circuit lecture

8/12/2019 Circuit Lecture

1/21

Capt (R) Faraz Ullah Khan

Muhammad Nasir Wattoo

The University Of Lahore

Basic Electrical Engg.

Chp 2.2Kirchoffs Laws


2/21

Charge Conservation

One Of The Fundamental ConservationPrinciples In Electrical Engineering:

CHARGE CANNOTBE CREATED NOR

DESTROYED


3/21

Node, Loops, Branches

NODE: Point Where Two,Or More, Elements Are

Joined (e.g., Big Node 1)

LOOP: A Closed PathThat Never Goes Twice

Over A Node (e.g., The BlueLine)

The redpath is NOT a loop (2x on Node 1)

BRANCH: a Component Connected

Between Two Nodes (e.g., R4 Branch)


4/21

Charge Conservation at Nodes

A Node Connects Several ComponentsBut It DOES NOT HOLD Any Charge

By The Conservation of Charge Principle

We Have Kirchoffs Current Law:

TOTAL CURRENT

FLOWING INTO THE NODE

MUST BE EQUAL TO THETOTAL CURRENT OUT OF

THE NODENO E


5/21

Kirchoffs Current Law (KCL)

Practical Restatement of KCL Sum Of Currents Flowing Into A Node Is

Equal To Sum Of Currents Flowing Out

Of The Node

Usual KCL Sign Convention

POSITIVE Direction INTO Node

-5A

+5A

NEGATIVE Direction OUT of Node


6/21

KCL Algebra

Two Equivalent KCL Statements Algebraic Sum Of Currents leaving

(Flowing OUT Of ) A Node Is ZERO

Algebraic Sum Of Currents entering(Flowing INTO) to A Node Is ZERO

Example: Use Any Sign

Convention

0

0

54321

tititititi

tiNodeINto


7/21

Supernodes or Closed Surfaces A Generalized Node Is Any Part Of A

Circuit Where There Is No Accumulation of

Charge. Set Of Elements Contained Within

The Surface That Are Interconnected

Suggests We Can

Make SUPERNODESBy Aggregating Nodes

0:Add

___________________________________

0:3Into

0:2Into

76521

7542

641

iiiii

iiii

iii


8/21

Supernodes cont.

INTERPRETATION: Sum Of Currents

Entering Nodes 2&3 is Zero

VISUALIZATION: We Can Enclose Nodes

2&3 Inside A Surface That Is Viewed As A

GENERALIZED Node (Or SUPERnode)

Supernode is Indicated as the GREEN

Surface on the Diagram; Write KCL Directly

00 76152 titititititiSuperNode

Same as Previous


9/21

KCL Problem Solving

KCL Can Be Used To

Find A Missing Current

(Currents INto Node-a) = 0

A5

A3

?XI

a

b

c

d

AIAAI XX 2or035 Which Way are

Charges Flowingin Branch a-b?

b

a

c

d

e2A

-3A4A

Ibe= ?

Iab= 2A

Icb= -3A

Ibd= 4A

Ibe= ?

Nodes = a,b,c,e,d

Branches = a-b, c-b,d-b, e-b

AIAAAI bebe 5or0432

Notation Practice


10/21

UnTangling

A node is a point of connection of two or morecircuit elements.

It may be stretched-out or compressed or Twisted or

Turned for visual purposesBut it is still a node

Equivalent

Circuits


11/21

KCL Alternate Sign Convention

KCL Works Equally Well When CurrentsOUT Are Defined as Positive

Write the +OUT KCL

1

2

3

4

5

Note That Node-5 Eqn is Redundant;

It Is The SUM of The Other 4


12/21

Example

Find Currents Use +OUT

1

2

3

4

KCL Depends Only On The Interconnection.The Type Of Component Is Irrelevant

KCL Depends Only On The Topology Of The Circuit


13/21

Example

Find Currents

Use +OUT

1

2

3

4

The Presence of the

Dependent SourceDoes NOT Affect KCL

KCL Depends Only On

The Topology

Again, Node-4 Eqn is

(Linearly) Dependent onthe Other 3


14/21

Example - Supernode

Supernodes Can EliminateRedundancies and Speed Analysis

Shaded Region =

Supernode, S

S

0602030404 mAmAmAmAImAI 704

The Current i5Becomes

Internal To The Node And It Is Not Needed!!!

Use +OUT

of Currents Leaving

Node-S = 0


15/21

KCL Convention: In = Out

An Equivalent Algebraic Statement ofCharge Conservation

NodeofOUTCurrentsNodeINTOCurrents

mAI 501 mAmAmAIT 204010

1IFind TIFind


16/21

Examples: In = Out

mAI

mAmAI

6

410

1

1

mAImA 412 1 mAII 321

1IFind 21 IandIFindmAImA 410 1

mAmAmAmAII

mAmAmAI

2686

8412

12

1


17/21

Find Ix

mAiimA

iimA

x

x

xx

41144

1044

mAimAi

mAmAii

x

x

xx

121089

01212010


18/21

Find Unknown Currents

The Plan Mark All Known Currents

Find Nodes Where All But

One Current is Known

Given I1= 2 mA

I2= 3 mA

I3= 6 mA

+

-

+ -

1I 22I

2I

3I

4I

5I6I

mA2

mA3

mA6

mAIIII 82 6216

1

2

3

1

2mAIIII 55526

3mAIIII 14435


19/21

Find Ix

At Node 2

mA1

mA4xI2

xI

mAI

mAmAI

3

014

1

1

1I

mAII

III

X

XX

3

2

1

1

mAmAIIc

mAImAIb

Xb

Xb

242)

21)

bI

1

2b

I1is Opposite

the AssumedDirection

At Node 1

Verification at

Nodes b & c

c


20/21

KCL & Direction Summary Demo

C

D

E

F

G

AIDE 10 AIEG

4

EFI

A5

xI

xI

__to__fromflowscurrentBDOn

EFI

__to__fromflowscurrentEFOn

A3

010)3()5( AAAIX

-8A

B D

0104 AAIEF

6A

E F

For Ixuse Iout= 0

Note Directions for IDE

and IEFand IEG

For IEFuse Iout= 0


21/21

Home Work Problem

Lets Work This

Problem

12 mA

3 mA

2 mA

4 mA

Ix

Iy

Iz

Find

zyx III

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