATE, ENTREPRENEUR FIELD QALIFICATIONS EU/NATIONAL QUALIFICATIONS KNOWLEDGE SKILLS COMPETENCES NEWCOMERSTUDENT ORIENTIATION GOVERNANCE Std. questionnaire ALUMNI, PARENTS ORIENTIATION STUDENT PROFILE Std. questionnaire FACULTY NGO STUDENT, ??? CIRCICULUM ??? INTRERNAL CONSTITUENT Std. questionnaire EXTRERNAL CONSTITUENT EXTRERNAL CONSTITUENT REQUIREMENTS EU/NATIONAL FIELD QUALIFICATIONS PROGRAM OUTCOMES QUESTIONNAIRES QUALITY IMP. TOOLS GOAL: NATIONAL/INTERNATIONAL ACCREDITION

  48. BLOOM’S TAXONOMYANDERSON AND KRATHWOHL (2001) !!Listening !! Doesn’t exits in the original!!! http://www.learningandteaching.info/learning/bloomtax.htm Ertuğrul Eriş

  49. ULUSAL LİSANS YETERLİLİKLER ÇERÇEVESİ BLOOMS TAXONOMY Ertuğrul Eriş

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