KIRCHHOFF CURRENT LAW

ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLESIN ELECTRICAL ENGINEERING

“CHARGE CANNOT BE CREATED NOR DESTROYED”

NODES, BRANCHES, LOOPS

NODE: point where two, or more, elementsare joined (e.g., big node 1)

LOOP: A closed path that never goestwice over a node (e.g., the blue line)

BRANCH: Component connected between twonodes (e.g., component R4)

The red path is NOT a loop

A NODE CONNECTS SEVERAL COMPONENTS.BUT IT DOES NOT HOLD ANY CHARGE.

TOTAL CURRENT FLOWING INTO THE NODEMUST BE EQUAL TO TOTAL CURRENT OUTOF THE NODE

(A CONSERVATION OF CHARGE PRINCIPLE)

NODE

KIRCHHOFF CURRENT LAW (KCL)

SUM OF CURRENTS FLOWING INTO A NODE ISEQUAL TO SUM OF CURRENTS FLOWING OUT OFTHE NODE

A5

A5

node the ofout flowing

negative the to equivalent is

node a into flowingcurrent A

ALGEBRAIC SUM OF CURRENTS FLOWING INTO ANODE IS ZERO

ALGEBRAIC SUM OF CURRENT (FLOWING) OUT OFA NODE IS ZERO

A node is a point of connection of two or more circuit elements.It may be stretched out or compressed for visual purposes…But it is still a node

A GENERALIZED NODE IS ANY PART OF A CIRCUIT WHERE THERE IS NO ACCUMULATIONOF CHARGE

... OR WE CAN MAKE SUPERNODES BY AGGREGATING NODES

0:

0:

7542

461

iiii

iii

3 Leaving

2 Leaving

076521 iiiii:3 & 2 Adding

INTERPRETATION: SUM OF CURRENTS LEAVINGNODES 2&3 IS ZERO

VISUALIZATION: WE CAN ENCLOSE NODES 2&3INSIDE A SURFACE THAT IS VIEWED AS AGENERALIZED NODE (OR SUPERNODE)

PROBLEM SOLVING HINT: KCL CAN BE USEDTO FIND A MISSING CURRENT

A5

A3

?XI

a

b

c

d

SUM OF CURRENTS INTONODE IS ZERO

0)3(5 AIA X

AIX 2

Which way are chargesflowing on branch a-b?

...AND PRACTICE NOTATION CONVENTION ATTHE SAME TIME...

?

4

3

,2

be

bd

cb

ab

I

AI

AI

AI NODES: a,b,c,d,eBRANCHES: a-b,c-b,d-b,e-b

b

a

c

d

e2A

-3A4A

Ibe = ?

0)2()]3([4 AAAIbe

WRITE ALL KCL EQUATIONS

THE FIFTH EQUATION IS THE SUM OF THE FIRST FOUR... IT IS REDUNDANT!!!

1 2 3( ) ( ) ( ) 0i t i t i t

1 4 6( ) ( ) ( ) 0i t i t i t

3 5 8( ) ( ) ( ) 0i t i t i t

FIND MISSING CURRENTS

KCL DEPENDS ONLY ON THE INTERCONNECTION.THE TYPE OF COMPONENT IS IRRELEVANT

KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CIRCUIT

WRITE KCL EQUATIONS FOR THIS CIRCUIT

•THE PRESENCE OF A DEPENDENT SOURCEDOES NOT AFFECT APPLICATION OF KCLKCL DEPENDS ONLY ON THE TOPOLOGY

•THE LAST EQUATION IS AGAIN LINEARLYDEPENDENT OF THE PREVIOUS THREE

Here we illustrate the use

of a more general idea of

node. The shaded surface

encloses a section of the

circuit and can be considered

as a BIG node

0NODE BIG LEAVING CURRENTS OF SUM

0602030404 mAmAmAmAI

mAI 704

THE CURRENT I5 BECOMES INTERNAL TO THE NODE AND IT IS NOT NEEDED!!!

