KIRCHHOFF’S CURRENT LAW, JUNCTION RULE,

1. SILVER OAK COLLEGE OF
ENGINEERING AND
TECHNOLOGY
TOPIC :- KIRCHHOFF’S CURRENT LAW
STUDENT NAME :- SIDDHI SHRIVAS
(130770107163)
DIVISON :- COMPUTER – C
GUIDED BY :-MR. BIKAS MUDULI, SOCET
1

2. KIRCHHOFF’S
CURRENT LAW (KCL)
This law is also called Kirchhoff's first law, Kirchhoff's
point rule, or Kirchhoff's junction rule (or nodal rule).
This fundamental law results from the conservation of
charge
2

3. KIRCHHOFF’S CURRENT LAW
IT STATES THAT :
“At any node (junction) in
an electrical circuit, the sum
of currents flowing into that node is
equal to the sum of currents flowing out
of that node, or The algebraic sum of
currents in a network of conductors
meeting at a point is zero.” 3

4. WHAT IS A JUNCTION OR NODE
4

5. A junction is an intersection of three or
more conductors.
5

6. KIRCHHOFF'S RULES: JUNCTION RULE
The algebraic sum of the currents into any junction is zero.
6
• I = 0 ( at any junction )
• I is + when it enters a junction.
• I is - when it enters a junction.
• I = 0
• I1 + I2 - I3 = 0
• I1 + I2 = I3

7. KIRCHHOFF'S CURRENT LAW
 It also states that the sum of the currents
entering a node must equal the sum of the
currents leaving a node. This law is a
consequence of the conservation of charge
(electrons) in electrical networks.
7I1 + I2 = I3 + I4 + I5

8. EXPLANATION OF
KIRCHHOFF'S LAW
WITH EXAMPLE
8

9. SOME FUNDAMENTAL ASSUMPTIONS – NODES
 A node is defined as a place
where two or more components
are connected.
 The key thing to remember is that
we connect components with
wires. It doesn’t matter how
many wires are being used; it
only matters how many
components are connected
together.
+
-
vA
RC
RD
iB
RF
RE
9

10. HOW MANY NODES?
 To test our
understanding of
nodes, let’s look at
the example circuit
schematic given here.
 How many nodes are
there in this circuit?
+
-
vA
RC
RD
iB
RF
RE
10

11. HOW MANY NODES – CORRECT ANSWER
 In this schematic, there
are three nodes. These
nodes are shown in
dark blue here.
 Some students count
more than three nodes
in a circuit like this.
When they do, it is
usually because they
have considered two
points connected by a
wire to be two nodes.
+
-
vA
RC
RD
iB
RF
RE
11

12. HOW MANY NODES – WRONG ANSWER
 In the example circuit
schematic given here, the
two red nodes are really
the same node. There
are not four nodes.
 Remember, two nodes
connected by a wire were
really only one node in the
first place.
+
-
vA
RC
RD
iB
RF
RE
Wire connecting two nodes
means that these are really a
single node.
12

13. SOME FUNDAMENTAL ASSUMPTIONS – CLOSED
LOOPS
 A closed loop can be defined
in this way: Start at any
node and go in any direction
and end up where you start.
This is a closed loop.
 Note that this loop does not
have to follow components.
It can jump across open
space. Most of the time we
will follow components, but
we will also have situations
where we need to jump
between nodes that have no
connections.
+
-
vA
RC
RD
iB
RF
RE
vX
+
-
13

14. HOW MANY CLOSED LOOPS
 To test our
understanding of
closed loops, let’s
look at the
example circuit
schematic given
here.
 How many closed
loops are there in
this circuit?
+
-
vA
RC
RD
iB
RF
RE
vX
+
-
14

15. HOW MANY CLOSED LOOPS –
AN ANSWER
 There are several closed
loops that are possible here.
We will show a few of them,
and allow you to find the
others.
 The total number of simple
closed loops in this circuit is
13.
 Finding the number will not
turn out to be important.
What is important is to
recognize closed loops when
you see them.
+
-
vA
RC
RD
iB
RF
RE
vX
+
-
15

16. +
-
vA
RC
RD
iB
RF
RE
vX
+
-
CLOSED LOOPS – LOOP #1
 Here is a loop we will
call Loop #1. The
path is shown in red.
16

17. +
-
vA
RC
RD
iB
RF
RE
vX
+
-
 Here is Loop #2. The
path is shown in red.
CLOSED LOOPS – LOOP #2
17

18. +
-
vA
RC
RD
iB
RF
RE
vX
+
-
CLOSED LOOPS – LOOP #3
 Here is Loop #3. The
path is shown in red.
 Note that this path is
a closed loop that
jumps across the
voltage labeled vX.
This is still a closed
loop.
18

19. +
-
vA
RC
RD
iB
RF
RE
vX
+
-
CLOSED LOOPS – LOOP #4
 Here is Loop #4. The
path is shown in red.
 Note that this path is a
closed loop that jumps
across the voltage
labeled vX. This is still a
closed loop. The loop
also crossed the current
source. Remember that
a current source can
have a voltage across
it.
19

20. +
-
vA
RC
RD
iB
RF
RE
vX
+
-
A NOT-CLOSED LOOP
 The path is shown in
red here is not
closed.
 Note that this path
does not end where it
started.
20

21. CURRENT POLARITIES
Here, the issue of the
sign, or polarity, or direction, of
the current arises. When we
write a Kirchhoff Current Law
equation, we attach a sign to
each reference current
polarity, depending on whether
the reference current is
entering or leaving the closed
surface. This can be done in
different ways.
21

22. KIRCHHOFF’S CURRENT LAW (KCL)
– AN EXAMPLE
 For this set of material, we will
always assign a positive sign
to a term that refers to a
current that leaves a closed
surface, and a negative sign to
a term that refers to a current
that enters a closed surface.
 In this example, we have
already assigned reference
polarities for all of the currents
for the nodes indicated in
darker blue.
 For this circuit, and using my
rule, we have the following
equation:
+
-
vA
RC
RD
iB
RF
RE
iA
iB
iC
iE
iD
0A C D E Bi i i i i     
22

23. KIRCHHOFF’S CURRENT LAW (KCL) –
EXAMPLE DONE ANOTHER WAY
 Some prefer to write this same
equation in a different way;
they say that the current
entering the closed surface
must equal the current leaving
the closed surface. Thus, they
write :
+
-
vA
RC
RD
iB
RF
RE
iA
iB
iC
iE
iD
0A C D E Bi i i i i     
A D B C Ei i i i i   
• Compare this to the
equation that we wrote in the
last slide:
• These are the same
equation. Use either method.
23

24. Example
Find the current I x.
4 A
2 A
-1 A 6 A
IX
9 A
Ans: IX = 22 A22 24

25. -8 A
-3 A
-5 A
-2 A
TRY BY YOURSELF
Example
Find the currents IW, I X, IY, IZ.

 
12 A
IX IY
IZ
6 A
9 A
2 A
IW
IW =
IX =
IY =
IZ =
25

26. Example
Find the currents IA, IB, and IC in the circuit below.
surface
1
surface
2
-2A
4A
IB IC
9A
2A IA
26

27. Solution for the above Example
surface
1
surface
2
-2A
4A
IB IC
9A
2A IA

node 1 node 2
At surface 1: IB = 2A
At node 1: Ic = 0A
At node 2: IA = 9A
27