kirchhoff current law
DESCRIPTION
KIRCHHOFF CURRENT LAW. ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLES IN ELECTRICAL ENGINEERING. “ CHARGE CANNOT BE CREATED NOR DESTROYED ”. (A CONSERVATION OF CHARGE PRINCIPLE). NODE. NODES, BRANCHES, LOOPS. A NODE CONNECTS SEVERAL COMPONENTS. BUT IT DOES NOT HOLD ANY CHARGE. - PowerPoint PPT PresentationTRANSCRIPT
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 flowingnegative 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 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
iiiiiii
3 Leaving2 Leaving
076521 iiiii:3 & 2 AddingINTERPRETATION: SUM OF CURRENTS LEAVINGNODES 2&3 IS ZEROVISUALIZATION: 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 XAI X 2
Which way are chargesflowing on branch a-b?
...AND PRACTICE NOTATION CONVENTION ATTHE SAME TIME...
?43
,2
be
bd
cb
ab
IAIAI
AI NODES: a,b,c,d,eBRANCHES: a-b,c-b,d-b,e-b
ba
c
d
e2 A
-3 A 4 A
Ibe = ?
0)2()]3([4 AAAIbe
WRITE ALL KCL EQUATIONS
THE FIFTH EQUATION IS THE SUM OF THE FIRST FOUR... IT IS REDUNDANT!!!
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 SOURCE DOES NOT AFFECT APPLICATION OF KCL KCL DEPENDS ONLY ON THE TOPOLOGY
•THE LAST EQUATION IS AGAIN LINEARLY DEPENDENT OF THE PREVIOUS THREE
Here we illustrate the useof a more general idea of node. The shaded surfaceencloses a section of thecircuit and can be consideredas 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 Find 21 I and I Find
mAimAii
x
xx
404410
01212010 mAmAii xx
+-
I5I1
I2
I3
I4
I2 = 6 m A, I3 = 8 m A, I4 = 4 m A
I1 = _ _ _ _ _ _ _
I5 = _ _ _ _ _ _ _
0345 III
mA5
0123 III
mA14
xi Find
+-
+ -
1I 22I
2I
3I
4I
5I
mAImAImAI 5,3,2 321
5
4
I
I
6I
mAIIII 802 6216
0625 III0534 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
xIxI FIND
0141 mAmAI
1I
021 XX III
mA3
mA3
bX
Xb
ImAImAImAI
4221
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( AAAI X
-8AB D
0104 AAIEF
6AE 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
BBV)( AB VVqW
q
abVa b
q
abqVW LOSES
cdVc d
q
cdqVW GAINS
AV
BBV
q
CV
ABV
BC
V
CAV
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
beR3R1
V VOLTAGETHE DETERMINEKNOWN 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
BN
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
VVV
064 acV V10
042 bdV
V6
_______________, ebad VV
06812 adV
01264 ebV
_______bdVFIRST VFIND MUST
1RVVVV RRR 1010112
111
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
Presistor 2k the
on disipatedPower VVDETERMINE
SAMPLE PROBLEM
We need to find a closed path where only one voltage is unknown
041240412
X
X
X
VVVV
VFOR
4V
2
2 0VVV
VVV
Xab
abX
-8V
Rememberpast topics
+-
+-
k10 k5
xV
V25 4
xV
1V
There are no loops with onlyone unknown!!!
The current through the 5k and 10k resistors is the same. Hence the voltage drop across the 5k is one halfof the drop across the 10k!!!
- Vx/2 +
][20
042
][25
VV
VVVV
X
XXX
][5
4
024
1
1
VVV
VVV
X
XX