ECE225Midterm2 Sol

7/25/2019 ECE225Midterm2 Sol

1/23

Exercise 3-11 Determine the Th evenin-equivalent circuit at terminals (a , b ) in Fig. E3-11.

Solution:

(1) Open-circuit voltage

We apply node voltage method to determine open-circuit voltage:

V 1

2 4 +

V 1 V 23

= 0,

V 2 V 13

+ 3 +V 2

5 = 0.

Solution gives: V 2 = 3 . 5 V.Hence,

V Th = V oc = 3 . 5 V .

(2) Short-circuit current

Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press


2/23

Because of the short circuit,V 2 = 0.

Hence at node V 1:

V 1

2 4 +

V 1

3 = 0

V 112

+ 13

= 4

V 1 = 24

5 V

I 1 = V 1

3 =

245 3

= 85

A ,

I sc = I 1 3 = 85

3 = 75

= 1 . 4 A

R Th = V Th I sc

= 3 . 5 1 . 4

= 2 . 5 .

Th evenin equivalent:



3/23

Exercise 3-12 Find the Th evenin equivalent of the circuit to the left of terminals (a , b ) in Fig. E3-12, andthen determine the current I .

Figure E3-12

Solution: Since the circuit has no dependent sources, we will apply multiple steps of source transformation tosimplify the circuit.



4/23

Across (a , b ),

V Th = V oc = 10 312 + 3

= 2 V

R Th = 3 12 + 0 . 6

= 3 123 + 12

+ 0 . 6 = 3

Hence,

I = 23 + 1 = 0 . 5 A .



5/23

Exercise 3-13 Find the Norton equivalent at terminals (a , b ) of the circuit in Fig. E3-13.

Figure E3-13

Solution: Th evenin voltage

At node 1: I = 2 A .

Hence,V Th = V oc = 10 I 3 3 I = I = 2 V .

Next, we determine the short-circuit current:



6/23

At node V 1:

2 3 I + V 1

10 +

V 1

3 = 0.

Also, I =

V 1

10 .

Hence,

2 3 I + I + 10

3 I = 0,

which gives

I = 1. 5 A , I 1 = 2 + 3 I I = 2 + 2 I = 5 A ,

I sc = 5 3 I = 5 4 . 5 = 0. 5 A .

R Th = V Th I sc

= 20 . 5

= 4 .

Norton circuit is:



7/23

Exercise 3-14 The bridge circuit of Fig. E3-14 is connected to a load R L between terminals (a , b ). Choose R L such that maximum power is delivered to R L . If R = 3 , how much power is delivered to R L?

Figure E3-14

Solution: We need to remove R L and then determine the Th evenin equivalent circuit at terminals (a , b ).Open-circuit voltage:

The two branches are balanced (contain same total resistance of 3 R ). Hence, identical currents will ow, namely

I 1 = I 2 = 243 R

= 8 R .

V oc = V a V b = 2 RI 1 RI 2 = RI 1 = R 8 R = 8 V .

To nd R Th , we replace the source with a short circuit:



8/23

R 2 R = R 2 R R + 2 R

= 23

R

Hence, R Th =

4 R

3 ,

and the Th evenin circuit is



9/23

For maximum power transfer with R = 3 , R L should be

R L = 4 R

3 =

4 33

= 4 ,

and

P max = 2s

4 R L = 82

4 4 = 4 W .



10/23

Exercise 4-7 Express o in terms of 1 , 2 and 3 for the circuit in Fig. E4-7.

Figure E4-7

Solution: Starting from the output of the second stage and moving backwards towards the inputs,

o = 10 10 3

5 10 3

3 103

0 . 5 103 1 +

3 103

10 3 2 +

3 103

2 103 3

= 12 1 + 6 2 + 3 3 .



11/23

Problem 4.10 In the circuit of Fig. P4.10, a bridge circuit is connected at the inputside of an inverting op-amp circuit.

