Chapter 4 - Circuit Theorems 2 Slides Per Page

8/3/2019 Chapter 4 - Circuit Theorems 2 Slides Per Page

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Chapter 4Circuit Theorems

Outline

Linearity Property

Superposition

Source Transformation

Thevenins Theorem

Nortons Theorem

Maximum Power Transfer

2


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Introduction

Discuss concepts of:

,

Source transformation, and

Maximum power transfer.

Since Kirchhoff's laws are unsuitable for analysis of complexcircuits, theorems such as Thevenins and Nortons theorems

will be discussed for analysis of complex circuits. These

theorems are applicable to linear circuits.

3

Linearity Property

Linearity is the property of an element describing a linear relationshipbetween cause and effect.

If the input (excitation) is multiplied by a constant, then the output (response)is multiplied by the same constant. For a resistor, for example, Ohms lawrelates the input ito the output v, v= iR.

If the current is increased by a constant, k, then the voltage increasescorrespondingly byk - Homogeneityproperty: kiR= kv.

Additivityproperty the response to a sum of inputs is the sum of theresponses to each input applied separately. If

4

( )

( )

1 1

2 2

1 2

1 2 1 2 1 2

and

then applying gives

v i R

v i R

i i

v i i R i R i R v v

=

=

+

= + = + = +


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Linearity Property

A resistor is a linear element because the voltage-currentrelationshi satisfies both the homo eneit and the additivitproperties.

In general, a circuit is linear if it is both additive andhomogeneous. A linear circuit consists of only linear elements,linear dependent sources and independent sources.

A linear circuitis one whose output is linearly related (or directly

. Note: sincep = i2R= v2/Ris a quadratic function, the

relationship between power and voltage (or current) is nonlinear.

5

Example 4.1

Consider the given linear circuit. The linear circuit has noinde endent sources inside it. It is excited b a volta e source v s

which serves as the input. The circuit is terminated by a load Rand the current ithrough Ris taken as an output.

Suppose vs= 10V gives i= 2A. According to the linearityprinciple, vs= 1V will give i= 2k (k = 1/10), i= 2(1/10) = 0.2A.Similarly, i= 1mA must be due to vs= 10k (k = 110

-3/2), vs=5mV.

6


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Example 4.2

For the given circuit, find v0when is= 15A and

s= .

7

Example 4.2

Solution:

( )1

21 2

0 2

2By current division:

2 6 4 6

2

4 4 6 3

When:

ss s

s

s

iRi i i

R R

i

v i i

= = =+ + +

= = = 6

4

i2

+

v2

i1

iS

8

( )

( )

0

0

15A, 15 10V 3

230A, 30 20V

3

This shows that when the current value is doubled, the voltage doubles.

s

s

i v

i v

= = =

= = =


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5

Example 4.3

Assume that V0 = 1V and use linearity toca cu ate t e actua va ue o 0 n t e g vencircuit.

9

Example 4.3

Solution:

1 1V0 1 1 1

1

0

, t en an .8 8 8

Since the 12 and 8 resistors are in parallel with the voltage source,

2.5V.

Assuming 1V gives 2.5V. Using the linearity propert

s

s

v

v v

v v

= = = = + = + =

= == = y, the actual source

10

10

(0 voltage of 10V will give 1 1 4 4V.2.5

v = = =

12

8

v1

+

vo5 VS = 10 V

+

i1


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Superposition When a circuit has two or more independent sources, one way to

determine the value of a specific value (voltage or current) is touse nodal or mesh analysis.

An alternative is to determine the contribution of eachindependent source to the variable and then add them up. This isknown as superposition. The idea of the superposition rests on thelinearity property.

The superpositionprinciple states that the voltage across (or

sum of the voltage across (or currents through) that element dueto each independent source acting alone.

11

Superposition

Steps to apply superposition principle:

Turn off all independent sources except one source. Find the output(voltage or current) due to that active source using the technique coveredpreviously.

Repeat previous step for each of the other independent sources.

Find the total contribution by adding algebraically all the contributionsdue to the independent sources.

Note:

Consider one inde endent source at a time while all other inde endentsources are turned off. This implies that every voltage source is replacedby 0V (or a short circuit), and every current source by 0A (or an opencircuit). This way we obtain a simpler and more manageable circuit.

Dependent sources are left intact because they are controlled by circuitvariables.

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Example 4.4

Use the superposition theorem to find vin thec rcu t.

