EECE210 Q2 Sp14 Solution

8/10/2019 EECE210 Q2 Sp14 Solution

1/8

American

University of Beirut

Department

of Electrical

and Computer Engineering

EECE 2t0

-

Electric

Circuits

Qurz

2

Closed Book

-

No Programmable

Calculators -

90

minutes

TEST ID:2000

April8, 2014

Name:

l


2/8

in

Fig.

1, find

power

aDsorr

--+

9+

I

)

l)

on)

,ll

Yq

)

3n

(,

)

t

?_

l.

absorbed by the2 A current source.

(b)

4w

(c)

12W

(d)

16w

(e)

None of the above

lvt

J

4

12.

;)

IAO

G

2h

---+

V,"r

=

(z

-?-L)

,s

=

-Vs

=-4V

?-

+ D

--VutZ

Lzh

vm = 8t,l

(abs

)

8+

4

+

?a

v*

Fig.

l:

Circuit

for Problem 1

2. If

Vs

=

36

V in

the circuit shown

in

Fig.

2, find the maximum

power

that can

be delivered to Rp.

(a)

9W

@r*

(c)

12W

(d)

0.7s

w

(e)

None

of

the above

12rz

BN

=

,{'r?

=7V

Fig. 2:

Circuit

for Problem

2

t

+sz

9o

*8)/f

/2

b2W

\,,

4v3

fi^

6(2

Kn

=(93

4O

=

352

652

If

Vs

=

8

V

in

the circuilshown

@srv

'E,yd

96v

t)

L

I

r.^*.;


3/8

3. In the

circuit

shown

in Fig.

3,

Rl

=

6 O,

R2

=

8

0,

and the

rnultiplying

constant

of

the

dependent

curent-controlled

voltage

source is

c

=

2 Y

lA.

Find the Th6venin

resistance

between terminals

a,b.

8a

G=-t,*ur*Zl,

a)

6O

@on

(c)

3O

(d)

so

ti,'"fQv'

(

2i,

I

it

=-V

+

=Y.Y*;e)

=

Y

r

Ei

.=

Eq

3+

6sz

(e)

None of the

above

Rrt

=

L/52

Fig. 3:

Circuit

of Problem 3

4. In the circuit

shown

in Fig.

4, the multiplying

constant of the dependent

voltage-conffolled cunent

source

is b

=

2

x

10-3

A/V.

Find

R1-

for maximum

power

transfer.

(a)

3.3

kf)

(b)

3.6

kCI

(c)

4.4

kf)

(tat)+.0

to

(e)

None

of

the above

1mA

Rru

Fig. 4: Circuit

of

Problem 4

. \\

=

VT

t*z

/

Vr

T

=

ZVt

=r3

(

I

Fn=

2ko

1ko

3ko

b.Vx

vr

=

tkQ

=)

ftr-r,

=

4Aa

T7


4/8

5.

In the op-amp circuit

shown in Fig. 5, Vcc

=

12Y, Rl

=

20 kO,

R2

=

150 kQ, Ra

=

10 kO,

Rb

=

20

kO, Va

=

400 mY and

Vb

=

300 mV.

Find VR.

tt

3o

3,

eV

@)t,zu

(b)

2.60 v

(c)

5.13

V

(d)

1.s4

v

{n':Vp= 2o

xool

2o+lo

Vn=

2C

+ lo

xoc3

=

I

o12s

VR

:)

VR:

V

3o

e)

None

of

the above

2c

+

tSo

lSoh

nz

?oh Rl

loR

Ra

Va

o.AV

Vb

O,?Y

;

Fig. 5: Circuit

of

Problem

5

In

the

circuit shown

in

Fig.

6,

r,2

:

5

V

and

Vcc

=

16 V. Find the maximum

value

of

v1

that

can be

used

such

that the op-amp operates

in its

linear region.

5

_V,

3+

5

:)

lo-5\

-fo

=

o

=')

{"

=

lo-Sq

=

-lb

5-tr"

-n:) U-54r2-6=e

-(J'r:

z5

+

2ok

nu

(a)

4.8 V

(b)

s.6 v

(c)

6.0

V

@s.zv

(e)

None of the above

s

ko

5/gV

2s ko

q

=

lo

+

/e

5

Fig. 6: Circuit

of

Problem

6


5/8

7.

Each

op-amp shown

in Fig.

kO.

Find

the

cunent

16.

(a)

54

mA

(b)

58

mA

$))38

mA

(d)

34

mA

(e)

None of the above

7

is operating

in the

linear region,

Vs

=

10

Y

Rr

=

20

kO,

and

Rr

=

4

V*

=

-4o

V5

=-LtoV

'

lo

Voz

=

-?o

Vot

=

$oV

To

=

*

?o

*

82-to

=

38

*ft

't

'fo

rg

'oV

l0

Ko

tloZ

:?

{30KA

I

l

Fig,

7:

Circuit

of Problem 7

8.

Find

the equivalent

capacitance

with respect to the

terminals a, b

in

the

network shown in

Fig.

8.

(a)

7

1tF

@t

uu

(c)

6

trrF

(d)

4

pF

(e)

None

of

the above

4fr

Fig.

8:

Circuit

of Problem 8

R.

-1

43

/xl7

=

8

Co-.

q

I

:)

cg

=

+

lt

5

q+8

=>

Cob=3+Z

=

S,,ttF

2,r

F


6/8

9.

In the circuit

shown

in

Fig.

