Exercise 3-11 Determine the Th´evenin-equivalent circuit ...zyang/Teaching/20122013Summer/Download… · Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology

Exercise 3-11 Determine the Thevenin-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:

V1

2−4+

V1−V2

3= 0,

V2−V1

3+3+

V2

5= 0.

Solution gives: V2 =−3.5 V.Hence,

VTh = Voc =−3.5 V.

(2) Short-circuit current

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

Because of the short circuit,V2 = 0.

Hence at node V1:

V1

2−4+

V1

3= 0

V1

(12

+13

)= 4

V1 =245

V

I1 =V1

3=

245×3

=85

A,

Isc = I1−3 =85−3 =−7

5=−1.4 A

RTh =VTh

Isc=−3.5−1.4

= 2.5 Ω.

Thevenin equivalent:


Exercise 3-12 Find the Thevenin 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.


Across (a,b),

VTh = Voc =10×312+3

= 2 V

RTh = 3 ‖ 12+0.6

=3×123+12

+0.6 = 3 Ω

Hence,

I =2

3+1= 0.5 A.


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

Figure E3-13

Solution: Thevenin voltage

At node 1:I = 2 A.

Hence,VTh = Voc = 10I−3×3I = I = 2 V.

Next, we determine the short-circuit current:


At node V1:

−2−3I +V1

10+

V1

3= 0.

Also,

I =V1

10.

Hence,

−2−3I + I +103

I = 0,

which gives

I = 1.5 A,

I1 = 2+3I− I = 2+2I = 5 A,

Isc = 5−3I = 5−4.5 = 0.5 A.

RTh =VTh

Isc=

20.5

= 4 Ω.

Norton circuit is:


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

Figure E3-14

Solution: We need to remove RL and then determine the Thevenin equivalent circuit at terminals (a,b).Open-circuit voltage:

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

I1 = I2 =243R

=8R

.

Voc = Va−Vb = 2RI1−RI2 = RI1 = R8R

= 8 V.

To find RTh, we replace the source with a short circuit:


R ‖ 2R =R×2RR+2R

=23

R

Hence,

RTh =4R3

,

and the Thevenin circuit is


For maximum power transfer with R = 3 Ω, RL should be

RL =4R3

=4×3

3= 4 Ω,

and

Pmax =υ2

s

4RL=

82

4×4= 4 W.


Exercise 4-7 Express υo in terms of υ1, υ2 and υ3 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,

υo =(−10×103

5×103

)[(− 3×103

0.5×103

)υ1 +

(−3×103

103

)υ2 +

(−3×103

2×103

)υ3

]= 12υ1 +6υ2 +3υ3.


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 Thevenin equivalent at terminals(a,b) for the bridge circuit.

(b) Use the result in (a) to obtain an expression forG = υo/υs.

(c) EvaluateG for R1 = R4 = 100Ω, R2 = R3 = 101Ω, andRf = 100 kΩ.

υo

υs

R2

b

a

Rf

R1

R4R3

+

_

+_

Figure P4.10: Circuit for Problem 4.10.

Solution: (a) The Thevenin equivalent circuit at(a,b):

vs

R2

b

a

R1

R4R3

+_

i1

i2

+

_

voc

υs+ i1(R1 +R2) = 0

ori1 =

−υs

R1 +R2.

Also,−υs+ i2(R3 +R4) = 0

andi2 =

υs

R3 +R4.

υTh = υoc = i1R2 + i2R4

=−υsR2

R1 +R2+

υsR4

R3 +R4=

[R4(R1 +R2)−R2(R3 +R4)]υs

(R1 +R2)(R3 +R4). (1)

Suppressingυs (by replacing it with a short circuit) leads to

RTh = (R1 ‖ R2)+(R3 ‖ R4)

=R1R2

R1 +R2+

R3R4

R3 +R4=

R1R2(R3 +R4)+R3R4(R1 +R2)

(R1 +R2)(R3 +R4).

(b) For the new circuit:

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vo

vTh

Rf

RTh

+

_

+

_

υo = −Rf

RThυTh (inverting amplifier) (3)

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

G =υo

υs=

−Rf [R4(R1 +R2)−R2(R3 +R4)]

R1R2(R3 +R4)+R3R4(R1 +R2)

(c) For R1 = R4 = 100Ω, R2 = R3 = 101Ω, andRf = 105 Ω,

G =−105[100(100+101)−101(100+101)]

100×101(100+101)+100×101(100+101)

= 4.9505≃ 5.


Problem 4.11 Determine the output voltage for the circuit in Fig. P4.11 and specifythe linear range forυs, given thatVcc = 15 V andV0 = 0.


vo

vsV0

100 kΩ

200 kΩ

2 kΩ

Vcc = 15 V+

+_

_

R1 R2

Inverting Amp

υ1

+R3

R4

υ2

_

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

R2 = 200 kΩ, R1 = 2 kΩ, R3 = 100 kΩ,

R4 = ∞, υ1 = υs, υ2 = V0 = 0.

