Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1 / 3 F.

Transcription

1 ECSE 10 NAME: Quiz 18 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. [3pt] (a) Find the current phasor I. [3pt] 1 / 3 F 1 / 6 H i i c cos3t V sin3t A _ ¼ i c Solution: 1 / 3 F A 1 / 6 H i i c cos3t V sin3t A _ ¼ i c cos 3t = 0 sin 3t = 1 π [0.5pt] Applying KCL at node A: Z L = jlω = j I c Z C = 1 jcω = j 1 4 jlω 1 π 0 1 jcω = I 1 I c j 1 π j = I [1pt] I = 1 j = π [1pt] 4 I = cos 3t π 4 [1pt] [0.5pt] 1

2 ECSE 10 Quiz 18 May 011. Consider the circuit diagram below. [7pt] (a) Assume that the circuit in steady state, find v(t) [7pt] 4Ω 10μF Ω 4 Ω sin5000t V cos10000t A v 3 V 1mH - Solution: The voltage and current sources have different frequencies, so we use the principle of superposition. Considering the DC voltage source and set other sources to zero (open circuit the current source and short circuit the voltage source). v = 3 4 = v [1pt] 4 Considering the AC voltage source and set other sources to zero (open circuit the current source and short circuit the voltage source). Applying KCL: v 4 j5 v v ( j) 4 j0 = 0 [1pt] v = 10 j8 66 j45 = [1.5pt] Considering the AC current source and set other sources to zero (short circuit the voltage sources). Applying superposition: v 4 j10 v v (1 j0) = 0 [1pt] 4 j10 v = = [1.5pt] v =.6cos(5000t 0.077) 1.76cos(10000t) [1pt]

3 Question 1: [Points: 100] The operational amplifier is ideal. Using phasor method, find the steady-state expression for V ( t ) V t = cos10 tv. when ( ) 6 in out Solution: 1 jωc ( )( ) ( ) = j100k VO VO = j VO 0 VO = 0 j j40 jv V = 0 O O 1 jv = j40 V O O O j40 = = 0 j0 = 8.8R135 1 j 6 ( ) = 8.8cos( ) V t t V

4

5

6

7 ECSE 10 NAME: Quiz September 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. 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) Find the value of R which makes the circuit behave as an oscillator. [pt] (c) Determine v 1 for t 0.[4pt] t=0 4V 1 F v 1-1 H _ 5i R i 3 Ω Solution: t=0 1 4V i1 1 F v 1 - i 1 H _ 5i R i 3 Ω 1

8 ECSE 10 NAME: Quiz September 011 ID: Applying KCL at node 1: v 1 = di dt Ri The characteristic equation: i i 1 i = 0 i = v 1 5i 3 i 1 = dv 1 dt i = v 1 i = v 1 dv 1 dt d v 1 dt (R 1 )dv 1 dt (1 R )v 1 = 0 [4pt] λ (R 1 )λ (1 R ) = 0 For an oscillator α should be 0 and ω > 0: [1pt] R = 1 [1pt] R = λ = ±j v 1(t) = A cos t B sin t [pt] Initial Conditions: [1pt] v 1 v 1 (0 ) = v 1 (0 ) = 4V i (0 ) = i (0 ) = 8A (0 ) dv 1 dt (0 ) = i (0 ) Applying initial conditions: A = 4 B = v 1 (t) = 4 cos t 4 3 sin dv 1 dt (0 ) = 6 3 t [1pt]

9 SoLUTI()r/~ NAME: _-'- ID#: _ - ECSE 10: Electric Circuits Quiz # Wednesday September 3 rd, 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!

10 Ifit/gnp/{ # JI (g PCI y = 6J<g ~ V:=: 10 j's-:j.j: (7) 'RorlJ-'Jl 8y-" 5"0"4 V I = /0 /(;J. ~~. -= /0M = "',.,f.[ d.f [tj. 5"1 ('i i) J(OT~rE By 'fad ~ I! V = 105'5.)" 'foe =' 10/143./ 0 & 1T,"-== 3L6o~ = J S t-j R!>98 = - 8J6 [O '~l ~= gl-:ltr,s." -::: ] 3CfI -cfg 06J.. -vi t-y z =- 9 gc{/ - JO 1{ 63 ~ 8 QO j- ;l Cf 'i 0 => V, CY~ = g. 'I c»s (R.t - -v: Elts IL t I:>0 # e vj I Til C IlLCI.J J../rTO R.

