Nodal Analysis.pptx

23ECE103 Fund of Electrical Engg 1
23ECE103:FUNDAMENTALS OF
ELECTRICAL ENGINEERING (3-0-0-3)
14 November 2023

Outline
• Objective
• Steps in Nodal Analysis
• Inspection method
• To solve
– DC circuits using Nodal analysis.
– AC circuits using Nodal analysis.
23ECE103 Fund of Electrical Engg 2
14 November 2023

Objective
14 November 2023 23ECE103 Fund of Electrical Engg 3
• To introduce Nodal analysis technique (Nodal Voltage method)
• To formulate Node equations
• To solve circuits using Nodal analysis
• To introduce nodal analysis by inspection method

Nodal Analysis
An analysis technique to solve electrical circuit where the node voltages are
used as the circuit variables
It involves systematic steps with an objective to solve the node voltages
Nodal Analysis method applies KCL to find the unknown voltages
With node voltages known, current in every branch can be calculated
6
2 7
1A 4A
Reference node
v1, v2 : node voltages
If v1 and v2 are known,
current in all branches
can be calculated

Nodal Analysis
Steps:
Step 1. Determine the reference node: typically the
one with the most branches (or ) bottom node
Step 2. Assign the rest of the nodes with node
voltages (referred to the reference node)
Step 3. Write down equations using KCL for every
non-reference node in terms of node voltages.
Step 4. Obtain the node voltages by solving the
simultaneous equations in step 3

Nodal Analysis- Example 1
6
2 7
1A 4A

Step 1. Determine the reference node: typically
the one with the most branches (or ) bottom
node
6
2 7
1A 4A

6
2 7
1A 4A
v1 v2

non-reference node in terms of node voltages
 if there are N nodes, there should be (N-1)
equations
6
2 7
1A 4A
v1 v2

 if there are N nodes, there should be (N-1)
equations
6
2 7
1A 4A
v1 v2
i1
i2
i3 i3
i4
i5

Step 3.
Apply KCL to Node 1
i1= i2 + i3
𝟏𝑨 =
(𝑽𝟏−𝟎)
𝟐
+
(𝑽𝟏−𝑽𝟐)
𝟔
---- (1)
6
2 7
1A 4A
v1 v2
i1
i2
i3 i3
i4
i5

Step 3.
Apply KCL to Node 2
i3 + i5 = i4
(𝑽𝟏−𝑽𝟐)
𝟔
+ −𝟒𝑨 =
(𝑽𝟐−𝟎)
𝟕
---- (2)
6
2 7
1A 4A
v1 v2
i1
i2
i3 i3
i4
i5

Step 4. Obtain the node voltages by solving the
simultaneous equations in step 3
𝟏𝑨 =
(𝑽𝟏−𝟎)
𝟐
+
(𝑽𝟏−𝑽𝟐)
𝟔
---- (1)
(𝑽𝟏−𝑽𝟐)
𝟔
+ −𝟒𝑨 =
(𝑽𝟐−𝟎)
𝟕
---- (2)
Rearrange and solve the equations,
V1=-2v & V2 = -14 V

Determine the current in every branch
9k 9k
3k
3k
4k
6k
+

12 V

node
9k 9k
3k
3k
4k
6k
+

12 V

Note: If a voltage source is placed near a node without resistance,
that node will have same voltage of voltage source.
9k 9k
3k
3k
4k
6k
+

12 V
v1 v2

9k 9k
3k
3k
4k
6k
+

12 V
v1 v2
i1
i2
i3 i4 i5

Step 3. KCL at node 1
𝐼1 = 𝐼2 + 𝐼3
(12−𝑉1)
9𝑘
=
(𝑉1−0)
6𝑘
+
(𝑉1−𝑉2)
3𝑘
---- (1)
9k 9k
3k
3k
4k
6k
+

12 V
v1 v2
i1
i2
i3 i4 i5

𝐼3 = 𝐼4 + 𝐼5
(𝑉1−𝑉2)
3𝑘
=
(𝑉2−0)
4𝑘
+
(𝑉2−0)
12𝑘
---- (2)
9k 9k
3k
3k
4k
6k
+

12 V
v1 v2
i1
i2
i3 i4 i5

Step 4:
(12−𝑉1)
9𝑘
=
(𝑉1−0)
6𝑘
+
(𝑉1−𝑉2)
3𝑘
---- (1)
(𝑉1−𝑉2)
3𝑘
=
(𝑉2−0)
4𝑘
+
(𝑉2−0)
12𝑘
---- (2)
Solving (1) & (2); v1=3V& V2=1.5V,
With these values find all branch currents.

v1=3V& V2=1.5V,
I1 = (12-V1)/9k = 9/9 =1 mA
I2= V1/6k = ½ mA
I3= (v1-V2) / 3k =1.5/3k =1/2 mA
9k 9k
3k
3k
4k
6k
+