13

Example 4.4

Solution:

1 2 1 2

1

Since there are only two sources, let where and are the contribution

due to the 6V voltage source and the 3A current source respectively.

To obtain , set the current source to

v v v v v

v

= +

( )

1 1 1

1 1

zeroKVL (loop 1): 6 8 4 0 0.5A

4 4 0.5 2V

i i i

v i

+ + = =

= = =

14

( )14

or using voltage division: 6 2V8 4

v = =+


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Example 4.4

Solution (contd.):

( ) ( )

2

2 22 3 2

3 2 3

o obtain , set the voltage source to zero.

KCL: 3 3 8V 8 4

or using current division:

83 2A and 4 4 2 8V

v

v vi i v

i v i

= + = + =

= = = = =

15

1 2Hence, 2 8 10V v v v = + = + =

Example 4.5

Use the superposition theorem to find v0.

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Example 4.5

Solution:

0 1 2 1 2

1

There are two sources. Let where and are the contribution

due to the 20V voltage source and the 8A current source respectively.

To obtain , set the current source to zero.

KVL

v v v v v

v

= +

( )

1 1 1 1

1 1

(loop 1): 20 2 3 5 0 2A

2 2 2 4V

i i i i

v i

+ + + = =

= = =

5 i13

17

( )12

or using voltage division: 20 4V2 3 5

v = =+ +

2 20 V+

+

v1

Example 4.5

Solution (contd.):

5 i2

+

i1

8 A

3 ( )

2

1 2 1 2

2 2

0 1 2

To obtain , set the voltage source to zero.

8Since 5 , using current division, 4A.

2

2 2 4 8V

Thus, 4 8 12V

v

R R i i

v i

v v v

= = = = =

= = =

= + = + =

2

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Example 4.6

Find vxusing superposition in the circuit.

19

Solution:

Example 4.6

1 2 1 2

The circuit involves a dependent (current) source which must be left intact.

Let , where and are the contribution due to the 10V voltage

source and the 2A current source respecti

xv v v v v = +

1

vely.

To obtain , set the current source to zero.

KCL: 0.1

v

i v i

+ =20 v1i1

1 11

1

100.1

20 4

2.5V

v vv

v

+ =

=

20

4 10 V + 0.1v1i2


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Example 4.6

Solution (contd.):

4 2 A

20

0.1v2

v2

2

2 3 4

2 22

2 2 2

To obtain , set the voltage source to zero.

KCL: 2 0.1

2 0.120 4

40 2 5

v

v i i

v vv

v v v

+ = +

+ = +

+ = +

i3

i4

21

2

2

1 2

40 4

10V

Hence, 2.5 10 12.5Vx

v

v

v v v

=

=

= + = + =

Example 4.7

Find Iin the circuit using superpositionpr nc p e.

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Solution:

Example 4.7

1 2 3 1 2 3

1

here are 3 sources. Let , where , and are the

contribution due to the 16V, 12V and 4A sources respectively.

To obtain , set all sources to zero except the 16V voltage sou

I i i i i i i

i

= + +

1

rce.

161A

16i = =

6

16V+

2 8 i1

2

2

To obtain , set all sources to zero except the 12V voltage source.

120.75A

16

i

i

= =

23

6

12V+

2 8 i2

Example 4.7

Solution (contd.):

( )

3

3

To obtain , set all sources to zero except the 4A current source.

2Using current division 4 0.5A

2 14

i

i

= =+

4A

6 2

8

i3

24

1 2 3Hence, 1 0.75 0.5 0.75AI i i i = + + = + =


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Source Transformation Another tool for simplifying circuits and follows the concept of equivalence.

A source trans ormationis the rocess of re lacin a volta e source vSin serieswith a resistor Rby a current source iSin parallel with a resistor R, or viceversa.

Source transformation of independent sources

e sources are urne o , e equ va ence res s ance a erm na a- n ocircuits is R. When terminal a-bare short-circuited, the short circuit currentflowing from ato bis iSC= vS/Rin the circuit on the left-hand side and iSC=iSfor the circuit on the right-hand side. In order for the two circuits to beequivalent, vS/R= iS. Hence, source transformation requires that:

25or SS S S

vv i R i

R= =

Source Transformation

The application of source transformation to dependent sources is similar tothat of independent sources.

Note:

Source transformation of dependent sources

The arrow of the current source is directed towards the positive terminalof the voltage source.

