Department of Computer Science and Engineering (CSE)

BRAC University

Fall 2023

CSE250 – Circuits and Electronics

NODAL ANALYSIS

PURBAYAN DAS, LECTURER

Department of Computer Science and Engineering (CSE)
BRAC University
 Ground
• Except for a few special cases, electrical and electronic systems are grounded for
reference and safety purposes.
• It is called ground since it is assumed to have zero potential.
• In general, the placement of the ground connection will not affect the magnitude
or polarity of the voltage across an element, but it may have a significant impact
on the voltage from any point in the network to ground.
• A reference node is indicated by any of the four symbols.

Signal ground Common ground Earth ground Chassis ground

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 Problem 1
For the series network shown below, determine,
i) The voltage 𝑉𝑎.
Hint: A node voltage is the potential
ii) The voltage 𝑉𝑏. difference between the given node and the
iii) The voltage 𝑉𝑎𝑏 reference node (ground in this case).

(𝑖𝑖𝑖) 𝑽𝒂𝒃 = 𝟎 𝑽
(𝑖𝑖) 𝑽𝒃 = 𝟏𝟔 𝑽
Ans: (𝑖) 𝑽𝒂 = 𝟏𝟔 𝑽

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 Problem 2
For the series network shown below, determine,
𝑖) The voltage 𝑉𝑎.
𝑖𝑖) The voltages 𝑉𝑏 and 𝑉𝑐
𝑖𝑖𝑖) The voltage 𝑉𝑎𝑏

(𝑖𝑖𝑖) 𝑽𝒂𝒃 = 𝟓𝟔 𝑽
(𝑖𝑖) 𝑽𝒃 = −𝟑𝟐 𝑽; 𝑽𝒄 = − 𝟒𝟖 𝑽
Ans: (𝑖) 𝑽𝒂 = 𝟐𝟒 𝑽

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Course Outline: broad themes
Nodal
Analysis

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 Nodal Analysis: General Approach
• Nodal analysis provides a general procedure for analyzing circuits using node
voltages as the circuit variables. Nodal analysis applies KCL to find unknown
voltages in a given circuit.
• A node voltage is the potential difference between the given node and some other
node that has been chosen as a reference node.
• Remember that applying KCL to n-1
nodes produces n-1 variables and n-1
equations. As you will see, it is not
necessary to apply KCL to every node
in a circuit. So, being a little discreet
can significantly reduce the number of
variables. See an example.
• But first, we need to look at four
cases.
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 Nodal Analysis General Approach: steps
Step 1: Identify all the nodes and Step 5: Solve the simultaneous equations
place a ground You should get a number of equations equal to the
Recall that, 'Node' is a connection point of two node variables in Step 4. Solve using a calculator.
or more branches. Make a node as the
reference node. Appropriate placement of
ground may provide advantage.
Step 4: Apply KCL
Apply KCL to the nodes where a variable is assigned.
Follow the four cases. Look for Supernodes. No need
to apply if a node voltage is known already.
Step 2: Look for voltage sources
directly connected to ground
If a voltage source is connected from a node to
the ground, voltage of that node is equal to Step 3: Selectively assign node variables
the value of the voltage source. Careful about Assign node variables only to those remaining nodes
the polarity of the node voltage. where more than two branches are connected.

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 Example 1
Use nodal analysis to determine the voltage across the 3 𝐴 current source. What is
the power of it? Is it absorbing or supplying?
Before solving the circuit using nodal analysis,
remember that "Current flows from a higher
potential to a lower potential in a resistor." This
is true since resistor is a passive element, by
the passive sign convention, current must
always flow from a higher potential to a lower
potential.

We can express this principle as,

𝑣ℎ𝑖𝑔ℎ𝑒𝑟 − 𝑣𝑙𝑜𝑤𝑒𝑟
𝑖=
𝑅
However, do we know which voltage is the
higher one beforehand?

