Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate).

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4 ECSE 10, Fall 013 NAME: Quiz #1 September 17, 013 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram below. The switch is opened at t = 0 after having been closed for a long time. [10pt] (a) Find the second-order differential equation governing v 1 (t) for t 0. [4pt] (b) Is the circuit over damped, critically damped, or under damped? Why? [1pt] (c) Determine v 1 (t) for t 0.[5pt] t=0 4V 1 H 1 F v 1 - _ 5i 1 / 8 Ω i 3 Ω Solution: t=0 1 4V i 1 H i1 1 F v 1 - _ 5i 1 / 8 Ω i 3 Ω 1

5 ECSE 10, Fall 013 NAME: Quiz #1 September 17, 013 ID: Applying KCL at node 1: v 1 = di 1 8 i The characteristic equation: i i 1 i = 0 i = v 1 5i 3 i 1 = dv 1 i = v 1 i = v 1 dv 1 d v dv v 1 = 0 [4pt] λ 3 8 λ = 0 λ 1, λ = ± j 16 Roots of the characteristic equation are two complex numbers, therefore the system is under damped. [1pt] ) v 1 (t) = e (A t cos 16 t B sin 16 t [1pt] Initial Conditions: [pt] v 1 (0 ) = v 1 (0 ) = 4V i (0 ) = i (0 ) = 3A v 1 (0 ) dv 1 (0 ) = i (0 ) Applying initial conditions: dv 1 (0 ) = 30 v 1 (t) = e (4 t cos A = 4 B = t sin 77 ) t [pt]

6 ECSE 10, Fall 013 NAME: Quiz #1 September 17, 013 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram below. The switch is opened at t = 0 after having been closed for a long time. [10pt] (a) Find the second-order differential equation governing v 1 (t) for t 0. [4pt] (b) Is the circuit over damped, critically damped, or under damped? Why? [1pt] (c) Determine v 1 (t) for t 0.[5pt] t=0 4V 1 H 1 F v 1 - _ 5i 1 / 8 Ω i 3 Ω Solution: t=0 1 4V i 1 H i1 1 F v 1 - _ 5i 1 / 8 Ω i 3 Ω 1

7 ECSE 10, Fall 013 NAME: Quiz #1 September 17, 013 ID: Applying KCL at node 1: v 1 = di 1 8 i The characteristic equation: i i 1 i = 0 i = v 1 5i 3 i 1 = dv 1 i = v 1 i = v 1 dv 1 d v dv v 1 = 0 [4pt] λ 3 8 λ = 0 λ 1, λ = ± j 16 Roots of the characteristic equation are two complex numbers, therefore the system is under damped. [1pt] ) v 1 (t) = e (A t cos 16 t B sin 16 t [1pt] Initial Conditions: [pt] v 1 (0 ) = v 1 (0 ) = 4V i (0 ) = i (0 ) = 3A v 1 (0 ) dv 1 (0 ) = i (0 ) Applying initial conditions: dv 1 (0 ) = 30 v 1 (t) = e (4 t cos A = 4 B = t sin 77 ) t [pt]

8 il ECSE 10 NAME: Quiz 1 9 May 011 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram below. (a) Find the value of L which makes the circuit have a critically damped natural response. [4pt] 10Ω n 1 8 Ω 1 / 8 F v C - Applying KCL at node n 1 : 1 8 Applying KVL at the left mesh: dv C v C 8 = i [1pt] L di 10i = v C [1pt] Combining the above equations and obtaining the equation for either v C or i: L d v C (10 L) dv C 18v C = 0 Or the similar equation for i: [1pt] L d i (10 L)di 18i = 0 The characteristic equation of the circuit is: Lλ (10 L)λ 18 = 0 1

9 ECSE 10 NAME: Quiz 1 9 May 011 ID: For the critically damped response, the characteristic equation should have a double root. Double root (L 10) 7L = 0 L = or L = 50 L = λ = 3 and L = 50 λ = 3 5 [1pt] Note that both values for L are acceptable, as the characteristic equation has double roots with negative values.