11 -----=f--"'"\--4--.,----r -t r~ Ecl. ftvesrlpn #... 7' PO /N"T>.. ' y-----.j 4CfJ.$1:. fa u(t).. _ J~Jt. 1i.,- Vc.{ 0) 71(0-) =tvc (ot) = g v. G

12 IE 1RIAL. F(5(J.1-'1 of F"R. CE:!:> ~"L!:!.: tv! ft) = IJ ~jni 1- g ClZ i: c..tl/ li { de :: A usl.t. 9.s in t ; d~vr- ~~.- _ A.5 in-{;- g C OJ 1:. ~ Sv8511TIJrlH G] d1j.j. d?lf 7/jINTO TfJ 7'H~ t3.d-g dt:l dt".j AM]) ];5(\//tTIHev ' COEFFI~ ~,.,r~ I ;(.S-'l.IO\'/t- 1'f./()68 =0 1)(10'1}. ~'1/0 " g =- 1-,.10 t" ~ 1:: / 6;<10 -S- ; 8 = 4. Vc f;t) -= _'.6 )f./os's/n. t it c~s t [v1 / f C ItfH.fK- rz /<J~ TIt: s 1.. f I'!/" G s f ~ "SX1011 =0 6/'1 ~oot3 ; S,:= S.<... = -S-OOI &CtO (Ef(lJAL ~ P. ltl-) ~ C f? I -n CR-LL'I J>A-MP E A [ I]

13 ECSE 10, Summer 010 Quiz 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 Fig. 1, V 1 = 7.81 of the input current source. 0.9 o V. Find the magnitude and angle Solution 1 j 1 5 I x? 10 V -j5 V 1 0 I x Figure 1: Problem 1 Summing the currents away from node 1 gives: V 1 V 1 j V 1 j5 V 1 0I x 5 = 0 (1) From the figure 1, I x is given by: Substituting I x and V 1 = 7.81 I x = V V 1 1 j 0.9 o = 68 j6 in eqn. 1 gives: () V = j16.80 Now summing the currents away from node yields: Substituting values of V 1 and V in eqn. 3 gives: I V 10 V V 1 1 j = 0 (3) I = o A 1

14 ECSE 10, Summer 010 Quiz Total Points: 10 Problem (5 points) The two loads shown in Fig. can be described as follows: L 1 absorbs an average power of 8kW at a leading power factor of 0.8. L absorbs 0kVA at a lagging power factor of 0.6. a)determine the power factor of two loads in parallel. b)determine the apparent power required to supply the loads. c)determine the magnitude of the current I s and the average power loss in the transmission line. d)given that the frequency of the source is 60Hz, compute the value of the capacitor that would correct the power factor to 1 if placed in parallel with the two loads j0.5 I s Vs 50 0 o V (rms) L1 I 1 L I Figure : Problem Solution a)the total complex power absorbed by the two loads is S = 50Is (4) = 50(I 1 I ) = 50I1 50I S = S 1 S S = P 1 jq 1 P jq (5) Now we know that P = S cos θ and Q = S sin θ, therefore eqn. 5 becomes Substituting values, S = P 1 j P 1 sin θ cos θ S cos θ j S sin θ Now using eqn. 4, S = 8000 j 8000(0.6) 0.8 I s = 0000(0.6) j0000(0.8) S = 0 j10 kva (6) 0000 j = 80 j40

15 ECSE 10, Summer 010 Quiz Total Points: 10 I s = 80 j40 = Therefore power factor of the combined loads: 6.57 o A pf = cos(θ v θ i ) = cos(0 6.57) = lagging b) The apparent power required to supply the loads: S = 0 j10 =.36 kva c)magnitude of the current: I s = 80 j40 = A Therefore, average power loss in the transmission line is: P line = I s R = (89.44) (0.05) = 400 W d)from eqn. 6, capacitor needs to supply 10 kvar to correct the pf to 1. i.e. Q = V rms X c Therefore, X c = = 6.5Ω C = 1 wx c = 1 π(60)( 6.5) = 44.4µF 3

16 ECSE 10: Electric Circuits Quiz # (January 3, 009) Name McGill ID # Question 1: (5 min, Marks) a) Determine the frequency and the phase angle between the two voltages: v ( t) = 10sin(500t 70 ) V v ( t) = 6cos(500t 30 ) V 1 f w 500 = = = π π (1 Mark) v ( t) = 6cos(500t 30 ) V = 6cos(500t 60 ) V = 6cos(500t 10 ) V = 6cos(500t 300 ) V Phase angle between two voltages: 70 ( 60) = 130 (1 Mark) Question : (8 marks, 5 min.) Find v(t) and i(t) for t>0 in this circuit. Solution) i) At t v(0 v(0 i(0 = 0 ) = 1V At t = 0 ) = v(0 ) = i(0 i(0 ) = 1V ) = 0 ) = 0 (1 Mark)

17 ECSE 10: Electric Circuits Quiz # (January 3, 009) ii) Applying KCL at node a v(0 ) i(0 ) = ic (0 ) 1 0 = ic (0 ) ic (0 ) = 6A dv(0 ) 6 = = 1 V/s dt v( ) = i( ) = 4V, i( ) = 4 = A (1 Mark) iii) Applying KCL at node v 1 dv i = (1) dt Left mesh di 4i 1 v = 0 () dt from(1)() a dv v dt 1 dv dt 1 d v v = 0 dt or 6v 5 dv dt d v dt = 0 ( Marks) characteristic equation s v v n f 5s 6 = 0 ==> s =, 3 the natural response ( t) = v( t) = v Ae n t ( t) = v( ) = 4 v f Be 3t the forced response the complete response = 4 Ae t Be (1 Mark) (1 Mark) 3t

18 ECSE 10: Electric Circuits Quiz # (January 3, 009) determine v(0) = 1 ==> taking the A and A B = 8 (1) derivative B using the dv t 3t = Ae 3Be dt t = 0 ==> A 3B = 1 () initial values from(1)() A = 1 v( t) = 4 1e v i = = 6e 1 = 6e B = 4 t t dv dt t e 4e 4e 3t 3t 1e A, 3t V, t t > 0 t > 0 6e 3t (1 Mark) (1 Mark)

19

20

21

22

23

24

25

26

27

28

29

30

31 ECSE 10 NAME: Quiz # September 7, 01 ID: Read each question and its parts carefully before starting. Show all your work. Give units with your answers (where appropriate). 1. Determine the frequency(in Hertz) of and the phase angle between the two following current waveforms. [pt] i 1 (t) = 4 cos(377t 65 ) i (t) = 3 sin(377t 45 ) Solution: freq = 377 π = 60Hz [0.5pt] i 1 (t) = 4 cos(377t 65 ) = 4 cos(377t (65 180) ) = 4 cos(377t 115 ) i (t) = 3 sin(377t 45 ) = 3 cos(377t (45 90) ) = 3 cos(377t 45 ) [0.5pt] [0.5pt] Phase difference= i 1 i = ( ) = 70 i 1 is lagging i by 70. [0.5pt] 1

32 ECSE 10 Quiz # September 7, 01. Consider the circuit diagram below. Assume that i s = 1 tu(t) A. [8pt] (a) Find the second-order differential equation governing v(t) for t 0. [3pt] (b) Find the value of α which makes the circuit behave as an oscillator. [1pt] (c) Determine v for t 0.[4pt] a 8 Ω 1 / F v - 4 Ω b i s a v _ 1 / 4 F v - 56 Ω Solution: Node voltage analysis: KCL at node a: i s = 1 dv dt (1 α)v v 8 KCL at node b: v = 4 dv dt (1 α)v 8i s [1pt] Therefore: v 56 1 dv 4 dt v αv v (1 α)v 4 8 = 0 [1pt] d v dt (1 α )dv dt (1 α )v = 4 3 i s di s dt [1pt] The characteristic equation: For an oscillator 1 α λ ( 1 α 4 97 α )λ ( ) = 0 should be 0 and ω > 0: α = = [1pt]

33 ECSE 10 Quiz # September 7, 01 α = λ = ±j 64 v(t) = v n (t) v f (t) Therefore according to the characteristic equation, the natural response,v n (t) is: And the forced response is: Finding C and D: Initial Conditions: [1pt] v n (t) = A cos t B sin 64 t [1pt] v f (t) = Ct D d v f dt v f = 97 3 i s di s dt 97 (Ct D) = (1 t) C = 13.4, D =.4 [1pt] 3 v(0 ) = v(0 ) = α = 57.94V v (0 ) = v (0 ) = 359.6V v = 4 dv 1 dt (0 ) (1 α)v(0 ) 8i s (0 ) Applying initial conditions [1pt]: dv dt (0 ) = 0 And finally: v(t) = A cos t B sin v(0 ) = A = 35.7 dv dt (0 ) = 0 B = t 13.4t v(t) = 35.7 cos t 13.4 sin t 13.4t