Source transformation is not possible when R= 0, which is the case withan ideal voltage source. However, for a practical, non-ideal source, Risnot equal to 0. Similarly, an ideal current source with R= cannot be

replaced by a finite voltage source.26


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Example 4.8

Find v0 using source transformation in the givenc rcu t.

27

Solution:

First, transform the current and volta e sources 3A4 = 12V and 12V/3

Example 4.8

= 4A).

Transform the 4 and 2 resistors in series which gives 6 and thenrans orm e vo age source = .

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Solution (contd.):

Combine the 3 and 6 resistors in arallel which ives 2 and combine the

Example 4.8

2A and 4A current sources (4A 2A = 2A).

Using current division:

(or further combine the resistors in parallel.)

29

( )

( )0

2 0.4A2 8

8 8 0.4 3.2V

i

v i

= =+

= = =

Example 4.9

Find i0 in the circuit using sourcetrans ormat on.

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Solution:

Combinin 6 3 ives 2 and transformin the 5A current source

Example 4.9

(5A2 = 10V). Also, combining the 1 and 4 resistors in series gives 5.

Combining the 10V and 5V voltage sources gives 15V and then transforming

10V+

2

7

+

5 3A

5 V

io

e source = . .

2 7.5A 7 5 3A

io

31

Solution (contd.):

Transformin the two current sources 7.5A+3A = 10.5A and also 2 5

Example 4.9

= 10/7.


(10/7) 10.5A 7

io

10

32

0 . .1077

= =+


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Example 4.10

Find ixusing source transformation in the givenc rcu t.

33

Solution:

Transform the CCVS 2ix/5 = 0.4ixA.

Example 4.10

Combining the two sources.

10 4 A 5

ix

0.4ix

ix


10 4 0.4ix A 5

34

( )5

4 0.410 5

1.176A

x x

x

i i

i

= +

=


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Provides a technique by which the fixed part of the circuit isre laced b an e uivalent circuit.

Thevenins Theorem

(a) Original circuit (b) Thevenin equivalent circuit

Thevenins theoremstates that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a voltage sourceVTh in series with a resistor RTh, where VTh is the open-circuit

voltage at the terminals and RTh is the input or equivalentresistance at the terminals when the independent sources areturned off.

35

VTh is the open circuit voltage across the terminals a-b.

Thevenins Theorem

Th sources are turned off.

Consider two cases in findingRTh: Case 1: If the network has no dependent sources, turn off all

independent sources. RTh is the input resistance of the networklooking between terminals a-b.

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Case 2: If the network has dependent sources, turn off all independentsources. As with superposition, dependent sources are not to be turned off

Thevenins Theorem

. Apply a voltage source vo at the terminals aand bto determine the

resulting current io to have RTh= vo/io. Alternatively, insert a current source io at terminals a-band find the

terminal voltage vo to have RTh= vo/io.

Both the approaches will give the same result and any value ofvo and iocan be used.

37

IfRThtakes a negative value (v = iR), it implies that the circuitis supplying power.

Thevenins Theorem

Consider a linear circuit terminated by a load RL. The current RLthrough the load and the voltage VLacross the

load are easily determined once the Thevenin equivalent of the

circuit at the load terminal is obtained.

38

ThL

Th L

LL L L Th

Th L

VI

R R

RV R I V

R R

=+

= =+


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Example 4.11

Find the Thevenin equivalent circuit of the circuit- .

the current though RL= 6, 16 and 36.

39

Example 4.11

Solution:

Find RThby turning off the 32V voltage source (replace it with a short-circuit)and the 2A current source (replace it with an open-circuit) as shown in Fig. a.

Consider Fig. bto find VTh. (Note: No current flows through the 1 resistor,therefore, it becomes redundant.)

( )4 124||12 1 1 3 1 4

4 12ThR = + = + = + =

+

(a) FindingRTH(b) FindingVTH40


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Solution (contd.):

Example 4.11

( ) ( )

( )

( ) ( )

( )

1 1 2 1 2

2

1 1

pp y ng mes ana ys s:

Loop 1: 32 4 12 0 16 12 32 1

Loop 2: 2A 2

Subs. 2 into 1 and solve for : 0.5A

Since the 1 resistor is redundant no voltage drop across it ,

i i i i i

i

i i

+ + = =

=

=

T

( ) ( )( )1 2

s equa to t e vo tage across t e res stor.