9,L1

=2H,Lz =

8

H, L3

=

equation around mesh I

(corresponding

to mesh-current I

(a)

6

-

24it

*

l4iz

-

4di1

ldt

-2di2/dt

*Tdlldt

=

0

(b)

6-

24\

*

l4h-

4dit

ldt

*

di2ldt

*5dlldr

=

0

@

-24it

*

t4b

-2dit

ldt

-2di2lclt

*4dhf

ctt

=

0

(d)

6

-

24i1*

t4l

-

8di1

I

dt

*

di2ldt

*Tdfuldt

=

0

(e)

None of the above

10

{}

5

H,

and

M

=

4 H.

Write the mesh-curent

-

6+zltit-

tq

tt

*

r#

(

i,

-L)

q,L

(ts-t,)

u

8* (tt-z-E)

Ji-

dt

q4-

(L:t-t'z)

-

o

da

r

+

-\.

)

-

6t

zvi,-tvQ

+

8f)

Z"lit

4T

I

alt3

=O

z't

oz

4t

4V

Fig.

9:

Circuit of Problem

9

10. The switch

in

the circuit shown in

Fig.

10 has been opened for

a

long time

and

is

closed at

/

:

0.

What

percentage

of the

initial

energy stored

in the inductor

has been dissipated

in

the

40

Q

resistor

when

,

:250

LLs.

(,i},rrr"

(b)

86.5Vo

(c)

39.3vo

(d)

95.0V0

(e)

None of the above

F

L

wtd

= t

LI;

_?E

+

Wr=*s*U-e_

L

'Z'40

t?

WB

=

t-e

Zor1il

vl(o)

)

)

-6

Sot

lo

dL

o,63L

63.1

al

/o

50

mA

40 O

,r(t)

/=0

f)roo

20

mH

i:

)


7/8

11, InthecircuitshowninFig.

11,Cl

=41tF,C2=3[tF,C3=5trrF,andC4=5,uF,Thecapacitorsare

initially

charged

to

Va(0)

=

-5

Y

Vc(0)

=

-15

V

and

Vd(0)

=

54.4Y. The

switch

is

closed at t

=

0.

Find the total

energy delivered

to the

resistor R.

(a)

4710.4

ytl

@ruou.o

ut

(c)

2649.6

1.rI

(d)

5961.6

prJ

(e)

None

of the above

(a)

3.24

ttl

(b)

s.76

ttJ

(c)

7.29

1tI

@o'rt

(e)

None

of the

above

l'vwh

.Is

5

\f-

{L

-25oot

=

/o,5L

L

=

t

+

{+

cg4g

V,

=

V6+l/a

-Vg

=

v{B

=

u(o)

-ryhl

/

5

+

CI=

Hfo

-f

+g4'Ll

+

t5

--

6tl'VV

=

t

cuvo'=

3eo6'A7F

2'

C2

C4

tfi

Fig.

1l:

Circuit of Problem

1 1

12. In

the

circuit

shown

in

Fig.

12,

Is

=

1.4

mA.

The

switch

has been

in position

a

for

a

long time.

At

t

=

0,

the switch is moved

to

position

b. Find the total energy

delivered

to the 5 kQ resistor.

+

^r

vd

Vo

=

l,4t/Z

=

/6'8V

f

=

0'25Ylf

6^

l'6Y

lo3

=

4xlf

a

s

ue

=

16,8

e-.'oou

v

:)

W

te kQ,

*1

a

r"5

sk()

3ko

0.25 pt

?//8

Fig.

12:

Circuit of Problem

12

=

/,6fr52

V._

=J

o

Oo

_a

I o,5

-50oo

t

e

)+

lo.5'

.1

wt*

Sooo

$ooo

*

$DoD

=

A'uy(


8/8

13.

In

the

circuit

shown

in

Fig. 13, both switches

(1

and 2)

have

been closed

for

a long time and

are

opened at

t

=

0. Assume that Vs

=

300

V

Rs

=

10

kQ, Rl

=

25 k0, R2

=

25

kQ,

R3

=

50

k0,

and

C

=

40

1tF.

Find the value of the

voltage Y2 at t

=

1.2

s.

(a)

60.24Y

(b)

2s.1s v

/6)+s.rs v

Vru.,o

u

(e)

None

of the above

@t.r*o

(b)

0.s10

A

(c)

1.872

A

(d)

0.702

A

(e)

None of the above

Vs

3oov

T

IIL-

ZH

t

i(t)

=

?,ue

r

=?atl

Lo

,E+

=

?,qA

9u

{2

,fu)

Vo

=

3oo, Rul

;

lO+

Re-q

:)

Vs

=

tfloV

3

T=

4oxlo-6x/ftlo

=

lt

R*q

=

?s/6

/so

:

lokt)

-t,2

V2

=

t|Oe-t

V

=)

af

t=t'?s Vz=t5oo

=

49'$V

Fig.

13:

Circuit

of

Problem

13

14. In the circuit shown in

Fig.

14, the switch

has been

in position

a for a long time. At /

=

0, the

swjtch

moves instantaneously

to

position

b.

Assume that 11(0-)

:0,

Vs

=

300

Y

Rs

=

125

Q,

Rx

=

600 Q,

Ry

=

200 O, Ll

=

6

H,

and

L2=3

H.

Findthe

vaiue

of the curent

i2(t)

at

/=

4

ms.

9,o

t

Z

,fq

=

-

qtDt

=z.ye

-VNf

.

-/

1z0e'

v

[29{z

ns

-P

i

(|)

aF

Fig. 14: Circuit

of Problem 14

L1

6H

L.Gl

=

I

3

L2

nt

-qro

x

I

-t?zoe

4t

+

",

o

_

t,6

-,

=

1,60

+o,5

c

t=0

L2

ln

|ul,,rt

I

i,(0

I

Rx

6ue

(u*s)

EECE210 Q2 Sp14 Solution

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