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2020 ENGG 225 Final Exam Solutions PNBQuestion I-
(a) With only source v,active
,I, and vz
are zeroed :
I OV
Gr- -
MI i,' must be Zero.
{ 8r FLY# Rz and Rs arei't 3122 in series but then#
in parallel withwith only Vz active . . .
a short circuit.All current will
bypass Rz and Rs
-
wrote Here,V,= 3v,so i " must be
§ TM-44ii.tf 7-r 3v I " = I = 0.2727 A
l l At llsI
1-=
Finally , by superposition , i = i 't I"
= 10.27277L
(b) At the top of the circuit,GF IF
Ceq, = Cale t C ,= 6×1 t 6
Cztcz 11
GF = 8.7272 F
TE 5F
Between terminals a and b,
Ceg = Ceorl Cs t CA = 8.7272×5 t I-
-
Cece + Cg 8.7272+5
(e) /Ceq=4.l788At DC
,all capacitors become open
circuit,inductors become short circuits .
p .2/13
ForThe circuit simplifies to :-Wh-
Total equivalent resistance across ¥①¥•.
the source is-m-
Reg = 4/14 = 22 Are
so i = vlreg =3 A /i=3(d)
Circuit simplifies to a series
combination of R ,and - jxg
Total impedance across the source
2-eg = R
,- jxs
i
| = 6 - jg
series.rs#tLC So I -- T,
= 2/-450N
- -€nat:jXz-j/ Zeq 6 - j9
Convert Zog to polar notation : Zeq = 10.8167/-5-6.310
so I = 2€50 = (2/10.8167)/45%-56.31010.8167/-5-6.310 = 0
.18491¥31
IIIt-o.net#
(ellwe are told that Iz = A- 1¥
.
By kaI
,+ Ia =3
so I,= 31-00 - the=3 tjo - ( 1.69 tj 3.625)= 1.3096 - j 3.625= 3.85A €70.11
Clearly , Ia Leads both I,and I
,while I
, tags both II and I .
P. 3/13Since all of the current source
, X ,and the series combination of Y and E
,
are in parallel , they will have the same voltage .
Since current heads voltage , this suggests that either of Y or Z is a
capacitor . That means X must be the opposite of this so that the
iinaginary parks of the total current cancel.
IXisan
(f) 20mA We have impedances
jwh ,= j 1001T (0.02) r
1.52= j 21T r = j 6.283N
so I = 2/-90-0I /jwc ,
= I /j ( 1001T x 0.0035)3.5mF = - j 0.909552
Total impedance Zeq = 1.5 t j 6.283 - j 0.9095 = 1.5 t j 5.3737= 5.579/79=09-0
Total current I = J/zeq = 21=9005.5791¥10-
= 0.3585 LEAT 10=-164.9-10-1
(g)Load impedance has PF = 0.7 leading
652-
f -- 2000 Hz A leading power factor means that the
power angle O is positive , implying thatIYR the current Leads the voltage
O = OI - Ou > 8 .Q must be negative .
Hence,circuit element Y must be a capacitor
O = Foo and tan O = XIR
so X = R tan O
X = 65 x tan (Foo ) -
- 59.59-2
pathsFor a capacitor X = llwc
, so C = l/wX= 1-
( = 1.459 x 10- b
f 2000×217×53.702
!C=l.A59##Question
(a) The Wheatstone bridge circuit "
gaon
280A 9902The bridge is balanced whengyu
Va = Vb .Resistor Radjust Can Ihor
be most easily calculated by71Or
maintaining the same proportionof resistor values in the bridge
That is : Rz = Ra so Radj = Rs Rsense- -
I-
Radj Rsense RA
Radj = 280×1202 a 33.939N
990A-
(b) The thermistor characteristics from Lab #z.
/Radj=33i931
For Rt = 3000C,the horizontal
line is drawn as shown .
The vertical line intersects
graph # to when × = 3.1A-
the:c:: the.nl::*:*:*. /So ¥ = 3. 1A conversion
of ok → 0C
Inverting Tx 103 K teC -
- 0.3185103 K - 273.15
|c=A9.3
p . 5/13
Questions
Note In the 02h quiz,T,= 2.0 secs
,and
times are randomized
ka) we have L -
- 0.2 H,I
,-- 2.0A
,
and t -- 0.20 secs
voltage : V -
- Lddtt The equation for the current in the
first interval is of the form in -- Mt
where m =- 8
,so in = - 1Gt
0.5 s-
and V= Ldi = 0.2 x m /V=-3.2dt
(b) Here,we need the equation of the line in the 3rd interval 1.costs less
For this question, L -
- 9 H,I
,-
- 10A,t = 1.25 .