12 12 0.5 2 30V

Alternatively, other methods such as nodal analysis

or source transformation can be used.

h

ThV i i= = =

Solution (contd.):

Example 4.11

The current through :

30

4

L

ThL

R

VI

R R R= =

+ +

Thevenin equivalent circuit

When:

306 3A10

3016 1.5A20

3036 0.75A40

L L

L L

L L

R I

R I

R I

= = =

= = =

= = = 42


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Example 4.12

Using Thevenins theorem, find the equivalentc rcu t to t e e t o t e term na s n t e c rcu tand then find I.

43

Solution:

To find RTh turn off the current source and the voltage source as shown in

Example 4.12

Fig. a.

Use source transformation to determine VThas shown in Fig. b.

Transform the 12V source (12V/6 = 2A).

6

4 (a)RTh

6 ( )

( )

6 6 ||4

12 4

12||4 312 4

ThR = +

= = = +

44

2A

6

4 (b)6 2A+

VTh


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Solution (contd.):

The two current sources combine to ive 4A.

Example 4.12

6 6

Transform the 4A current source (4A6 = 24V).

Using voltage division:

Thevenin equivalent circuit:

24 V+

(c)

4 +VTh

( )4

24 6V 6 6 4

ThV = =+ +

45

3

6V

61.5A

3 1Th

Th L

VI

R R= = =

+ +

Example 4.13

Find the Thevenin equivalent circuit of thec rcu t e ow to t e e t o t e term na s.

46


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Solution:

To find RTh, set the independent source equal to zero but leave the dependent source.

Example 4.13

Excite the network with v0 connected to the terminal a-b. Lets use v0 = 1V for ease andthen obtain RTh= v0/i0 = 1/i0 . (A current source i0 can be used instead ofv0 to get RTh= v0/i0 = v0/1 . )

1.5Ix v0 = 1V+

3

5

a

b

4

Ix

i0

rans orm e . x . x .

47

5

7.5Ix v0 = 1V+

3 a

b

4

Ix

i0

i1 i2+

Example 4.13

Solution (contd.):

( )

( ) [ ]

( )

( )

1 1 1 2

0 1 2 0

0

2 1

Loop 1: 7.5 5 3 4 0

7.5 5 3 4 0; ;

4.5 4 0 1

Loop 2: 1+4 0

x

x x x x x

x

I i i i i

I I I I i i I i i

I i

i i

+ + + =

+ + + + = = =

+ =

=

48

( )

( ) ( )

0 1 2 0

0

+ ; ;

4 4 1 2

Solve 1 and 2 simultan

x x

x

i i i i

I i

= = =

+ =

0

0

0

eously : 2A and 2.25A

1=0.444

2.25

x

Th

I i

vR

i

= =

= =


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Solution (contd.):

To find VTh, consider the following circuit and apply source transformation.

Example 4.13

Transform the 6V voltage source (6V/5 = 6/5A).

6 V

5 Ix

4 +

1.5Ix

3

+

VTh

b

a

49

+

VTh

1.5Ix

3

5

a

b

4

Ix

6/5 A

Example 4.13

Solution (contd.):

Combine the current sources (6/5+1.5Ix) and transform it [(6/5+1.5Ix)5 =6+7.5Ix(V)].

5 Ix

46 + 7.5Ix

3

+

VTh

b

a

+

( )

6 7.5 6 7.5

5 3 4 12

12 6 7.5

1.33A

4 1.33 4 5.33V

x xx

x x

x

Th x

I IVI

R

I I

I

V I

+ += = =

+ += +

=

= = =

50

Thevenin equivalent circuit:0.444

5.33V


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Example 4.14

Obtain the Thevenin equivalent of the givenc rcu t.

51

Solution:

To find a l 1V at the terminals :R v =

Example 4.14

( ) [ ]

( )( )

( )

1 1 1 2 1

1 1 1 1 2

1 2

2 1

,

Applying mesh analysis:

Mesh 1: 5 10 4 15 0; 5

5 10 4 5 15 15 010 15 0 1

Mesh 2: 1+15 0

x xi i v i i v i

i i i i i i i

i i

+ + + = =

+ + + =

=

= 15 5 io

10 4v

x

+

+

vx vo = 1Vi1 +

i2

( ) ( )

1 2

Solve 1 and 2 s

=

1 2

0 2

0

0

1 2imultaneously : A and A5 15

2 A15

1 15= 7.5

2 215

Th

i i

i i

vR

i

= =

= =

= = = 52


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Example 4.14

( )Solution contd. :

15 5

10 4vx+

+

vx i

+

vTh

[ ]

o n , cons er t e o ow ng c rcu t.