10 ECSE 10 Quiz 1 9 May 011. Consider the circuit diagram below. [6pt] (a) Find the second-order differential equation governing v(t). [3pt] (b) Is the system over-damped, under-damped, or critically damped? Why? [1pt] (c) Find the initial values for v and i (v, dv di, i, ), and determirne v for t 0. [pt] 1H i Ω 5V t=0 ½ Ω 1F v - Solution: Applying KCL at the node containing capacitor: Applying KVL at the left mesh: i = v dv di L i = v [1pt] Combining above equations together: d v 4dv 5v = 0 Characteristic equation of the circuit: [1pt] [1pt] λ 4λ 5 = 0 λ = ± j under-damped [1pt] Initial conditions: [1pt] i(0 ) = 10; v(0 ) = 5 i(0 ) = 10; v(0 ) = 5 i(0 ) = v(0 ) dv(0 ) dv(0 ) = 0 3

11 ECSE 10 Quiz 1 9 May 011 di(0 ) i(0 ) = v(0 ) di(0 ) According to the characteristic equation, v is: = 5 Applying initial conditions: [1pt] v = e t (A cos t B sin t) v(0 ) = 5 A = 5 And for t 0: dv(0 ) = 0 B = A = 10 v = e t (5 cos t 10 sin t) 4

12 ECSE 10, Summer 013 NAME: Quiz # 1 May 10, 013 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram below. [10pt] (a) Which circuit analysis method (node voltages or mesh currents) will result in a fewer number of equations? Justify your answer. [pt] (b) Find the second-order differential equation governing i L (t). [8pt] i C i L Super node v 1 v - i L L R 1 i C C R i s 1

13 ECSE 10, Summer 013 Quiz # 1 May 10, 013 Solution: There are 5 meshes and the current of meshes are known. Therefore, there will be 3 mesh equations and constraint equations for dependent sources. On the other hand, there is 1 super node and constraint equations for dependent sources. [pt] Applying KCL at the super node: For the capacitor: i L v 1 R 1 i C v 1 i L R = i s [pt] C dv = i C, v = v 1 i L i C = C d(v 1 i L ) The first equation can be rewritten as: [1pt] i L v 1 R 1 C dv 1 C di L v 1 i L R = i s [pt] For the inductor: L di L = v 1 Combining the two equations together: ( LC d i L L L C R 1 R ) dil [1pt] (1 R ) i L = i s [pt]

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17 Question 1 Find current I for t>0 if the circuit is in steady-state at t=0 -. Solution: Initial conditions I (0 ) = 8A L V (0 ) = 0V C After the switch is opend, KCL at V1 V1 V1 V 8 = V V V = = 1 1 4V = V 4 V V 4 6 KCL at V V V 1 1 V dv C 0 3 L = Sub V1

18 V V dv V C = 0 3 L V 3 dv V 6 V 3C 0 4 L = 1 dv V V V C L = 1 dv 3V 4 V 1C 0 L = 4 dv V 8 V 4C = 0 L Differentiate V w.r.t. time d V dv 4 4C V = 8 L d V dv V = 8 Characteristic equation 0.s s 0.8 = 0 s = 1, 4 Therefore, the total response is I() t = Ae Ae A t 4t I( ) A = 3 = 8= A 1 3 I(0) = A A = 8 A (1) 1 di t 4t VC = L = 5( Ae 1 4Ae ) di VC (0) = L = 5( A1 4A) = 0 () Solve (1) and () A = 8 1 A = Therefore, t 4t It () = 8e e Α

19 ECSE 10 NAME: Quiz #1 September 0, 01 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram below. Assume that the circuit is at steady state at t = 0. [10pt] (a) Find the second-order differential equation governing v c (t) for t 0. [3pt] (b) Is the system over-damped, under-damped, or critically damped? Why? [pt] (c) Determine v c for t 0. [3pt] (d) Determine v 1 for t 0. [pt] t=0 Ω 6V 4 Ω v 1-1 H 1 / 13 F v c - 1

