2. Kirchhoff’s Current Law
Statement of Kirchhoff’s current law at a node:
At any node the algebraic sum of the ammeter readings equals zero.
where ij(t) is the jth ammeter reading and N is the number of branches
connected to the node.
In mathematical form the law appears as
N
j
j t
i
1
0
)
(
Note that since we are free to connect the ammeters in any direction we
choose, the sign of the current ij(t) will depend on the reference direction
(i.e. ammeter connections) chosen for it, and which way we define as
positive and which way as negative.
i1
i2
i3
Figure 1.24
3. Kirchhoff’s current law at a node(continued)
To understand the use of this law, consider the node shown
in Figure 1.25, where we have represented ammeters with
arrows on the schematic diagram. The reference direction for
the ammeters( i.e. the direction of the arrows) follows the
conventional view of current flow, that is, current flows into
the ammeter via the positive terminal and leave via the
negative terminal.
i1(t)
i2(t)
i3(t)
i4(t)
i5(t)
Thus, if we define the positive direction for the arrow
as pointing into the node and the negative direction
as pointing outward from the node, then we can
write Kirchhoff’s current law for the node as
0
)
(
)
(
)
(
)
(
)
( 5
4
3
2
1
t
i
t
i
t
i
t
i
t
i
Figure 1.25
4. Kirchhoff’s current law at a node (continued)
Alternatively, if we define the positive direction for
the arrow as pointing outward from the node and
the negative direction as pointing into the node,
then we can write Kirchhoff’s current law for the
node as
i1(t)
i2(t)
i3(t)
i4(t)
i5(t)
Figure 1.25
5
1
5
4
3
2
1 0
)
(
)
(
)
(
)
(
)
(
)
(
j
j t
i
t
i
t
i
t
i
t
i
t
i
5. Kirchhoff’s current law at a node (continued)
Example
Solve for the current i4(t) in Figure 1.26.
i1(t) = 5 A
i2(t) = 2 A
i3(t) = 3 A
i4(t)
4
1
4
3
2
1 0
)
(
)
(
)
(
)
(
)
(
j
j t
i
t
i
t
i
t
i
t
i
Figure 1.25
Solution
Defining current into the nodes as positive,
Substituting values for i1, i2, and
i3 and then solving for i4, we
obtain
0
3
2
5 4
i
or
i4 = 6 A
6. Note
Although the directions of the currents were given in the
example, there are times when they will not be specified.
In that case we simply choose the directions and solve
the necessary KCL equations. Should a mathematical
solution indicate an unknown current with a minus sign,
the unknown current has a direction opposite to that
initially chosen.
Kirchhoff’s current law at a node (continued)
7. It is possible to generalise Kirchhoff’s current
law to include a closed surface. By a closed
surface we mean some set of elements
completely contained with the surface that
are interconnected. Since the current
entering each element within the surface is
equal to that leaving the element (i.e., the
element stores no net charge), it follows that
the current entering an interconnection of
elements is equal to that leaving any surface
enclosing that interconnection. Therefore,
Kirchhoff’s current law can also be stated:
Kirchhoff’s current law for a surface
4 A
i1
2 A
2 A 8 A
3 A
i2
Closed
surface
Figure 1.26
The algebraic sum of currents entering any closed surface is zero.
8. 0
2
3
8
4
2 2
i
To illustrate this generalisation of Kirchhoff’s
current law stated earlier, we need only
consider the multiple-node arrangement
shown in Figure 1.26.
Now we apply Kirchhoff’s current law to the
surface. Assuming that currents entering the
surface are positive and those leaving are
negative, we can write
Kirchhoff’s current law for a surface (continued)
4 A
i1
2 A
2 A 8 A
3 A
i2
Closed
surface
Figure 1.26
Thus,
A
9
2
i
Note that we did not need to solve for i1 to determine i2.
9. Exercise
Find the currents i1, i2, and i3 in the network in Figure 1.26
Kirchhoff’s current law for a surface (continued)
Figure 1.26
i1
2 A
i3
-2 A
1 A
i2
4 A
4 A 2 A
3A
4 A
Ans: i1 = - 3 A, i2 = - 2 A, i3 = 3 A.