KVL:

5 10 4 15 0; 5

5 10 20 15 0

10 0

0

Th

x xi i v i v i

i i i i

i

i

+ + + = =

+ + =

=

=

53

15 15 0 =0V ThV i= =

Nortons theorem states that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a current source IN in

Nortons Theorem

N, N -the terminals and RN is the input or equivalent resistance at theterminals when the independent sources are turned off.

RN is the same as RTh: RN= RTh

Original circuit

54

Norton equivalent circuit


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Nortons Theorem IN is the short-circuit current from terminal ato b: IN= iSC

Dependent and independent sources are treated the same way as in

Finding Norton current

Thevenins theorem. Relationship between Nortons and Thevenins theorem.

This is essentially source transformation and for this reason, sourcetransformation is often called Thevenin-Norton transformation. 55

and Th N Th N Th

VR R I

R= =

Nortons Theorem

Since VTh, INand RThare related, to determine the Thevenin orNorton equivalent circuit requires:

the open circuit voltage VOCacross terminals aand b.

the short circuit current ISCat terminals aand b.

the equivalent or input resistance RIN

at terminals aand bwhen all independent sources are turned off.

Any two of the three can be calculated using the method thattakes the least effort and the third can be found using Ohmslaw. Since

the open-circuit and short-circuit tests are sufficient to find anyThevenin or Norton equivalent circuit.

56

, , OCTh OC N SC Th N SC

VV V I I R R

I= = = =


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Example 4.15Solution (contd.):

Alternativel I can be determined b V R . Obtain V as the, .

open-circuit voltage across terminals a-b.

Norton equivalent circuit:

IN = 4.5A RN = 3a

59

Example 4.16

Find the Norton equivalent circuit of the givenc rcu t.

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Solution:

To obtain R set the inde endent current source to zero and connect

Example 4.16

,

a voltage ofv0 = 1V to the terminals and RN= v0/i0 = 1/i0.

2 6

2vx

+

+

vx 1V

+

ix

i01

61

[ ]0

0

KVL: 2 6 0; 1V

2 6 1 0 0.5A

Apply KCL at node 1: 0.5 0.5 1A2

11

1

x x x x

x x

xx

oN

o

v i v v v

i i

vi i

vR

i

+ = = = + = =

= + = + =

= = =

Example 4.16

Solution (contd.):

To obtain I short-circuit terminals a-band find I . I nore the 2 , .resistor since it is short-circuited. Also, since vx= 0, the dependent

voltage source gives 0V. Since the 6 resistor is short-circuited: ISC=IN= 10A.

2 6 10 A

2vx

+

+

vx

Isc

62

Norton equivalent circuit:

b

IN = 10A RN = 1 a


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Maximum Power Transfer Aim is to transfer/deliver maximum power to a load in a system with

known internal losses.

The Thevenin equivalent is useful in finding the maximum power a

linear circuit can deliver to a load.

For the given Thevenin equivalent, with adjustable load resistance RL,

the power delivered to the load is 22 Th

L LTh L

Vi R R

R R

= = +

63

Maximum Power Transfer

For a given circuit, VThand RThare fixed. By varying the load

resistance RL, the power delivered to the load varies as shown below.

The power is small for small or large values ofRLbut maximum for

some value ofRLbetween zero and infinity.

Maximum power is transferred to the load when the load resistance

equals the Thevenin resistance as seen from the load (RL= RTh).

64


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Maximum Power Transfer Note:

Maximum ower transferred R = R

When RLRTh

2

max4

Th

Th

Vp

R=

2Th

LV

p R

=

65

Th L

Example 4.17

Find the value ofRLfor maximum powertrans er n t e g ven c rcu t. n t e max mumpower.

66


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Solution:

Example 4.17

Th VThacross the terminals a-b. To get RTh , use the circuit below:

67

( )2 3 6||12 5 96 12

ThR = + + = + = +

Solution:

Example 4.17

Th,

1 2

2

1 1

Mesh analysis:

Mesh 1: 12 18 12 0

Mesh 2: 2A2Solve for , A3

Analyze mesh 2 to determine :Th

i i

i

i i

V

+ =

= =

68

( ) ( )( ) ( )2 1 2 212 3 0 12 2 3 2 0322V

For max. power transfer: 9

and the max. power

Th Th

Th

L Th

i i i V V

V

R R

+ + = + + =

=

= =

( )

2 2

max22

is: 13.44W4 4 9

Th

L

Vp

R= = =

Chapter 4 - Circuit Theorems 2 Slides Per Page

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