20 ECSE 10 Quiz #1 September 0, 01 Solution: Applying KVL at the right mesh: For the capacitor: 4i i di v c = 0 [1pt] 1 dv c 13 = i Combining the two equations together: [1pt] d v c 6dv c 13v c = 0 Characteristic equation of the circuit: [1pt] λ 6λ 13 = 0 λ = 3 ± j under-damped [pt] According to the characteristic equation, v c is: v c = e 3t (A cos t B sin t) [1pt] Initial conditions: [1pt] v c (0 ) = v c (0 ) = 6; i(0 ) = i(0 ) = 0; 1 dv c 13 (0 ) = i(0 ) = 0; dv c (0 ) = 0; Applying initial conditions: v c (t) = e 3t (A cos t B sin t) And for t 0: dv c (0 ) v c (0 ) = 6 A = 6 = 0 3A B = 0 B = 9 v c = e 3t (6 cos t 9 sin t) [1pt] In order to find v 1 : [pt] v 1 = 4i

21 ECSE 10 Quiz #1 September 0, 01 i = 1 dv c 13 = e 3t ( 39 sin t) v 1 = 1e 3t sin t 3

22 ECSE 10 NAME: Quiz 1 15 September 011 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Question 1 [4pt] (a) Describe what a second order circuit means by using an example. [1pt] (b) What are the three possible solution types of the natural response of a second order circuit? How are they related to the exponential damping coefficient, undamped natural frequency and natural frequencies? [3pt] Solution: (a) A second-order circuit is characterized by a second-order differential equation. It consists of the equivalent of two energy storage elements. [1pt] R L C (b) natural responses: [3pt] Overdamped: α > ω 0 Unequal and real natural frequencies Critically damped: α = ω 0 Equal and real natural frequencies Underdamped: α < ω 0 Complex conjugate natural frequencies 1

23 ECSE 10 Quiz 1 15 September 011. Consider the circuit diagram below. The switch is closed at t = 0 after having been opened for a long time. [6pt] (a) Find the second-order differential equation governing v(t) for t 0. [3pt] (b) Is the system over-damped, under-damped, or critically damped? Why? [1pt] (c) Determine v for t 0. [pt] 3 / 13 Ω 1V t=0 1 / 6 H F v - Solution: Applying KVL at the right mesh: For the capacitor: 3 13 i 1 di 6 v = 0 [1pt] dv = i Combining the two equations together: [1pt] d v 6dv 13v = 0 Characteristic equation of the circuit: [1pt] λ 6λ 13 = 0 λ = 3 ± j under-damped [1pt]

24 ECSE 10 Quiz 1 15 September 011 Initial conditions: [0.5pt] v(0 ) = v(0 ) = 1; i(0 ) = i(0 ) = 0; dv (0 ) = i(0 ) = 0; dv (0 ) = 0; According to the characteristic equation, v is: v = e 3t (A cos t B sin t) [0.5pt] Applying initial conditions: [1pt] v(0 ) = 1 A = 1 And for t 0: dv(0 ) = 0 B = 3 A = 18 v = e 3t (1 cos t 18 sin t) 3

25 NAME: 10#: S()A.l/T/()N~ ECSE 10: Introduction to Electronics Quiz #1 Wednesday September 16 th, 009, I - Write all of your solutions directly on the question sheet - Use the back of the pages if you need more room - The quiz is 30 minutes and consists of two problems - The exam is out of 10 points. Good luck!