Form of equation : y = Mit b
slope : M =5 = 30 . Intercept : b -- y - Mx
1.5 - I.Os = 5 - 304.5)=- AO
so I = 30T - AO
At t -- 1.2ns,
i = 304.23 - to = - AA
Energy W,
e'zLi2 = I (9)GAY = 72T /wL=72T!-
(c) For this question , L -
- 5 It,
I,-
- AA,
and t -- 1.35 s
we are operating once again the interval to Ct Clios,and
y ' mxtb : m=2-C = 12,
and b = 2 - 124.530.5 = - 16
so i = 12T - 16
P. 6/13Therefore , at t= 1.35
,
I = 124.35) - 16 = 0.2A
The voltage : V - Ldi = 5×12 = 60 v
dt
Power p e vi /p=12
(d) For this part, C -
- IOF,volts =
-5v,too .
Energy : we = { Cv? = { ( 10K-55 /we=125
(e) Note: There was an error in the 1322 quiz for this question .
We 'll answer here anyway !
For this part, C = 2E,I
,= 6.0A
,and Vo = 3v , and t - 0.55
Voltage on the capacitor in the first interval O - tf 0.5 s
t
left ) = { Jilt)dt t Uco) where m ==- 12
o 0.5- O
0.5 and ich = - 12T
so 46.5) - tf - l2tdt t 3
0.5 0.5
= - bftdt +3 = -6 [ It ]o t 3O
= - 3 ( 0.55 t 3-
= 2.25 v /V=2r25I
(f) Here,we are using ( = 5.5 F
,I,= 5A
,Velo) = Gv
,and t -- 0.650 see
This places us in the second interval 0.5ate i.Os .
We will need to repeat part Ce) with different numbers to find
the capacitor voltage at the end of the first interval.
With I,
= 5A,
the equation of the line in interval 1 is :
m = - 5 - O-
=- 10
,so Ict) = - lot
0.5 0.5
The voltage volt) =g÷f- lotdt t Velo)
O
as p . 7113'
Velho) = -¥0,11 Eh! ) t 6 = - 0.9091 ( 0.25) t 62
o= 5.773 u
In the second interval Ict ) = - 5A ( constant)
0.65
so Velo . 650 ) = If f-5)dt t 5.773 v5.5of
=
-55,3 It !!'
t 5.773 = -0.9091 (0.65 - 0.5) t 5.773
UeCo.65o)=5.6363
Question
Foor(a)
910Y V2
Zu 1950A710N
630C
stage 580N Stage
The first stage is a comparator circuit . As usual,no current will flow
through R,and Rz so the voltage Vt = Ov and V- =2v .
The
op-amp is being run open-top
Ideally ,U,= -Aud ,
where Vd -
-LV) - (Vt) = 2 - o = 2v .
So,with A =
-[huge- number) x Vd,
V,=- [ bigger-huge-number] v
The op-amp is only being powered by t lov , so this will cause U, to be
clipped at - lov .
The op-amp is saturated .
V,=- lov
.
Stage 2 is an inverting amplifier configuration .
Node Vz : Vz - tout t Vz - V , = O-
-
-
Foo 1950
p . 8/13where Vz = 0
,and O - V'out +
-V
= O-
1-
400 1950
So Vout = - Vix Aoe = 10x 0.20513
1950
fvou+5
(b)390N
5002
660A Vs750 Yf
-2. v ↳for
VG
v color stage 2 Stage 3 590N
stagey-
-
Stageianalysis
Noddy V3-Y-tvz-V2-fvz.UA =0,and vz
-- O
660 750 390
with U,-- Vz = -Zu
, 390-4 t 391 Vz = - VA660 750
0.59091 V , t 0.52 Vz = - Va
( 0.59097 t 0.52) x -2 = - VA
VA = 2.222 V
Stagezanalysis
Node 5 Vs + Us - VG = 0,
and Vg = Vq-
--
610 500
501 x 2.222 t 2.222 = Vb610
VG = A. 09-298 V
Stagesanalysis
This is a voltage - follower circuit . Just by observingthe properties on an ideal op-amp , Vo = Vo
I✓out = 4.04298 v-
P. 9/13
Questions
(a)
Ci ) For this question, i. Ct) = 2.08 Siri ( 90ft t 9.630) A
0
Using a cosine instead,I,Ct) = 2.08 cos (gott t 9.63 - 900 )
0
Phaser notation : I = 2.08 ¥37 A
Adding the magnitude and phase : Ans = 2.08 - 80.37
|Ans=78o20
(Ii ) Here,I,Ct) = 15.23 cos (596T t 19.670) A
,and C
,= 886mF
The radian frequency is 596 racks . Ze = lljwc
Ec =-1 = - j 0.00189352j (59676.886)
or in phaser notation Zc = 0.001893
Adding the magnitude and phase : Ans = - got 0.001893
|Ans=-89.998
(iii) we have I. Ct ) = 6.80 cos (537T - 30.710) and R,= 607A .
The voltage up,across this resistor is siinply
Vp, Ct) = Ict) R , = 6.80 x 607 cos (537T - 30.710 )
Note that since R,is a real constant
,so UCH will be
P. 10/13
exactly lipase with i. Ct) . So yet) leads 4G) by 00
lphasedifferene.TO/
(b)
+ v -
Rlt
va
-
£,
Ci ) The mesh equation for mesh II :
- T t Ie (Rat za) t VI = 0 where Rz= 21h2-<= j 9.02
- 251¥20 t ITCH t j9.02) t 251¥20 = 0 VT = 251¥20Ta = 251¥20
Impedance term : X = 21 t j 9.02ISum=3O.
( ii) KVL around loop A gives : Y - Va t TIR , ,where UT -- II. 1€80I,= 5/-0
So Ja = UT t IR,= 11.11¥ t 5×82 R
,= 82N
T2.287 - j 10.760
Ta = 2.287 - j 10.760 t 4- to
= A- 12.287 - j 10.760
In polar notation : Ta = 412.4247¥950
IIV-al-9-12.AM#-/(iii ) Apparent power can be computed using : Papp -- Vigne
p- 11/13we know I = 5.8/-57-70 and Ze = -jar
Papp = (5.8/5-5)"
= 0.801 VA
21
IPapp=Oa80l
Question
(a) With Vt = 170 v and Re = Gor,
Pr. = Vfl RE - 1702/60 /ff=A8l.67
(b) With IL = 150 A,
RF = 5-Or, Vt = 230
,and Ra = Oullr
we need Ia .
IA = IL - IF = 150 - 230/50 = 24-5.7 A
and PA = In? Rn. = ( 29-5.772×0.112
IPA=2325r5W_
(C) Now using Vt = 220W, IL = 100A, RA a 0.052 , RF = 80N,
we first need Ia as in part (b)
Ia = IL - VTIRF = 100 - 220/80 = 97.25
Then,KVL around the whole outer loop :
- UF t IARA t EA = 0,
so EA ? Up - IARAEA = 220 - 97.25 ( 0.05)-
|Ea=2l5nl38#
(d) With new values Vi = 270 u, Ih = 120A
,Ra = 0.07N
, Rf = look,and Tout = 24-0 Nm
,we will first need the machine constant kg .
p.CZ/Bk0/--Toot/IaNote : since the rotational losses
are negligible , Tout = Tdev .
As before,IA = 120 - 29-0/100 = 117.6
so Kol = 29-0/117.6 = 2.09-8
Also as before, EA = ZAO - 117.6 ( 0.07 ) = 231.78 u
And rotation speed Wm = Ea Ikf , so Am = 231.78/2.09-8=113.17 radlse
Finally , nm = Wm x 60121T = 1080.67 rpm-
|nm=l08O.67rp#/,
(e) Now,let Vt = 190 v
,Ih -
- 70A,RA = 0.12k
, RF = odor,the total
input power will be simply
Pin = VTI, = 190 v x 70 A = 13,300 W
Once again , we are allowed to assume that rotational losses are
insignificant . We can find Pout as follows :
Pout = Pin - Pe - Pa
Pr. = Ugh / Rp = 190490 = 9-01.11 W
PA = In? Ra = ( Ia - Ip)'Ra -
- CIL - VIRE )-
RA= ( 70 - 190/90 ) ' x 0.12 = 583.28W
Hence,
Pout = 13,300 - 401.11 - 583.28= 12,355.6 w
Efficiency : Y = Pout x 100% = 12,355.62 x 100%
Pin 13,300-11of = 92.9%I
(f) Finally , let Ut = 190 u, Ia - 80A , Ra a 0.09 Ah
,RF = 502
,and
Toot = 290 N- m .The machine now has rotational losses Trot = 19.0 Nim
.
Total developed torque Tdeu = Tout t Trot = 290+19 = 309 Nm
p. 13/13As before
, EA = Vt - Ia Ra = 190 - (80-190/50)= 183.142 V
Total developed power : Poder = IAEA = 76.2 x 183.142
= 13,955W
Using T -
- Palm ,Om = Pdevltdev = 13,955/309
= 4-5.16 radlsec
Finally , nm = 45.16 x 60121T rpm -
|Mm=A31.28rpm#