10. Kirchhoff’s Voltage Law
Kirchhoff’s voltage law results from the law of conservation of energy
which requires that the work done must be equal to the energy put in.
Kirchhoff’s law for voltage can be stated as follows:
In any closed loop the algebraic sum of the voltmeter readings
equals to zero.
Kirchhoff’s voltage law can is stated mathematically as
N
j
j t
v
1
0
)
(
where N is the number of voltmeter readings in the loop.
11. Kirchhoff’s voltage law (continued)
Notes
1. The summation of the loop voltages is an algebraic one; that is the
polarities of the voltages must be taken into account in the summation
process. In this course, the sign of a voltage term in the equation is
determined by the sign of the reference polarity that we first encounter
for the voltage as we traverse the loop in one given direction. The sign
of the voltage term will be the opposite if the loop is traversed the other
way around.
2. If there are n elements then n voltages occur and n-1 values must be
known in order to use KVL to find the other value of voltage.
12. Kirchhoff’s voltage law (continued)
Example
Find the voltage across element # 3 in Figure x.
2 V
5 V
A
#1
#2
#3
13. Kirchhoff’s voltage law (continued)
Solution
In order to apply KVL we must:
2 V
5 V
A
#1
#2
#3
1. Assign a voltage across each element
where voltage is unspecified. Which
terminal we assume to be positive is
arbitrary.
2. Choose a direction to traverse the closed
path and mark this on the drawing. (It is
usually convenient and practical to traverse
all paths in the clockwise direction.
14. Kirchhoff’s voltage law (continued)
2 V
5 V
A
#1
#2
#3
3. As we start and traverse the path in the
direction chosen, enter the voltages for
which we encounter the positive terminal
first as positive voltages in our equation,
enter all values for which we encounter the
negative terminal first as negative.
15. Kirchhoff’s voltage law (continued)
Solution
In Figure x the direction has been changed
to counterclockwise (ccw). Again, starting at
point A and using KVL we obtain
v – (5 V) – (2 V) = 0
or
v = 7 V
2 V
5 V
v
A
16. Kirchhoff’s voltage law (continued)
Solution
Finally, in Figure x the connection of the
voltmeter measuring the voltage v is reversed.
Starting at A and going clockwise, KVL gives
us
(2 V) + (5 V) + v = 0
or
v = - 7 V
2 V
5 V
v
A
17. Kirchhoff’s voltage law (continued)
Example involving the arrow notation
For the circuit shown in Figure x, find v1 and v2.
# 1
# 2
# 3
# 4
14 mV
21 mV
v1
- 8 mV
A
v2
18. Example involving the arrow notation (continued)
Solution
To apply KVL to this problem we recall that in the arrow notation
the arrowhead indicates where the (+) terminal of the measuring
voltmeter is connected, and the tail indicates where the (-)
terminal is connected. Thus, starting at point A and going
clockwise around the perimeter of the circuit, KVL yields
- (21 mV) – (- 8 mV) + v1 + (14 mV) = 0
or
v1 = - 1 mV
19. Example involving the arrow notation (continued)
To find v2, we can sum voltages in the loop involving elements 1
and 2, or alternatively, we can sum voltages in the loop involving
elements 2 and 3. Thus, in the former, starting at point A and
going clockwise, KVL yields
v2 + v1 + (14 mV) = 0
or
v2 = - v1 – (14 mV)
= - (- 1 mV) – (14 mV)
= - 13 mV
20. Example involving the arrow notation (continued)
For the latter, starting at point A and going clockwise, KVL yields
- (21 mV) – (- 8 mV) – v2 = 0
or
v2 = - (21 mV) + (8 mV)
= - 13 mV
21. Exercise on use of the double subscript notation
For the circuit shown in x, find vAF, vAD, vBC, and vAC, given that
vDF = 10 V, vDE = 6 V
vBE = 5 V, vAE = - 2 V
and, vFC = 4 V.
A
E
F
B
C
D
Figure 2
22. Solution
i. To find vAF, consider the loop EAFD. Starting at point E and
traversing the loop clockwise, application of KVL to the loop
leads to the equation
vEA + vAF + vFD + vDE = 0
Using the fact that vEA = - vAE, vFD = - vDF, we can rewrite
the above equation as
-vAE + vAF - vDF + vDE = 0
Substituting the given voltages into the above equation, we
obtain
- ( - 2 V) + vAF – (10 V) + ( 6 V) = 0
or
vAF = 2 V
23. ii. To find vAD, consider the loop DEA. Starting at point E and traversing
the loop clockwise, application of KVL to the loop leads to the
equation
vEA + vAD + vDE = 0
Using the fact that vEA = - vAE we can rewrite the above equation as
-vAE + vAD + vDE = 0
Substituting the given voltages into the above equation, yields
- ( - 2 V) + vAD + ( 6 V) = 0
giving
vAD = - 8 V
24. iii. To find vBC, consider the loop BCFAE. Starting at point B and
traversing the loop clockwise, application of KVL to the loop leads to
the equation
vBC + vCF + vFA + vAE + vEB = 0
Using the fact that vCF = - vFC , vFA = - vAF, and vEB = - vBE, we can
rewrite the above equation as
vBC - vFC - vAF + vAE - vBE = 0
Substituting the known voltages into the above equation, yields
vBC – (4 V) – (2 V) + (- 2 V) – ( 5 V) = 0
giving
vBC = 13 V
25. The device in Figure 5 consists of five circuit elements. Use KVL to
find the unknown voltage drop for each unspecified element.
Exercise
26. Exercise
For the circuit shown in Figure 3, use KCL and KVL to determine i1, i2,
vAD, and vX.
3 V
20 V
8 V
E A
D
C
B
10 A 8 A
vX
i2 i1
- 3 A
vAE = - 5 V
27. Power Balance Equation
Conservation of Energy Law
Conservation of energy requires that the power absorbed by an
element be supplied from some other source: perhaps chemical energy
from a battery or mechanical energy driving a generator. This
observation gives us the power balance equation:
Power Balance Equation
The algebraic sum of the instantaneous electric power
absorbed by all components in a network is zero:
0
1
n
i
i
p
where the sum is over all the components in the network, including sources.
28. where Pi is the average power of the ith component, we can write
0
1
0 1
dt
p
T
T n
i
i
The above expression can be rewritten as
0
1
1 0
n
i
T
idt
p
T
Integration of Eq.() over one period T and dividing the result by T, we
obtain
Since
i
T
i P
dt
p
T
0
1
The result says that the algebraic sum of the average electric power
absorbed by all components in a network also must be zero.
0
1
n
i
i
P
29. Example
Compute the power delivered by each of the element in the following
circuit and show that the algebraic sum of the power delivered is zero.
10 V - 4 V
2 A
6 V
# 2
# 1 # 3
30. Solution
The 2 A current enters element #1 via its + terminal; hence, the power
delivered by Element #1 is
10 V - 4 V
2 A
6 V
# 2
# 1 # 3
p1 = vi = (10)(2) = 20 W
Since the 2 A current enters element #2 via its
+ terminal, the power delivered by Element #1
is
p2 = - vi = -(6)(2) = -12 W
For element #3 the 2 A current flows in via its – terminal; hence, the
power delivered by Element #3 is
p3 = vi = (-4)(2) = -8 W
31. Solution (continued)
10 V - 4 V
2 A
6 V
# 2
# 1 # 3
Algebraic sum of the power delivered,
0
)
8
(
)
12
(
20
3
2
1
3
1
p
p
p
p
i
i
32. Exercise
Is element E in the network in Figure x absorbing or supplying power, and
how much?
16 V
E
10 V 8 V
6 V
3 A
A
B C D
-9 A
33. Note
To solve this problem we need to assign first reference directions for the
voltage and current at element E before we can work out their numerical
values.