26 Question # 1-(3 points) a) ( pts) for the following second-order homug~nel~us differential e~uat~~n. ~eri~',e the characteristic equation, assuming that the trial form for the solution IS.r - Ke _ d!x d'(! I cit., Clearlv state your reasoning in the final step. - l - l x=o b) (I pt) What does the time-constant of a circuit signify? -x.:= J<. est: &:. = Kses-t ; c{<'; = j(s" e S i:. di.,z Pl()6Jq/lyq II'/TO (J) / K est. (s e. «, S - q c) ::::0 est :f. 0 -' t ~o IJ-H~ )( = 0 OTHE'je WI.5 E HENCE / 5~ f a J S -j- a.; = c:.j-i-hi<i}et ~/~T1C. fivfl-tltjh. lf1 T! PJ - u u l,st I'J-NT of,:}- C/,RCI/IT CtfHRI'Jc1CR./f:n rss:!<j:j-r 1fT WHICff r s e tlh-tlirij.-l- R&SPONS -OP ti rw'op.~~ TO z Ro. ) WI TH EHERB] Y Src;RIT61, LEMEl'lr- f)e:cftys,/

27 Question # (7 points) Consider the second-order circuit. Switch S is initially closed. The circuit is assumed to be in steady state at I < O. At I = 0, switch S is opened. For R = 40 Q, L = 10 mh, and C = 5 I!F, answer the following questions. t=o L 100m_A. v(t) c R a) (I pt) Determine the steady-state voltage v(ol and i(o-). b) ( pts) Find the second-order homogeneous differential equation governing v(l). Express your answers in terms of R, L, and C. c) (I pt) Find the roots of the characteristic equation? d) (I pt) Is the system overdamped, underdamped, or critically damped? Why? e) ( pt) Determine v(l) for I:=:: O. IOOWl A r ) i (0-)..'. <- (0-) = '00 m It 7J (0-) -:;: 1<. z{o) (),s-] i: >0, L C. 1)[t) - R j<c L _ /)((1-) = 4 V ~ i(i) ~ -,.. V (t) =1. r!ici:) R i {t) -- ~ _ CD

28 iet)= -edv _-- [I]. A-ts,o/ d~v L ~ i := o, k- Pi Vh ~ C cit /fe-nef/ d v of.e. &/11 --L V = 0 [.] di:l L cc gj ~ JJ-n-AAc. icri.f. TIC. ; 5 ~ -f f<./t.. ~ Yll: =0 [0 ' s-] i?\jprr/oh :9 5 e J /- «0/ ~/ {7-(76 = 0!<.()O!S.' S'll. = -, tjcjo :t j 4&rro [0 ' 5J E C1fIH<'IfCTEP/~TIC /<{:JoTS f}r Cf!7t'/JP~ i' Ct»ITv(J,/J-TES [0 "~J i H-EHCC/ T Hs: S Y~ rzr«i s ()H oer Dn tt?oe» ('t''> I) [&>.~J (It cos wt- IS Sin. wt.) vj)t p.~ ~\IR.;:::= ojjw.: 'V (1::) = e-l1oot (A- c..s 4IJrlO t 1S.sir'l. J,ooot) [0 11'1/11 ItL Co H D ] rr 0 H x : V (,0 ) = 1/Co-y :: A- = 4V rlrv I. - = '-I O(JO~-'<'eoo It d i: o. (0"') =,(0-) =-5f-'1;t = l19om, 'V (t) = e : , A-= (f)orrt. - =;> E= -g. [o ~] - [0 s-] 4 CD> 4t'JtJo-c -3 Sin 4O()() r [v] -:.000 t( ) i: ;;,. 0.

29 ECSE 10, Summer 010 Quiz 1 Total Points: 10 NAME: Write your solutions directly on the question sheet. need more room. SID: Use the backs of the pages if you Problem 1 (5 points) For the circuit shown in figure 1, v(0 )=1V, and i L (0 )=30mA. a) Find the initial current in each branch of the circuit. (1pt) b) Find the initial value of dv. (1pt) c) Find the expression for v(t). (pt) d) By how does the resistance need to be changed to obtain a critically damped voltage response. (1pt) ic il ir 0.uF 50mH 00 v Solution a) Figure 1: Problem 1 Applying KCL, b) Since Therefore, i L (0 ) = i L (0) = i L (0 ) = 30mA i R = V R = 1 00 = 60mA i C (0 ) = i L (0 ) i R (0 ) = 90mA dv(0 ) i C = C c) The roots of the characteristic equation are: = ( ) dv = 450kV/s λ 1, λ = α ± α ω o where α = 1 RC ω o = 1 LC 1