mAI 501 mAmAmAIT 204010

0410 1 ImAmA

01241 mAmAI03 12 ImAI

1I FindTI Find

1I Find21 I and I Find

mAi

mAii

x

xx

4

04410

01212010 mAmAii xx

+

-

I5I1

I2

I3

I4

I2 = 6mA, I3 = 8mA, I4 = 4mA

I1 = _______

I5 = _______

0345 III

mA5

0123 III

mA14

xi Find

+

-

+ -

1I22I

2I

3I

4I

5I

mAImAImAI 5,3,2 321

5

4

I

I

6I

mAIIII 802 6216

0625 III

0534 III

mA8mA5

mA5

mA2

THE PLAN

MARK ALL THE KNOWN CURRENTS

FIND NODES WHERE ALL BUT ONE CURRENTARE KNOWN

DETERMINE THE CURRENTS INDICATED

mA1

mA4xI2

xI

xI FIND

0141 mAmAI

1I

021 XX III

mA3

mA3

bX

Xb

ImAI

mAImAI

42

21

ONVERIFICATI

bI

A

B

C

D E

F

G

AIDE 10AIEG 4

EFIA5

xI

xI__ to __ from flowscurrent BD On

EFI

__ to __ from flowscurrent EF On

A3

This question tests KCL and

convention to denote currents

Use sum of currents leaving node = 0

010)3()5( AAAIX

-8A

B D

0104 AAIEF

6A

E F

KCL

KIRCHHOFF VOLTAGE LAW

ONE OF THE FUNDAMENTAL CONSERVATION LAWSIN ELECTRICAL ENGINERING

THIS IS A CONSERVATION OF ENERGY PRINCIPLE“ENERGY CANNOT BE CREATE NOR DESTROYED”

KIRCHHOFF VOLTAGE LAW (KVL)

KVL IS A CONSERVATION OF ENERGY PRINCIPLE

A POSITIVE CHARGE GAINS ENERGY AS IT MOVESTO A POINT WITH HIGHER VOLTAGE AND RELEASESENERGY IF IT MOVES TO A POINT WITH LOWERVOLTAGE

AV

BB

V)(AB

VVqW

q

ab

V

a b

q

abqVW LOSES

cd

V

c d

q

cdqVW GAINS

AV

BB

V

q

CV

AB

V

B

CV

CA

V

ABqVW

BCqVW

CAqVW

A “THOUGHT EXPERIMENT”

IF THE CHARGE COMES BACK TO THE SAMEINITIAL POINT THE NET ENERGY GAIN MUST BE ZERO (Conservative network)

OTHERWISE THE CHARGE COULD END UP WITHINFINITE ENERGY, OR SUPPLY AN INFINITEAMOUNT OF ENERGY

0)( CDBCAB VVVq

KVL: THE ALGEBRAIC SUM OF VOLTAGEDROPS AROUND ANY LOOP MUST BE ZERO

A B V

A B )( V

DROP NEGATIVEA

IS RISE E A VOLTAG

0321

RRRS VVVV

VVR 181

VVR 122

LOOP abcdefa

THE LOOP DOES NOT HAVE TO BE PHYSICAL

beV

0][3031

VVVV RbeR

PROBLEM SOLVING TIP: KVL IS USEFULTO DETERMINE A VOLTAGE - FIND A LOOP INCLUDING THE UNKNOWN VOLTAGE

be

R3R1

V VOLTAGETHE DETERMINE

KNOWN AREV V:EXAMPLE ,

BACKGROUND: WHEN DISCUSSING KCL WE SAW THAT NOT ALL POSSIBLE KCL EQUATIONSARE INDEPENDENT. WE SHALL SEE THAT THESAME SITUATION ARISES WHEN USING KVL

THE THIRD EQUATION IS THE SUM OF THEOTHER TWO!!

A SNEAK PREVIEW ON THE NUMBER OFLINEARLY INDEPENDENT EQUATIONS

BRANCHES OF NUMBER

NODES OF NUMBER

DEFINE CIRCUIT THE IN

B

N

EQUATIONSKVL

TINDEPENDENLINEARLY

EQUATIONSKCL

TINDEPENDENLINEARLY

)1(

1

NB

N

EXAMPLE: FOR THE CIRCUIT SHOWN WE HAVE N = 6, B = 7. HENCE THERE ARE ONLY TWO INDEPENDENT KVL EQUATIONS

ecae VV , VOLTAGESTHE FIND

GIVEN THE CHOICE USE THE SIMPLEST LOOP

DEPENDENT SOURCES ARE HANDLED WITH THESAME EASE

______

______

______

bd

ac

ad

V

V

V

064 acV V10

042 bdV

V6

_______________, ebad VV

06812 adV

01264 ebV

_______bdV

FIRST VFIND MUST1

R

VVVV RRR 1010112111

V11

DEPENDENT SOURCES ARE NOT REALLY DIFFICULT TO ANALYZE

REMINDER: IN A RESISTOR THE VOLTAGE ANDCURRENT DIRECTIONS MUST SATISFY THEPASSIVE SIGN CONVENTION

V

V

+

-

+

-

a

b V4 xV

1V

2V

kR 2

VVVV 4,12 21

2k

ab

x

P

resistor 2k the

on disipatedPower

V

V

DETERMINE

SAMPLE PROBLEM

We need to find a closed path where only one voltage is unknown

04124

0412

X

X

X

V

VVV

VFOR

4V

2

2 0

VVV

VVV

Xab

abX

-8V

Remember

past topics

+

-

+

-

k10 k5

xV

V25 4

xV

1V

There are no loops with only

one unknown!!!

The current through the 5k and 10k

resistors is the same. Hence the

voltage drop across the 5k is one half

of the drop across the 10k!!!

- Vx/2 +

][20

042

][25

VV

VVVV

X

XXX

][5

4

024

1

1

VV

V

VVV

X

XX