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#