Transcript No Slide Title

KIRCHHOFF CURRENT LAW
ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLES
IN ELECTRICAL ENGINEERING
“CHARGE CANNOT BE CREATED NOR DESTROYED”
NODES, BRANCHES, LOOPS
A NODE CONNECTS SEVERAL COMPONENTS.
BUT IT DOES NOT HOLD ANY CHARGE.
TOTAL CURRENT FLOWING INTO THE NODE
MUST BE EQUAL TO TOTAL CURRENT OUT
OF THE NODE
(A CONSERVATION OF CHARGE PRINCIPLE)
NODE: point where two, or more, elements
are joined (e.g., big node 1)
LOOP: A closed path that never goes
twice over a node (e.g., the blue line)
The red path is NOT a loop
BRANCH: Component connected between two
nodes (e.g., component R4)
NODE
KIRCHHOFF CURRENT LAW (KCL)
SUM OF CURRENTS FLOWING INTO A NODE IS
EQUAL TO SUM OF CURRENTS FLOWING OUT OF
THE NODE
5A

 5A
A GENERALIZED NODE IS ANY PART OF A
CIRCUIT WHERE THERE IS NO ACCUMULATION
OF CHARGE
... OR WE CAN MAKE SUPERNODES BY
AGGREGATING NODES
A current flowing into a node
is equivalent to the negative
flowing out of the node
ALGEBRAIC SUM OF CURRENT (FLOWING) OUT OF
A NODE IS ZERO
ALGEBRAIC SUM OF CURRENTS FLOWING INTO A
NODE IS ZERO
Leaving 2 : i1  i6  i4  0
Leaving 3 :  i2  i4  i5  i7  0
Adding2 & 3 : i1  i2  i5  i6  i7  0
INTERPRETATION: SUM OF CURRENTS LEAVING
NODES 2&3 IS ZERO
VISUALIZATION: WE CAN ENCLOSE NODES 2&3
INSIDE A SURFACE THAT IS VIEWED AS A
GENERALIZED NODE (OR SUPERNODE)
PROBLEM SOLVING HINT: KCL CAN BE USED
TO FIND A MISSING CURRENT
SUM OF CURRENTS INTO
NODE IS ZERO
b
IX  ?
c
5A
WRITE ALL KCL EQUATIONS
5 A  I X  (3 A)  0
I X  2 A
a
Which way are charges
flowing on branch a-b?
3A
d
...AND PRACTICE NOTATION CONVENTION AT
THE SAME TIME...
I ab  2 A,
I cb  3 A
I bd  4 A
I be  ?
NODES: a,b,c,d,e
BRANCHES: a-b,c-b,d-b,e-b
d
c
a
-3A
2A
4A
b
Ibe = ?
e
I be  4 A  [(3 A)]  (2 A)  0
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 LAST EQUATION IS AGAIN LINEARLY
DEPENDENT OF THE PREVIOUS THREE
•THE PRESENCE OF A DEPENDENT SOURCE
DOES NOT AFFECT APPLICATION OF KCL
KCL DEPENDS ONLY ON THE TOPOLOGY
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
SUM OF CURRENTSLEAVINGBIG NODE  0
I 4  40mA  30mA  20mA  60mA  0
I 4  70mA
THE CURRENT I5 BECOMES INTERNAL TO THE
NODE AND IT IS NOT NEEDED!!!
Find I1
Find I T
I1  50mA
Find I1
10mA  4mA  I1  0
IT  10mA  40mA  20mA
Find I1 and I2
I 2  3mA  I1  0
I1  4mA  12mA  0
Find ix
10i x  i x  44mA  0
i x  10i x  120mA  12mA  0
i x  4mA
I 3  I 2  I1  0
I1
I3
I5  I 4  I3  0
I5
+
-
I2
I2 = 6mA, I3 = 8mA,
I4
I4 = 4mA
mA
I1 = 14
_______
5mA
I5 = _______
DETERMINE THE CURRENTS INDICATED
I3
I1
+
-
5mA
I 4  2mA
+
-
I4
2I 2
I5
I6
I 2 8mA
I1  2mA , I 2  3mA , I 3  5mA
I 6  I1  2 I 2  0  I 6  8mA
I5  I 2  I6  0
I 4  I3  I5  0
I 5  5mA
THE PLAN
MARK ALL THE KNOWN CURRENTS
FIND NODES WHERE ALL BUT ONE CURRENT
ARE KNOWN
FIND I x
Ix
 3mA
I X  I1  2 I X  0
I1  4mA  1mA  0
I1  3mA
VERIFICATI ON
I b  1mA  I X  2mA
1mA
2 I X  4mA  I b
Ib
2I x
4mA
This question tests KCL and
convention to denote currents
Use sum of currents leaving node = 0
A
I X  (5 A)  (3 A)  10 A  0
5A
F
I EF
B
Ix
D
E
I DE  10 A
I EF  4 A  10 A  0
I EG  4 A
3A
C
I x  -8A
G
On BD current flows fromB
__ to D
__
I EF  6A
OnEF current flows from__
E toF__
KCL
KIRCHHOFF VOLTAGE LAW
ONE OF THE FUNDAMENTAL CONSERVATION LAWS
IN ELECTRICAL ENGINERING
THIS IS A CONSERVATION OF ENERGY PRINCIPLE
“ENERGY CANNOT BE CREATE NOR DESTROYED”