(a) Obtain the Th evenin equivalent at terminals (a , b ) for the bridge circuit.(b) Use the result in (a) to obtain an expression for G = o/ s.(c) Evaluate G for R 1 = R 4 = 100 , R 2 = R 3 = 101 , and R f = 100 k .

o

s

R 2

b

a

R f

R 1

R 4 R 3

+

_

+ _

Figure P4.10: Circuit for Problem 4.10.

Solution: (a) The Th evenin equivalent circuit at (a , b ):

v s

R 2

b

a

R 1

R 4 R 3

+ _

i1

i2

+

_

v oc

s + i1( R 1 + R 2) = 0

ori1 =

s R 1 + R 2

.

Also, s + i2( R 3 + R 4) = 0

andi2 =

s R 3 + R 4

.

Th = oc = i 1 R 2 + i2 R 4

= s R 2 R 1 + R 2

+ s R 4

R 3 + R 4=

[ R 4( R 1 + R 2) R 2( R 3 + R 4)] s( R 1 + R 2)( R 3 + R 4)

. (1)

Suppressing s (by replacing it with a short circuit) leads to

R Th = ( R 1 R 2) + ( R 3 R 4)

= R 1 R 2 R 1 + R 2

+ R 3 R 4

R 3 + R 4=

R 1 R 2( R 3 + R 4) + R 3 R 4( R 1 + R 2)( R 1 + R 2)( R 3 + R 4)

.

(b) For the new circuit:

All rights reserved. Do not reproduce or distribute. c 2013 National Technology and Science Press


12/23

v o

v Th

R f

R Th

+

_

+_

o = R f R Th

Th (inverting amplier) (3)

Inserting Eqs. (1) and (2) into (3) leads to

G = o

s = R f [ R 4( R 1 + R 2) R 2( R 3 + R 4)] R 1 R 2( R 3 + R 4) + R 3 R 4( R 1 + R 2)

(c) For R 1 = R 4 = 100 , R 2 = R 3 = 101 , and R f = 105 ,

G = 105[100 (100 + 101 ) 101 (100 + 101 )]

100 101 (100 + 101 ) + 100 101 (100 + 101 )

= 4.9505 5.



13/23

Problem 4.11 Determine the output voltage for the circuit in Fig. P4.11 and specifythe linear range for s , given that V cc = 15 V and V 0 = 0.


v o

v sV 0

100 k

200 k 2 k

V cc = 15 V+

+ _

_

R 1 R 2

Inverting Amp

1

+ R 3

R 4

2

_

Solution: The given circuit is the same as the difference amplier circuit of Table4-3, with:

R 2 = 200 k , R 1 = 2 k , R 3 = 100 k , R 4 = , 1 = s , 2 = V 0 = 0.

Applying the difference amplier equation given by Eq. (4.41),

o = R 4

R 3 + R 4

R 1 + R 2 R 1

2 R 2 R 1

1

=

200 103

2 103

s = 100

s.

Since |( o)max | = 15 V, the linear range of s is

| s | 15100

= 150 mV ,

or 150 mV s 150 mV.



14/23

Problem 4.23 For the circuit in Fig. P4.23, obtain an expression for voltage gainG = o/ s.

+_ 6 k

10 k

4 k

s

+_ 0

5 k


Solution: By voltage division,

p = s6

4 + 6 = 0.6 s .

n = p = 0.6 s. n s

10k +

n o5k

= 0.

Simplication leads to

o= 0.4

s.

Hence,G =

0

s= 0.4.



15/23

Problem 4.24 Find the value of o in the circuit in Fig. P4.24.

+_ 0

2 A

5 V

2 k

6 k6 k

+_

4 k


Solution: Converting the input current source into a voltage source leads to

+_ o n 1

4 V

5 V

2 k6 k

6 k

+

_

4 k

+_

Fig. P4.24(a)

Apply nodal analysis:

p = n = 5 V.

@ n : n 1

6 k +

n o6 k

= 0,

@ 1 : 1 4

2 k +

1 o

4 k +

1 n

6 k = 0.