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 Case 1: resistor between nodes
There are four scenarios that we may encounter while writing currents in terms of
node voltages throughout the nodal analysis procedure. We will arbitrarily choose the
direction of the current flowing through a wire.
■ Case 1 In case of only a resistor connected between two nodes of voltages 𝑉𝑎 and
𝑉𝑏 , the current, assumed to be flowing in a particular direction, can be written as,

𝑉𝑎 − 𝑉𝑏 𝑉𝑏 − 𝑉𝑎
𝐼= 𝐼=
𝑅 𝑅

The actual direction of the current can be known after solving for the node voltages.
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 Case 2: current source between nodes
■ Case 2 In case of a current source connected between two nodes of voltages 𝑉𝑎 and
𝑉𝑏 , current flowing between the nodes will be equal to the current supplied by the
current source.

𝐼 = 𝑖𝑥 𝐼 = −𝑖𝑥
If any other elements are connected in series with a current source, the current
between the nodes will still be equal to the current supplied by the source.

𝐼 = 𝑖𝑥 𝐼 = 𝑖𝑥
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 Case 3: resistor & voltage source in series
between nodes
■ Case 3 In case of a resistor and a voltage source in series connected between two nodes
under consideration, the current, assumed to be flowing in a particular direction, can be
written as,

𝑉𝑎 − 𝑉𝑏 − 𝑉𝑠 𝑉𝑎 − 𝑉𝑏 + 𝑉𝑠
𝐼= 𝐼=
𝑅 𝑅

This is how we might perceive the scenario. We'll assume the current flows from 𝑉𝑎 to 𝑉𝑏 .
Given that voltage sources tend to produce power, we add 𝑉𝑠 with the term (𝑉𝑎 − 𝑉𝑏 ) in the
numerator if the current contributed by the source (indicated in black arrow) is in the same
direction (from 𝑉𝑎 to 𝑉𝑏 ), otherwise we deduct 𝑉𝑠 .
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 Case 4: voltage source between nodes
■ Case 4 Because Ohm's Law cannot be applied in the absence of a resistor, in the case of
a voltage source linked between two nodes, we don’t know the current of a voltage
source in advance. This is a unique case in which the condition is known as a Supernode.
This is handled differently, as demonstrated by an example later. We may still write KVL
equation as,

𝐼 =? 𝐼 =?

𝑉𝑎 − 𝑉𝑏 = 𝑉𝑠 𝑉𝑎 − 𝑉𝑏 = −𝑉𝑠

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 Example 1: General Approach (step 1)
Node 1 Node 2
First identify all the nodes in this circuit.
Recall that, A node is the point of connection
between two or more branches. A node is
an equipotential portion of a circuit. Node 3

There are 7 nodes as identified in the circuit.

Make one of the nodes as the reference
node. It is most convenient (not mandatory) Node 5 Node 6
Node 4
to choose the node that has the maximum
number of circuit elements connected to it.
Let's assign the node 7 as the reference Node 7
node. 0𝑉
Place a ground to the reference node.
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 Example 1: step 2 Node 1 Node 3

The 2nd step is to assign node

variables to the remaining nodes. Node 2

There are 6 nodes apart from the ground.

We don't need to apply KCL separately to Node 5
all the remaining nodes. Node 4 Node 6

One thumb rule is that, assign node

variable (apply KCL) to the nodes where
at least three or more branches are 0𝑉
connected, if the node voltage is not
already known.
This enables us to put the nodes 4, 5, and 6 out of consideration. Assign variables
𝑉1, 𝑉2, and 𝑉3 to the nodes 1, 2, and 3 respectively.
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 Example 1: step 3 𝒊𝟏
The 3rd step is to apply KCL separately 𝒊𝟐
to each of the nodes in consideration.
Applying KCL to the node 1 𝑉2
𝑉1 𝑉3
Let's add currents to all the wires (4 𝒊𝟑
wires) connected to node 1. The direction 𝒊𝟒
of the currents are taken arbitrarily.
According to the KCL,
𝑖1 + 𝑖2 + 𝑖4 = 𝑖3
0𝑉
Sum of currents Sum of currents
entering the node leaving the node