34 ECSE 10 NAME: Quiz 18 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. [3pt] (a) Find the current phasor I. [3pt] 1 / 3 F 1 / 6 H i i c cos3t V sin3t A _ ¼ i c 1

35 ECSE 10 Quiz 18 May 011. Consider the circuit diagram below. [7pt] (a) Assume that the circuit in steady state, find v(t) [7pt] 4Ω 10μF Ω 4 Ω sin5000t V cos10000t A v 3 V 1mH -

36 ECSE 10, Summer 013 NAME: Quiz # May 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. [10pt] (a) Find the equivalent impedance (Z eq ) seen from ab. [5pt] (b) Is it capacitive or inductive? Why? [pt] (c) Find the phasor of the input current, i. [pt] (d) Find i(t). [1pt] i(t) a 1 / 3 H 1 / 6 F 0.5 H 1 / 18 F 1 Ω 5sin(6t45º) V 1 Ω 3 Ω 9 Ω 1 / 1 F b / 3 H Z eq Solution: I a j Ω -j1 Ω j3 Ω -j3 Ω 1 Ω 5 45 V 1 Ω 3 Ω 9 Ω -j Ω b j4 Ω Z eq Z eq = ( j1 1) [(3 j3) (j ( j3) (1 9 j)) j4] 1

37 ECSE 10, Summer 013 NAME: Quiz # May 17, 013 ID: Z eq = 1.5 j0.8ω [5pt] The imaginary part of the impedance is negative, therefore the impedance is capacitive. [pt] v(t) = 5 sin(6t 45 ) = 5 cos(6t ) = 5 cos(6t 45 ) [1pt] I = V Z = j0.8 = A [1pt] i(t) =.94 cos(6t )A [1pt]

38 ECSE 10, Fall 013 NAME: Quiz # September 4, 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. 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? [1pt] (c) Determine v c (t) for t > 0. [6pt] t=0 1 6 Ω 6V 3t V 4 Ω 1 H 1 / 13 F v c - 1

39 ECSE 10, Fall 013 Quiz # September 4, 013 Forcing Function Ke at Table 1: Forced Responses Assumed Response Ce at Kt n, n = 0, 1,,.. C n t n C n 1 t n 1... C 1 t C 0 K cos at C 1 cos at C sin at K sin at C 1 cos at C sin at Ke at cos bt e at (C 1 cos bt C sin bt) Ke ( at sin bt e at (C 1 cos bt C sin bt) n ) (( n ) ( n K i t i e at cos bt e at C i t i cos bt ( i=1 i=1 i=1 n ) (( n ) ( n K i t i e at sin bt e at C i t i cos bt i=1 Solution: Applying KVL at the right mesh: i=1 ) ) D i t i sin bt ) ) D i t i sin bt i=1 For the capacitor: 6i di dt v c = 3t [1pt] 1 dv c 13 dt = i Combining the two equations together: [1pt] d v c dt 6dv c dt 13v c = 39t Characteristic equation of the circuit: [1pt] λ 6λ 13 = 0 λ = 3 ± j under-damped [1pt] v c (t) = v cn (t) v cf (t) According to the characteristic equation, v n is: And the forced response is: v cn (t) = e 3t (A cos t B sin t) [1pt]

40 ECSE 10, Fall 013 Quiz # September 4, 013 t=0 1 i 6 Ω 6V 3t V 4 Ω 1 H 1 / 13 F v c - v f = Ct D [1pt] Finding C and D: Initial conditions: [1pt] d v cf dt 6 dv c f dt 13v cf = 39t 6C 13Ct 13D = 39t C = 3, D = v c (0 ) = v c (0 ) = 6; [1pt] i(0 ) = i(0 ) = 0; 1 dv c 13 dt (0 ) = i(0 ) = 0; dv c dt (0 ) = 0; Applying initial conditions: [1pt] v c (t) = e 3t (A cos t B sin t) 3t v(0 ) = 6 A = 6 A = dv(0 ) = 0 3A B 3 = 0 B = 49 dt 6 3

41 ECSE 10, Fall 013 Quiz # September 4, 013 And for t 0: v = e 3t ( cos t sin t) 3t [1pt] 4