Simplify:

16k

( 1 + o) = 106 k

,

12k

+ 14k

+ 16k

1 14k

o = 176000

,

16k

16k

1112k

14k

1

o

=106k

176k

,

1

o =

327

387

.

o = 5. 429 V .



16/23


4 6

2 n

+_ o

5 4

+_+_ 9 V3 V

p



p = 9 46 + 4

= 3.6 V

n = p

n 35k

+ n 9

2k +

n o4k

= 0.

Solution gives o = 6.72 V .



17/23

Problem 4.38 Determine o

and the power dissipated in RL

in the circuit of Fig. P4.38.

2

7

3 4 V

2 V

+_

4 2

5

+_

+ _

o

R L

3

R eq

a p

n



p = 4 3k

2k + 3k =

125

= 2.4 V

n 27k

+ n a

5k = 0

n =

p =

2.4 V .

Solution gives a = 2.686 V .

The 2-k and 4-k output resistors are equivalent to a single resistor

R eq = 2 42 + 4

k = 86

k .

By voltage division,

o = a R eq

3k + R

eq

= 2.686 86 k

3k + 86 k

= 0.826 V .

P RL = 2o R L

= (0.826 )2

2k = 0.34 mW .



18/23


+_

o+_

9 V 5 V 8 k3 k

4 k

3 k

6 k

+_

+_

8 k n2

a n1

p2

p1

i 1

i 2

i 3


Solution: For the rst stage:

p1 = 0 n1 = 0 , n1 9

8k +

n1 a6k

+ n1 o

3k = 0 ,

which simplies to8 o + 4 a + 27 = 0 . (1)

For the second stage:

p2 =5 83 + 8

=4011

V,

n2 = p2 =4011

V.

Since i n2 = 0, a = n2 = 4011 V.

Hence,

0 = 27 4 a

8

=18

27 4 4011

= 5 . 19 V .



19/23

Exercise 5-9 Determine C eq and V eq(0) at terminals (a , b ) for the circuit in Fig. E5-9, given thatC 1 = 6 F, C 2 = 4 F and C 3 = 8 F, and the initial voltages on the three capacitors are 1(0) = 5 V and 2(0) = 3(0) = 10 V.

Figure E5-9

Solution:

C eq = C 1(C 2 C 3)C 1 + C 2 + C 3

= C 1(C 2 + C 3)C 1 + C 2 + C 3

= 6 10 6(4 10 6 + 8 10 6)

(6 + 4 + 8) 10 6 = 4 F ,

V eq(0) = 1(0) + 2(0) = 5 + 10 = 15 V .



20/23

Problem 5.19 For the circuit in Fig. P5.19, nd C eq

at terminals (a , b ). Assume allinitial voltages to be zero.

Solution:

a

b

c

d

5 F 3 F 5 F

5 F3 F

6 F 6 F

a

b

5 F

6 F ( )13 + 13 16+ 65= F1

a

b

5 F

a

b

6 + = F65

365

C eq = 5 3655 + 365

= 180

61 = 2. 95 F

Figure P5.19



21/23

Problem 5.20 Find C eq

at terminals (c , d ) in the circuit of Fig. P5.19.

Solution:

a

b

c

d

5 F 3 F 5 F

5 F3 F

6 F 6 F

c

d

5 F

5 F

6 F

c

d

5 F

5 F

c

d

( )13 + 13 16+ 65= F1

( )15 15 536+ + = 1.86 F1

6 + = F65

365

Figure P5.20



22/23

Exercise 5-13 Determine L eq at terminals (a , b ) in the circuit of Fig. E5-13.

Figure E5-13

Solution:

Leq = 2 mH + ( 6 mH 12 mH )

= 2 + 6 12

6 + 12 mH

= 6 mH .



23/23

Problem 5.30 All elements in Fig. P5.30 are 10-mH inductors. Determine Leq

.

Solution:

L

L

L L L

L

L

L

L

L 2 L2 L L eq

L eq

L

L

L L eq

L eq L eq = 2.5 L = 25 mH

( )1 L L212 L 12 L+ + =1

Figure P5.30

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ECE225Midterm2 Sol