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 Example 1: step 3 (continued … 2)
𝒊𝟏
𝑖1 + 𝑖2 + 𝑖4 = 𝑖3
Now express the unknown currents in i2
terms of node voltages and resistances
using Ohm's law and recall the cases. 𝑉2
𝑉1 𝑉3
𝑉1 − 𝑉3 𝑉1 − 0 − 15 𝑉2 − 𝑉1
3+ + − =0 𝒊𝟑
10 20 5 𝒊𝟒

𝑖1 𝑖2 𝑖4 𝑖3
𝑐𝑎𝑠𝑒 2 𝑐𝑎𝑠𝑒 1 𝑐𝑎𝑠𝑒 3 𝑐𝑎𝑠𝑒 2
Simplifying the equation yields, 0𝑉
7𝑉1 − 4𝑉2 − 2𝑉3 = − 45 (𝑖)

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 Example 1: step 3 (continued … 3)
In a similar way, apply KCL to node 2

𝑉2 − 𝑉1 𝑉2 − 0 − 4𝑖0 𝑉2 − 𝑉3 𝑉2
+ + =0 𝑉1
5 5 5 𝑉3
where, all the currents are assumed to be
leaving the node 2 (arbitrary assumption)
Due to the gain (4𝑖0 ) of the dependent
source, the parameter 𝑖0 is present in the
equation. We need to replace 𝑖0 in terms
of the node voltages. 𝑖0 can be written as, 0𝑉
𝑉1 − 𝑉3
𝑖0 = [𝑠𝑒𝑒 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖0 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑑𝑖𝑎𝑔𝑟𝑎𝑚]
10
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 Example 1: step 3 (continued … 4)
𝑉 −𝑉
Replace 𝑖0 in the equation for node 2 by 110 3
𝑉1 − 𝑉3
𝑉2 − 𝑉1 𝑉2 − 4 10 − 0 𝑉2 − 𝑉3
+ + =0 𝑉2
5 5 5 𝑉1 𝑉3
Simplifying the equation yields,
7𝑉1 − 15𝑉2 + 3𝑉3 = 0 (𝑖𝑖)
Next, apply KCL to node 3,
𝑉3 − 𝑉2 𝑉3 − 𝑉1 𝑉3 − 0 + 10
+ + =3
5 10 15
Or, 0𝑉
3𝑉1 + 6𝑉2 − 11𝑉3 = −70 (𝑖𝑖𝑖)
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 Example 1: step 3 (continued … 5)
We have derived the three node equations
consisting of three variables.
7𝑉1 − 4𝑉2 − 2𝑉3 = − 45 𝑉2
7𝑉1 − 15𝑉2 + 3𝑉3 = 0 𝑉1 𝑉3
3𝑉1 + 6𝑉2 − 11𝑉3 = −70

Solving the three simultaneous equations

yields,
𝑉1 = −7.19 𝑉
𝑉2 = −2.78 𝑉 0𝑉
𝑉3 = 2.89 𝑉

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 Example 1: power of 3A source
Determining the voltage and power of the 3 𝐴 − 𝟏𝟎. 𝟎𝟖 (𝑽) +
source.
The voltage across the voltage 3 𝐴 source is
either 𝑉1 − 𝑉3 or 𝑉3 − 𝑉1 . With 𝑉3 > 𝑉1 , we
calculate the voltage as a positive quantity to be, 𝑉2
𝑉1 𝑉3
𝑉3 − 𝑉1 = 2.89 − −7.19 = 10.08 𝑉
The polarity of the voltage is such that 𝑉3 is at a
higher potential than 𝑉1 , as shown in the figure.
According to the passive sign convention, the
power supplied by the 3A source is thus, 0𝑉
𝑝 = −10.08 × 3 = −30.24 (𝑊𝑎𝑡𝑡)

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 Nodal Analysis: Format Approach
• Nodal analysis using Format approach allows to write nodal equations rapidly and in
a form that is convenient for the use of determinants.
• The first node equation from Example 1 can be written in this form,
𝑉1 − 𝑉3 𝑉1 − 0 − 15 𝑉2 − 𝑉1
3+ + − = 0 (𝑓𝑟𝑜𝑚 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 1)
10 20 5

1 1 1 15 𝑉2 𝑉3
𝑉1 + + − − − +3=0
20 5 10 20 5 10
• Note that, each node voltage variable is multiplied by the sum of the conductances
(reciprocal of 𝑅) attached to that node. Note also that the other nodal voltages
within the same equation are multiplied by the negative of the conductance
between the two nodes. The current sources are represented to the same side of the
equals sign with a positive sign if they leaves the node and with a negative sign if
they draw enter to the node. So, to summarize the procedure …
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 Format Approach: procedure
■ Steps
1. Choose a reference node and assign a subscripted voltage label to all the (𝑁 − 1)
remaining nodes of the network.
2. The number of equations required for a complete solution is equal to the number of
subscripted voltages (𝑁 − 1). Column 1 of each equation is formed by summing the
conductances (reciprocal of 𝑅) tied to the node of interest and multiplying the result by
that subscripted nodal voltage.
3. We must now consider the mutual terms, which, as noted in the preceding slide, are always
subtracted from the first column. It is possible to have more than one mutual term if the
nodal voltage of current interest has an element in common with more than one other
nodal voltage. This is demonstrated in an example to follow. Each mutual term is the
product of the mutual conductance and the other nodal voltage, tied to that conductance.
4. A current source is assigned a positive sign if it draws current from a node and negative sign
if it supplies current from the node.
5. Solve the resulting simultaneous equations for the desired voltages.
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 Example 1: Format Approach
Node 1, 𝑉1 Node 3, 𝑉3
Identify all the nodes and label them
(with ground being the 0th node).
Write the component equations for all Node 3
the voltage sources (voltage difference =
labeled variable). 𝑉2

𝑉4 = 15 𝑉 −−−− −(𝑖)
Node 5 𝑉5 𝑉6
Node 4 𝑉4 Node 6
𝑉1 − 𝑉3
𝑉5 = 4𝑖0 = 4 ×
10
 4𝑉1 − 4𝑉3 − 10𝑉5 = 0 −−−− −(𝑖𝑖)
Node 7
𝑉6 = −10 𝑉 −−−− −(𝑖𝑖𝑖) 0𝑉

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Example 1: Format Approach … (2)
Node equation formation.
𝑵𝒐𝒅𝒆 𝟏, 𝑽𝟏 : There are 3 resistors ( 20 ,
5 , 10 ) connected to 𝑉1. We write,
1 1 1 𝑽𝟏 𝑽𝟑
𝑉1 + + …=0 𝑽𝟐
20 5 10
The other end of the 20 , 5 , and 10 
resistors are connected to the nodes 𝑉4 , 𝑉2 , and
𝑉3 respectively. So, we subtract,
1 1 1 𝑉4 𝑉2 𝑉3 𝑽𝟒 𝑽𝟓 𝑽𝟔
𝑉1 + + − − − …=0
20 5 10 20 5 10
Finally, we subtract any currents entering to
that node (or add if leaving),
𝟎𝑽
1 1 1 𝑉4 𝑉2 𝑉3
𝑉1 + + − − − +3=0
20 5 10 20 5 10