12 V
v1 v2
i1
i2
i3 i4 i5

v1=3V& V2=1.5V,
I3= (v1-V2) / 3k =1.5/3k =1/2 mA
I4 = V2/4k = 1.5/4k = 4.5/12 mA
I5 =V2/12k = 1.5/12 mA
9k 9k
3k
3k
4k
6k
+

12 V
v1 v2
i1
i2
i3 i4 i5

Determine the unknown voltages
3kΩ

node
3kΩ

Step 2. Assign the rest of the nodes with node voltages
(referred to the reference node)
Note: If a voltage source is placed near a node without resistance, that
node will have same voltage of voltage source.
3kΩ

i1
i2 i4
3kΩ
i3
i5 i6

𝐼1 = 𝐼2
(12−𝑉1)
3𝑘
=
(𝑉1−𝑉2)
9𝑘
---- (1)
i1
i2 i4
3kΩ
i3
i5 i6

𝐼2 = 𝐼3 + 𝐼4
(𝑉1−𝑉2)
9𝑘
=
(𝑉2−0)
6𝑘
+
(𝑉2−𝑉3)
3𝑘
---- (2)
i1
i2 i4
3kΩ
i3
i5 i6

𝐼4 = 𝐼5 + 𝐼6
(𝑉2−𝑉3)
3𝑘
=
(𝑉3−0)
4𝑘
+
(𝑉3−0)
12𝑘
--- (3)
i1
i2 i4
3kΩ
i3
i5 i6

Step 4:
(12−𝑉1)
3𝑘
=
(𝑉1−𝑉2)
9𝑘
---- (1)
(𝑉1−𝑉2)
9𝑘
=
(𝑉2−0)
6𝑘
+
(𝑉2−𝑉3)
3𝑘
---- (2)
(𝑉2−𝑉3)
3𝑘
=
(𝑉3−0)
4𝑘
+
(𝑉3−0)
12𝑘
--- (3)
Solving (1), (2) & (3); v1=? ,V2=? & V3=?

Use Nodal analysis to find node voltages in the
given circuit. Also find the power drop across
2 ohm resistor.
Ans: V1=20V, V2=13.33V; P2ohms=?

given circuit. Also find V0.

given circuit. Also find V0.
Ans: V1=30V, V2=20V;
V0=?

given circuit.
1. How many node voltages need to be found
to solve the circuit?
a) 5
b) 2
c) 3
d) 4
2. What will be the voltage at the reference
node?
a) 0V
b) 12V
c) voltage across 20 ohms
d) voltage across 10 ohms
3. What is the equation of current through the 4
ohm resistor?

Nodal Analysis – Inspection method
Steps:
Step 1: First, convert all the voltage sources to equivalent current
sources.
Step 2: The conductances of all branches connected to node ‘i’ are
added and denoted by Gii. Gii is called the self conductance of node ‘i’.
Step 3: All the conductances connected to nodes ‘i’ and ‘j’ are added
and denoted by Gij. Gij is called mutual conductance of nodes ‘i’ and
‘j’. This Gij is written with negative sign. If no conductance is
connected between nodes ‘i’ and ‘j’ then Gij = 0, Gij = Gji.
Step 4: Ii denotes the value of the current source to node ‘i’ and is
written on the right hand side of the equation. The sign Ii is positive if it
is flowing towards node ‘i’, otherwise it is negative. If no current
source is connected to node ‘i’, then Ii = 0.

1
2
+
1
6
−
1
6
−
1
6
1
6
+
1
7
𝑉1
𝑉2
=
1
−4
6
2 7
1A 4A
v1 v2

1
2
+
1
6
−
1
6
−
1
6
1
6
+
1
7
𝑉1
𝑉2
=
1
−4
6
2 7
1A 4A
v1 v2
solve the equations,
V1=-2v & V2 = -14 V

2 ohm resistor.

1
4
+
1
6
−
1
4
−
1
4
1
4
+
1
2
𝑉1
𝑉2
=
5
5

2 ohm resistor.
Ans: V1=20V, V2=13.33V; P2ohms= V2
2 /R = ?

Nodal Analysis for AC circuits -
Example 1
If ZL = – j 2Ω, find VL in the circuit using nodal
analysis.

Example 1
Step 1: Let us mark nodes along with the node
voltages V1 and V2 in the given figure.
Step 2:
At node (1):
At node (2):
i.e.

Example 1
Comparing:

Example 1
To calculate V2
Load voltage

Example 2
Use Nodal Analysis to calculate V2

Example 2
Calculating the parallel branches of j3 and –j5 ohms, resulting j7.5 Ohms.

Example 2

Nodal Analysis.pptx

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