q
C
W  q (VB  VA )
W  qVAB
V B
A POSITIVE CHARGE GAINS ENERGY AS IT MOVES
TO A POINT WITH HIGHER VOLTAGE AND RELEASES
ENERGY IF IT MOVES TO A POINT WITH LOWER
VOLTAGE
B VB
AB
KVL IS A CONSERVATION OF ENERGY PRINCIPLE
A “THOUGHT EXPERIMENT”
V
KIRCHHOFF VOLTAGE LAW (KVL)
q
W  qVBC

VA  VCA 
W  qVCA
B
VB
VC
IF THE CHARGE COMES BACK TO THE SAME
INITIAL POINT THE NET ENERGY GAIN
MUST BE ZERO (Conservative network)

VA
OTHERWISE THE CHARGE COULD END UP WITH
INFINITE ENERGY, OR SUPPLY AN INFINITE
AMOUNT OF ENERGY
q
 Vab 

q
q(VAB  VBC  VCD )  0
a
b
 Vcd 

c
d
LOSES W  qVab
KVL: THE ALGEBRAIC SUM OF VOLTAGE
DROPS AROUND ANY LOOP MUST BE ZERO
GAINS W  qVcd

V 
A
B
 (V ) 
A
A VOLTAG E RISE IS
A NEGATIVE DROP
B
PROBLEM SOLVING TIP: KVL IS USEFUL
TO DETERMINE A VOLTAGE - FIND A LOOP
INCLUDING THE UNKNOWN VOLTAGE
THE LOOP DOES NOT HAVE TO BE PHYSICAL

Vbe

 VS  VR  VR  VR  0
1
2
3
VR  12V
2
VR  18V
1
EXAMPLE : VR1, VR3 ARE KNOWN
DETERMINE THE VOLTAGE Vbe
VR  Vbe  VR  30[V ]  0
1
LOOP abcdefa
3
BACKGROUND: WHEN DISCUSSING KCL WE SAW
THAT NOT ALL POSSIBLE KCL EQUATIONS
ARE INDEPENDENT. WE SHALL SEE THAT THE
SAME SITUATION ARISES WHEN USING KVL
A SNEAK PREVIEW ON THE NUMBER OF
LINEARLY INDEPENDENT EQUATIONS
IN THE CIRCUIT DEFINE
N
B
N 1
NUMBER OF NODES
NUMBER OF BRANCHES
LINEARLY INDEPENDEN T
KCL EQUATIONS
B  ( N  1) LINEARLY INDEPENDEN T
KVL EQUATIONS
EXAMPLE: FOR THE CIRCUIT SHOWN WE HAVE
N = 6, B = 7.
HENCE THERE ARE ONLY TWO INDEPENDENT
KVL EQUATIONS
THE THIRD EQUATION IS THE SUM OF THE
OTHER TWO!!
FIND THE VOLTAGES Vae ,Vec
GIVEN THE CHOICE USE THE SIMPLEST LOOP
DEPENDENT SOURCES ARE HANDLED WITH THE
SAME EASE
Vad  ______
10V
Vac  ______
Vac  4  6  0
6V
Vbd  ______
11V
Vbd  _______
Vbd  2  4  0
MUST FIND VR FIRST
 12  VR  1  10VR  0  VR  1V
1
1
1
1
DEPENDENT SOURCES ARE NOT REALLY
DIFFICULT TO ANALYZE
Veb  4  6  12  0
REMINDER: IN A RESISTOR THE VOLTAGE AND
CURRENT DIRECTIONS MUST SATISFY THE
PASSIVE SIGN CONVENTION
Vad  12  8  6  0

V
Vad  _______, Veb  ________

 V 
SAMPLE PROBLEM
 4V 

V1

b  Vx 

R  2k
+
-
+
-
V1  12V , V2  4V a
V2

DETERMINE
Vx 
4V
Vab  -8V
Power disipated on
the 2k resistor
P2k 
Remember
past topics
We need to find a closed path where only one voltage is unknown
FOR VX
VX  V2  V1  4  0
VX  4  12  4  0
VX  V2  Vab  0
Vab  VX  V2
5k
10k

25V

 Vx 
+
-
V1
There are no loops with only
one unknown!!!
Vx/2 +
+
-

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
4
VX VX
V1 

0
4
2
VX VX
 25[V ]  VX 

0
V
2
4
V1   X  5[V ]
4
VX  20[V ]