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

υo =

(

R4

R3 +R4

)(

R1 +R2

R1

)

υ2−

(

R2

R1

)

υ1

= −

(

200×103

2×103

)

υs = −100υs.

Since|(υo)max| = 15 V, the linear range ofυs is

|υs| ≤15100

= 150 mV,

or−150 mV≤ υs ≤ 150 mV.


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

+

_ 6 kΩ

10 kΩ

4 kΩ

υs

+_

υ0

5 kΩ


Solution: By voltage division,

υp = υs6

4+6= 0.6υs.

υn = υp = 0.6υs.

υn−υs

10k+

υn−υo

5k= 0.

Simplification leads toυo = 0.4υs.

Hence,

G =υ0

υs= 0.4.


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

+_

υ0

2 A

5 V

2 kΩ

6 kΩ

6 kΩ

+

_

4 kΩ


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

+_

υoυnυ1

4 V

5 V

2 kΩ

6 kΩ

6 kΩ

+

_

4 kΩ

+

_

Fig. P4.24(a)

Apply nodal analysis:

υp = υn = 5 V.

@ υn :υn −υ1

6 kΩ+

υn −υo

6 kΩ= 0,

@ υ1 :υ1 −42 kΩ

+υ1 −υo

4 kΩ+

υ1 −υn

6 kΩ= 0.

Simplify:

16k

(υ1 +υo) =10

6 kΩ,

(

12k

+1

4k+

16k

)

υ1 −14k

υo =17

6000,

[ 16k

16k

1112k − 1

4k

][

υ1

υo

]

=

[106k

176k

]

,

[

υ1

υo

]

=

[327

387

]

.

υo = 5.429 V.

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Problem 4.36 Find the value of υo in the circuit in Fig. P4.36.

4 Ω

6 Ω

2 Ω υn

+_

υo

5 Ω 4 Ω

+

_

+

_ 9 V3 V

υp



υp = 9×4

6+4= 3.6 V

υn = υp

υn −35k

+υn −9

2k+

υn −υo

4k= 0.

Solution givesυo = −6.72 V.


Problem 4.38 Determine υo and the power dissipated in RL in the circuit ofFig. P4.38.

2 Ω

7 Ω

3 Ω4 V

2 V

+_

4 Ω 2 Ω

5 Ω

+

_

+_

υo

RL

3 Ω

Req

υaυp

υn



υp = 4×3k

2k+3k=

125

= 2.4 V

υn −27k

+υn −υa

5k= 0

υn = υp = 2.4 V.

Solution givesυa = 2.686 V.

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

Req =2×42+4

k =86

kΩ.

By voltage division,

υo = υaReq

3k+Req

=2.686×

(

86 k

)

3k+ 86 k

= 0.826 V.

PRL =υ2

o

RL=

(0.826)2

2k= 0.34 mW.


Problem 4.52 Find the value ofυo in the circuit in Fig. P4.52.

+

_

υo+_

9 V 5 V 8 kΩ

3 kΩ

4 kΩ

3 kΩ

6 kΩ

+

_

+_

8 kΩυn2

υa

υn1

υp2

υp1

i1

i2

i3


Solution: For the first stage:

υp1= 0 υn1 = 0,

υn1 −98k

+υn1 −υa

6k+

υn1 −υo

3k= 0,

which simplifies to8υo +4υa +27= 0. (1)

For the second stage:

υp2=

5×83+8

=4011

V,

υn2 = υp2=

4011

V.

Sincein2 = 0, υa = υn2 = 4011 V.

Hence,

υ0 =−27−4υa

8

=18

(

−27−4×4011

)

= −5.19 V.


Exercise 5-9 Determine Ceq and Veq(0) at terminals (a,b) for the circuit in Fig. E5-9, given thatC1 = 6 µF, C2 = 4 µF and C3 = 8 µF, and the initial voltages on the three capacitors are υ1(0) = 5 V andυ2(0) = υ3(0) = 10 V.

Figure E5-9

Solution:

Ceq =C1(C2 ‖C3)C1 +C2 +C3

=C1(C2 +C3)C1 +C2 +C3

=6×10−6(4×10−6 +8×10−6)

(6+4+8)×10−6 = 4 µF,

Veq(0) = υ1(0)+υ2(0) = 5+10 = 15 V.


Problem 5.19 For the circuit in Fig. P5.19, findCeq 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

= F

−1

a

b

5 F

a

b

6 + = F 65

365

Ceq=5× 36

5

5+ 365

=18061

= 2.95 F

Figure P5.19


Problem 5.20 Find Ceq 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

= F

−1

( )15

15

536

+ + = 1.86 F

−1

6 + = F 65

365

Figure P5.20


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

Figure E5-13

Solution:

Leq = 2 mH+(6 mH ‖ 12 mH)

=(

2+6×126+12

)mH

= 6 mH.


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 2L2LLeq

Leq

L

L

LLeq

Leq Leq = 2.5L = 25 mH

( )1

L

L

2

1

2L

1

2L+ + =

−1

Figure P5.30


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Exercise 3-11 Determine the Th´evenin-equivalent circuit ...zyang/Teaching/20122013Summer/Download… · Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology