EECE 260
Exercise
• Find the equivalent resistance, REQ for the
following resistor combination circuit.
•
Solution
Exercise
Exercise
Solution
Circuit Terminologies
• Node: An electrical connection between two or more elements.
• Ordinary node: An electrical connection node that connects to only two elements.
• Extraordinary node: An electrical connection node that connects to three or more elements.
• Branch: Trace between two consecutive nodes with only one element
between them.
• Path: Continuous sequence of branches with no node encountered more than
once.
• Extraordinary path: Path between two adjacent extraordinary nodes.
• Loop: Closed path with the same start and end node.
• Independent loop: Loop containing one or more branches not contained in any
other independent loop.
• Mesh: Loop that encloses no other loops.
In series: Elements that share the same current. They have only ordinary nodes
between them.
In parallel: Elements that share the same voltage. They share two extraordinary
nodes.
1.10 For the circuit in Fig. P1.10:
(a) Identify and label all distinct nodes.
(b) Which of those nodes are extraordinary nodes?
(c) Identify all combinations of 2 or more circuit elements that are connected in series.
(d) Identify pairs of circuit elements that are connected in parallel.
(a) Nodes identified in Fig. P1.10(a).
(b) Nodes a and b are extraordinary.
(c) Series connections: 8 Ω and 6 Ω.
(d) Parallel connections: 12 V and 4 Ω
4 Ω and short circuit
(a) Nodes identified in Fig. P1.10(a).
(b) Nodes b and c are extraordinary.
(c) Series connections: 8 Ω and 6 Ω;, 12 V and 10 Ω
(d) Parallel connections: none
Kirchhoff’s Laws for DC Circuits
Kirchoff’s first
first law:
law: The
The sum
sum of
of the
the currents
currents
Kirchoff’s
entering aa junction
junction isis equal
equal to
to the
the sum
sum of
of the
the
entering
currents leaving
leaving that
that junction.
junction.
currents
Junction Rule: I (enter) = I (leaving)
Kirchoff’s
law:sum
The sum
of the
emf’s
Kirchoff’s
second second
law: The
of the
emf’s
around any
closed loop
must
theloop
summust
of the
IR the
drops
around that
around
anyequal
closed
equal
sum
same loop.
or the
sum ofaround
the product
the current
of the
IR drops
that of
same
loop. and resistance
in a closed mesh plus the sum of the e.m.f in that mesh is zero
Voltage Rule:
E = IR
 IR +  e.m. f = 0
Sign Conventions for Emf’s
▪ When applying Kirchoff’s laws you must
assume a consistent, positive tracing direction.
▪ When applying the voltage rule, emf’s are
positive if normal output direction of the emf is
with the assumed tracing direction.
▪ If tracing from A to B, this
emf is considered positive.
▪ If tracing from B to A, this
emf is considered negative.
A
+
B
E
+
A
E
B
Signs of IR Drops in Circuits
▪ When applying the voltage rule, IR drops are
positive if the assumed current direction is
with the assumed tracing direction.
▪ If tracing from A to B, this
IR drop is positive.
▪ If tracing from B to A, this
IR drop is negative.
A
I
+
B
+
A
I
B
Kirchhoff’s first law or Point law or Current
law (KCL)
I = 0
I1 + I 2 + I 3 − I 4 − I 5 = 0
Example
Find the value of I from the circuit in the figure below.
Solution:
0.2 A – 0.4 A + 0.6 A – 0.5 A + 0.7 A – I = 0
1.5 A – 0.9 A – I = 0
0.6 A – I = 0
I = 0.6A
example
Solution
We can calculate the current leaving
the point a (I) using Ohm’s law:
The current is entering the resistor
because the positive sign of the 2Ω
resistor is facing the point a. Thus, the
circuit becomes
Kirchoff’s Laws: Loop I
1. Assume possible consistent
flow of currents.
2. Indicate positive output
R1
directions for emf’s.
3. Indicate consistent tracing
direction. (clockwise)
Junction
Junction Rule:
Rule: II22 =
= II11 +
+ II33
Voltage
Voltage Rule:
Rule: E
E=
= IR
IR
EE11++EE22== II11RR11 ++ II22RR22
R3
+
I1
Loop I
R2
E2
E1
I2
I3
E3
Kirchoff’s Laws: Loop II
4. Voltage rule for Loop II:
Assume counterclockwise
positive tracing direction.
Voltage
Voltage Rule:
Rule: E
E=
= IR
IR
Bottom Loop (II)
R1
EE22++EE33=
=II22RR22 +
+ II33RR33
Would the same equation
apply if traced clockwise?
Yes!
-- EE22 -- EE33=
= -I-I22RR22 -- II33RR33
R3
I1
Loop I
R2
E2
E1
I2
I
Loop II
3
+
E3
Kirchoff’s laws: Loop III
5. Voltage rule for Loop III:
Assume counterclockwise
positive tracing direction.
Voltage
Voltage Rule:
Rule: E
E=
= IR
IR
Outer Loop (III)
R1
EE33–– EE11== -I-I11RR11 ++ II33RR33
Would the same equation
apply if traced clockwise?
Yes!
EE33-- EE11==II11RR11 -- II33RR33
R3
I1
Loop I
R2
E2
E1
I2
I
Loop II
3
+
E3
Four Independent Equations
6. Thus, we now have four
independent equations
from Kirchoff’s laws:
I2 = I1 + I3
Outer Loop (III)
R1
E1+ E2 = I1R1 + I2R2
E2 + E3 = I2R2 + I3R3
E3 - E1= -I1R1 + I3R3
R3
I1
Loop I
R2
E2
E1
I2
I
Loop II
3
+
E3
Example 4. Use Kirchoff’s laws to find the
currents in the circuit drawn to the right.
+
Junction
Junction Rule:
Rule: II22 +
+ II33==II11
I1 5 
Consider Loop I tracing
clockwise to obtain:
Voltage Rule: E = IR
Loop I 12 V
10 
12 V = (5 )I1 + (10 )I2
I2
Recalling that V/ = A, gives
55II11 +
+ 10
10II22 =
= 12
12 AA
I3
20 
6V
Example 5 (Cont.)
Finding the currents.
Consider Loop II tracing
clockwise to obtain:
Voltage Rule: E = IR
I1 5 
12 V
6 V = (20 )I3 - (10 )I2
10 
Simplifying: Divide by 2
and V/ = A, gives
I2
Loop II 20 
I3
10
10II33 -- 55II22 =
= 33 AA
6V
Example 5 (Cont.)
Three independent equations
can be solved for I1, I2, and I3.
(1)
(1) II22 +
+ II33=
= II11
(2)
(2) 55II11 +
+ 10
10II22 =
= 12
12 AA
(3)
(3) 10
10II33 -- 55II22=
=33 AA
I1 5 
12 V
10 
Substitute Eq.(1) for I1 in (2):
5(I2 + I3) + 10I3 = 12 A
Simplifying gives:
55II22 +
+ 15
15II33 =
= 12
12 AA
I2
Loop II 20 
I3
6V
Example 5 (Cont.)
Three independent
equations can be solved.
(1)
(1) II22 +
+ II33=
= II11
(3)
(3) 10
10II33 -- 55II22 =
= 33 AA
(2)
(2) 55II11 +
+ 10
10II22 =
= 12
12 AA
15
15II33+
+ 55II22== 12
12 AA
Eliminate I2 by adding equations above right:
Putting I3 = 0.6 A in (3) gives:
10I3 - 5I2 = 3 A
10(0.6 A) – 5I2 = 3 A
15I + 5I = 12 A
3
2
25I3 = 15 A
I3 = 0.600 A
II22=
= 0.600
0.600 AA
Then from (1):
II11=
= 1.20
1.20 AA
Summary of Formulas:
Rules
Rules for
for aa simple,
simple, single
single loop
loop circuit
circuit
containing
containing aa source
source of
of emf
emf and
and resistors.
resistors.
Re = R
Current :
I = E
R
Voltage Rule:
E = IR
C
Single Loop
2
3
- 3V+
A
-
18
V
+
Resistance Rule:
D
B
Summary (Cont.)
For resistors connected in series:
For
For series
series
connections:
connections:
II =
= II11=
= II22=
= II33
VVTT=
= VV11 +
+ VV22 +
+ VV33
RRee=
= RR11+
+ RR22+
+ RR33
RRee =
= R
R
2
3 1
12 V
Summary (Cont.)
Resistors connected in parallel:
For
For parallel
parallel
connections:
connections:
VV =
= VV11=
= VV22=
= VV33
IITT=
= II11 +
+ II22 +
+ II33
1 = N 1

Re i=1 Ri
Parallel Connection
R1
R2
R3
R1R2
Re =
R1 + R2
VT
2
12 V
4
6
Summary Kirchoff’s Laws
Kirchoff’s first
first law:
law: The
The sum
sum of
of the
the currents
currents
Kirchoff’s
entering aa junction
junction isis equal
equal to
to the
the sum
sum of
of the
the
entering
currents leaving
leaving that
that junction.
junction.
currents
Junction Rule: I (enter) = I (leaving)
Kirchoff’s
law:sum
The sum
of the
emf’s
Kirchoff’s
second second
law: The
of the
emf’s
around any
closed loop
must
theloop
summust
of the
IR the
drops
around that
around
anyequal
closed
equal
sum
same loop.
or the
sum ofaround
the product
the current
of the
IR drops
that of
same
loop. and resistance
in a closed mesh plus the sum of the e.m.f in that mesh is zero
Voltage Rule:
E = IR
Common DC Circuit Theory Terms:
Circuit – a circuit is a closed loop
conducting path in which an electrical
current flows.
Path – a single line of connecting elements or
sources.
Mesh – a mesh is a single open loop that does
not have a closed path. There are no
components inside a mesh.
Branch – a branch is a single or group of
components such as resistors or a source
which are connected between two
nodes.
Node – a node is a junction, connection
or terminal within a circuit were two or
more circuit elements are connected or
joined together giving a connection
point between two or more branches.
Loop – a loop is a simple closed path in a A node is indicated by a dot.
circuit in which no circuit element or
node is encountered more than once.
Kirchhoff’s second law or Mesh law or
Voltage law (KVL).
 IR +  e.m. f = 0
Determination of Sign
• In applying Kirchhoff’s laws to specific problems, particular
attention should be given to the algebraic signs of voltage drops
and e.m.f.s; otherwise the results will be wrong.
• 1. arise in voltage is +ve
• 2. fall in voltage is -ve
• Note: the sign of the battery is independent of the direction of
the current through the branch. Current flows from higher
potential to lower potential.
Determination of Sign
• Consider the closed path
ABCDA in the figure below
• -I1R1
• -I2R2
• +I3R3
• -I4R4
• +E1
• -E2
− I1 R1 − I 2 R2 + I 3 R3 − I 4 R4 − E2 + E1 = 0
I1 R1 + I 2 R2 − I 3 R3 + I 4 R4 = E1 − E2
Kirchhoff's Voltage Law – Series Circuit
VTotal + (− IR1 ) + (− IR2 ) = 0
VTotal = IR1 + IR2 , VTotal = IRTotal
IRTotal = IR1 + IR2
(“I” Cancels)
n
RTotal = R1 + R2 → RTotal( series ) =  Ri
i
Here we see that applying Kirchhoff's Voltage Law to this loop produces the
formula for the effective resistance in a series circuit. The word effective or
equivalent means the same thing as the TOTAL.
Kirchhoff’s Current Law
“The sum of the currents flowing into a junction is equal to the sum of
the currents flowing out.”
When two resistors have BOTH ends connected
together, with nothing intervening, they are
connected in PARALLEL. The drop in potential
when you go from X to Y is the SAME no
matter which way you go through the circuit.
Thus resistors in parallel have the same potential
drop.
Examples
• Find the current flowing in the R3
solution
• The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.
• Using Kirchhoff’s Current Law, KCL the equations are given as:
• At node A : I1 + I2 = I3
• At node B : I3 = I1 + I2
• Using Kirchhoff’s Voltage Law, KVL the equations are given as:
• Loop 1 is given as : 10 = R1 I1 + R3 I3 = 10I1 + 40I3
• Loop 2 is given as : 20 = R2 I2 + R3 I3 = 20I2 + 40I3
• Loop 3 is given as : 10 – 20 = 10I1 – 20I2
• As I3 is the sum of I1 + I2 we can rewrite the equations as;
• Eq. No 1 : 10 = 10I1 + 40(I1 + I2) = 50I1 + 40I2
• Eq. No 2 : 20 = 20I2 + 40(I1 + I2) = 40I1 + 60I2
• “Simultaneous Equations” can be reduced to give us the values of I1 and I2
• Substitution of I1 in terms of I2 gives us the value of I1 as -0.143 Amps
• Substitution of I2 in terms of I1 gives us the value of I2 as +0.429 Amps
• As : I3 = I1 + I2
• The current flowing in resistor R3 is given as : -0.143 + 0.429 = 0.286 Amps
• and the voltage across the resistor R3 is given as : 0.286 x 40 = 11.44 volts
Applying Kirchhoff’s Laws
Goal: Find the three unknown currents.
First decide which way you think the
current is traveling around the loop. It
is OK to be incorrect.
Red Loop → V + (− I 3 6) + (− I1 4) = 0
24 = 6 I 3 + 4 I1
Using Kirchhoff’s Voltage Law
Blue Loop → V + (− I 2 2) + (− I 3 6) = 0
12 = 2 I 2 + 6 I 3
I1 + I 2 = I 3
Using Kirchhoff’s Current Law
Applying Kirchhoff’s Laws
24 = 6 I 3 + 4 I1
12 = 2 I 2 + 6 I 3
I 3 = I1 + I 2
24 = 6( I1 + I 2 ) + 4 I1 = 6 I1 + 6 I 2 + 4 I1 = 10 I1 + 6 I 2
12 = 2 I 2 + 6( I1 + I 2 ) = 2 I 2 + 6 I1 + 6 I 2 = 6 I1 + 8I 2
24 = 10 I1 + 6 I 2 → −6(24 = 10 I1 + 6 I 2 )
12 = 6 I1 + 8 I 2 → 10(12 = 6 I1 + 8 I 2 )
− 144 = −60 I1 − 36 I 2
− 24 = 44 I 2
I 2 = -0.545 A
120 = 60 I1 + 80 I 2
A NEGATIVE current does NOT mean you are wrong. It means
you chose your current to be in the wrong direction initially.
Applying Kirchhoff’s Laws
12 = 2 I 2 + 6 I 3 → 12 = 2(−0.545) + 6 I 3
I 3 = 2.18 A
24 = 6 I 3 + 4 I1 → 24 = 6(?) + 4 I1
I1 = 2.73 A
Instead of :
I 3 = I1 + I 2
It should have been : I1 = I 2 + I 3
2.73 = 2.18 + 0.545
Find the current through each resistor and
the current drawn from the battery.
1. Draw and label the current
flow in the circuit
2. Apply Kirchoff’s Rules
12
I3
I
I2
2
4
Loop1
+ 12 − V2 − V4 = 0
12 − I 2 R2 − I 2 R4 = 0
12 − 2 I 2 − 4 I 2 = 0
I2 = 2A
12 V
Kirchoff’s Junction I = I 2 + I 3
Rule
= 2 +1 = 3A
Kirchoff’s
Loop
Rule
Loop2
+ 12 − V12 = 0
12 − I 3 R12 = 0
12 − 12 I 3 = 0
I 3 = 1A
For the circuit below, all resistors have the same value, R (10 )
a) Draw and label the current flow
b) Calculate the equivalent resistance of the circuit
c) Calculate the total current provided by the battery.
For the circuit below, all resistors have the same value, R (10 )
a) Draw and label the current flow
b) Calculate the equivalent resistance of the circuit
c)
Calculate the total current provided by the battery.
Req = 2 8/11 R = 27.3
I = 3A
Find the currents in the circuit.
b
• I1 = 0.381 A
• I2 = -0.814 A
• I3 = 1.20 A
• Find the currents in the circuit.
Source Transformation
An electrical source transformation is a method for simplifying circuits by replacing a
voltage source with its equivalent current source, or a current source with its equivalent
voltage source. We transform a current source into a voltage source by using ohm's law. A
voltage source can be changed into a current source by using ohm's formula, V= IR.
Example
1. Transform the circuit from current source into a voltage source
Solution: V= IR, which is V= 2A x 3Ω = 6V.
2. Transform the circuit from voltage source into a current source
Solution: .
I=
V 20
=
= 4A
R 5
Exercise
Solution
For the two loops, KVL gives
25 – 15I1 - 10 + 3I2 – 3I1 = 0 20 312 + 31, - 2l2 = 0
which when solved yield 12 = 5
A and Vab = (5)(2) = 10 V.
Solution
NODAL ANALYSIS
Nodal Analysis or Node Voltage Mode is a systematic method used for analyzing circuits using
node voltage as circuit variables. In simple words, this method is used for determining the
voltage (potential difference) between nodes. Nodes are the points where branches or elements
connect with each other.
• Procedure of Nodal Analysis
• The following steps are to be followed while solving any electrical circuit using nodal analysis:
• Step 1: To identify the principal nodes and select one of them as a reference node. This
reference node will be treated as the ground.
• Step 2: All the node voltages with respect to the ground from all the principal nodes should be
labelled except the reference node.
• Step 3: The nodal equations at all the principal nodes except the reference node should have a
nodal equation. The nodal equation is obtained from Kirchhoff’s current law and then from
Ohm’s law.
• Step 4: To obtain the node voltages, the nodal equations can be determined by following Step 3.
• Hence, for a given electrical circuit, the current flowing through any element and the voltage
across any element can be determined using the node voltages.
Types of Nodes in Nodal Analysis
There are two types of nodes in nodal analysis:
• Non-reference node
• Reference node
• Non-Reference Node
• The node with a definite node voltage is a non-reference node.
• Reference Node
• The node that acts as a reference point to all the other nodes is known as
the reference node, which is also known as the datum node.
There are two types of reference nodes, and
they are:
•Chassis ground
A reference node that acts as a common
node for more than one circuit is called
chassis ground.
•Earth ground
In any circuit where the earth’s potential
is used as a reference, it is known as the
earth’s ground.
Properties of Nodal Analysis
• The properties of nodal analysis have been stated in these points:
• Nodal analysis is an application of Kirchhoff’s current law, used for
the calculation of voltage.
• If there are ‘n’ number of nodes present in the given circuit, then there
will be ‘n-1’ number of equations formed to solve. For example, if
there are 10 nodes, 10-1 = 9 number of equations are required to solve.
• The number of non-reference nodes is equal to the number of nodal
equations to be solved. For example, if there are 10 nodes, there will
be 1 reference node and 10-1=9 non-reference nodes which is equal to,
10-1 = 9 number of equations are required to solve.
Super Node
• What is a Super Node?
• The super node is defined as a voltage source connected between the two
non-reference nodes such that these two nodes form a generalized node.
• Properties of Super Node
• The following are the properties of super nodes:
• The difference between the voltage of two non-reference nodes can be
determined at the super node.
• A super node does not have its own voltage.
• Kirchhoff’s current law and Kirchhoff’s voltage law are both applied for
solving the super node.
super node
Whenever there is a voltage source connected between two unknown
voltage or non-reference nodes, the two nodes are joined to form a
generalized node, which is known as a super node.
In the above image, the 5V
source is connected between two
non-reference nodes that are
node 2 and node 3. Therefore,
node 2 and node 3 combine to
form a super node.
Properties of Super node
• Some key properties of super node are discussed below.
• A super node itself does not have a well-defined voltage.
• To solve for the voltages within a super node, both Kirchhoff's Current
Law (KCL) and Kirchhoff's Voltage Law (KVL) must be applied.
• Any circuit element, such as a resistor, can be placed in parallel within a
super node since the nodes are effectively shorted together.
• KCL is satisfied at a super node just like at a regular node.
• Within a super node, the voltage difference between any two nodes is
zero.
Examples
• Example 1. Apply analysis to find node voltage V in the following
circuit.
Solution 1
• Example 2. Use analysis to compute the voltage across the 18 A
current source in the circuit of this figure:
Solution 2
• By replacing the parallel
conductance with their
equivalents, the circuit simplifies
as follows:
Applying analysis at Nodes 1, 2, and 3 we get:
Node 1: 16v1−12v2=12
Node 2: −12v1+27v2–15v3=−18
Node 3: −15v2+21v3=24
On simplifying the equations above, we get:
4v1–3v2=3
…1
−4v1+9v2–5v3=−6
…2
−5v2+7v3=8
…3
Add the equation 1 and 2 and group with 3.
We get:
6v2–5v3=−3
−5v2+7v3=8
For the problem, v2 = v_{18A}