30 ECSE 10, Summer 010 Quiz 1 Total Points: 10 Using the above equations, we get λ 1 = 5000rad/s λ = 0000rad/s Because roots are real and unequal, we know that the response is overdamped and of the form: v(t) = C 1 e λ1t C e λt (1) Applying boundary conditions of v(o ) and dv(0 ) Therefore, 1 = C 1 C = 5000C C C 1 = 14V, C = 6V to the equation 1, we get: v(t) = 14e 5000t 6e 0,000t V, t 0 () d) From part c, ω o = Therefore, for critical damping, α = 10 4 = 1 RC or 1 R = = 50Ω Hence, R needs to be increased by 50Ω. Problem (5 points) a)a sinusoidal voltage source has a maximum amplitude of 110V. The voltage passes through one complete cycle in 0ms. The magnitude of voltage at zero time is 55V. Write the expression for v(t) using the sin function. Express the phase in degrees. (pt) b)find the phasor representation for the following expression: (pt) 0 cos(ωt 30 o ) 40 cos(ωt 60 o ) c) In a circuit, 5 identical resistors of values 10Ω and 1 capacitor of 10µF are connected in parallel with a sinusoidal source of frequency 60Hz. What is the total admittance of the circuit? (1pt) Solution a) Since the voltage is nonzero at zero time, we can write it in the cosine form, i.e. v(t) = V max cos(ωt φ) (3)

31 ECSE 10, Summer 010 Quiz 1 Total Points: 10 Now, Therefore, 3 becomes: ω = π T = π = 100π rad/s To find φ, we will evaluate the equation 4 at zero time: Therefore 4 becomes Now converting equation 5 into sin form: b) v(t) = 110 cos(100πt φ) V (4) v(0) = 55 = 110 cos φ φ = 60 o v(t) = 110 cos(100πt 60 o ) V (5) v(t) = 110 sin(100πt 60 o 90 o ) = 110 sin(100πt 150 o ) V c) 0 cos(ωt 30 o ) 40 cos(ωt 60 o ) = 0 30 o o = (17.3 j10) (0 j34.64) = 37.3 j4.64 = o 44.7 cos(ωt o ) Y = 5 R jπfc Y = 5 jπ = 0.5 j S 3

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34 ECSE 10, Summer 014 May 9 th, 014 Quiz #1 NAME: ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram shown. At t = 0, the independent current source I s steps up from 0 amperes to 1 ampere and the independent voltage source V s steps down from 5V to 0V. I K is the current passing through R. [10 points] (a) Find the second-order differential equation governing I L (t). [ points] (b) Find the characteristics equation and its roots. [3 points] (c) Is the circuit over-damped, critically damped, or under-damped? Why? [1 point] (d) What is the equation for the natural response? What are the initial conditions? What are the values of the coefficients of the natural response equation? [3 points] (e) Determine I L (t) for t.[1 point] 0A t=0 1A IS C=1µF IL R1=50Ω L=1mH - R=0Ω IK 30 IK - 5V VS 0V t=0 1

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36 ECSE 10, Summer 014 May 9 th, 014 Quiz #1 NAME: ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Consider the circuit diagram shown. At t = 0, the independent current source I s steps up from 0 amperes to 1 ampere and the independent voltage source V s steps down from 5V to 0V. I K is the current passing through R. [10 points] (a) Find the second-order differential equation governing I L (t). [ points] (b) Find the characteristics equation and its roots. [3 points] (c) Is the circuit over-damped, critically damped, or under-damped? Why? [1 point] (d) What is the equation for the natural response? What are the initial conditions? What are the values of the coefficients of the natural response equation? [3 points] (e) Determine I L (t) for t.[1 point] 0A t=0 1A IS C=1µF IL R1=50Ω L=1mH - R=0Ω IK 30 IK - 5V VS 0V t=0 1

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