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Example 1: Format Approach … (3)
Substituting 15 𝑉 for 𝑉4 from equation 𝑖 ,
1 1 1 15 𝑉2 𝑉3
𝑉1 + + − − − +3=0
20 5 10 20 5 10
1 1 1 𝑉3 𝑉2 9 𝑽𝟏 𝑽𝟑
 𝑉1 + + − − =− 𝑽𝟐
20 5 10 10 5 4
−−−−− −(𝑖𝑣)
Node 3, 𝑽𝟑 : Similarly, 10 , 5 , and 15 
resistors are connected between 𝑉3 and 𝑉1 , 𝑉3
and 𝑉2 , and 𝑉3 and 𝑉6 respectively. Also, the 3 𝐴 𝑽𝟒 𝑽𝟓 𝑽𝟔
current is entering to 𝑉3 .
1 1 1 𝑉1 𝑉2 𝑉6
𝑉3 + + − − − −3=0
10 5 15 10 5 15
Substituting −10 𝑉 for 𝑉6 from equation 𝑖𝑖𝑖 , 𝟎𝑽
1 1 1 𝑉1 𝑉2 7
𝑉3 + + − − = −−−− −(𝑣)
10 5 15 10 5 3
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Example 1: Format Approach … (4)
Node 2, 𝑽𝟐 : Similarly, three 5  resistors are
connected between 𝑉2 and 𝑉1 , 𝑉2 and 𝑉3 , and 𝑉2
and 𝑉5 respectively. So,
1 1 1 𝑉1 𝑉3 𝑉5 𝑽𝟏 𝑽𝟑
𝑉2 + + − − − =0
5 5 5 5 5 5 𝑽𝟐
From equation (𝑖𝑖),
4𝑉1 − 4𝑉3
𝑉5 =
10
Substituting for 𝑉5 from equation (𝑖𝑖), 𝑽𝟒 𝑽𝟓 𝑽𝟔
1 1 1 𝑉1 𝑉3 4𝑉1 − 4𝑉3
𝑉2 + + − − − =0
5 5 5 5 5 10 × 5
1 1 1 1 2 1 2
𝑉2 + + − 𝑉1 + − 𝑉3 − =0 𝟎𝑽
5 5 5 5 25 5 25
−−−−−−−− −(𝑣𝑖)
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Example 1: Format Approach … (5)
We got three equations with three variables.
1 1 1 𝑉3 𝑉2 9
𝑉1 + + − − =−
20 5 10 10 5 4
𝑽𝟏 𝑽𝟑
1 1 1 𝑉1 𝑉2 7
𝑉3 + + − − = 𝑽𝟐
10 5 15 10 5 3
1 1 1 1 2 1 2
𝑉2 + + − 𝑉1 + − 𝑉3 − =0
5 5 5 5 25 5 25
Solving the three equations we get, 𝑽𝟒 𝑽𝟓 𝑽𝟔
𝑉1 = −7.19 𝑉
𝑉2 = −2.78 𝑉
𝑉3 = 2.89 𝑉 𝟎𝑽
𝑉1 − 𝑉3
𝑖0 = = −1.008 𝐴
10

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 Problem 3
• Find all the node voltages.

Ans: 𝟎 𝑽; 𝟑𝟎 𝑽; − 𝟐. 𝟓 𝑽

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 Problem 4
• Find the voltage of node 𝑎 using nodal analysis.
Note that, the problem does not
have a specific answer as the node
voltage depends on the placement
of ground.
• If the ground is placed on node
𝑎, then 𝑉𝑎 = 0 𝑉.
• If the ground is placed on the
bottom-most node, then 𝑉𝑎 =
50 𝑉.
So, node voltages depend on the
position of the ground, however,
elemental voltages do not.
Wherever the ground is placed,
voltage across the elements and
their currents will be the same.

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 Problem 5
• Find 𝑖0 using nodal analysis. Determine the current supplied by the 60 𝑉
source.

Ans: 𝒊𝟎 = 𝟏. 𝟕𝟑 𝑨; 𝟏. 𝟐𝟔𝟐 𝑨

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 Problem 6
• Find the node voltages.

Ans: 𝟎 𝑽; 𝟑𝟐 𝑽; − 𝟐𝟓. 𝟔 𝑽; 𝟔𝟐. 𝟒 𝑽;

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 Problem 7
• Determine the current through the source resistor 𝑅𝑠 using nodal analysis.

Ans: 𝒊𝒔 = 𝟏. 𝟏𝟖 𝑨

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 Problem 8
Use nodal analysis to determine the voltage across the 3 𝐴 current source. What is
the power of it? Is it absorbing or supplying?

Ans: Node voltages = 𝟎 𝑽; 𝟏𝟎 𝑽; 𝟒. 𝟗𝟑𝟑 𝑽; 𝟏𝟐. 𝟐𝟔𝟕 𝑽

𝒗𝟑𝐀 = ± 𝟐. 𝟐𝟔𝟕 𝑽
𝑷𝟑𝐀 = −𝟔. 𝟖𝟎𝟏 𝑾, 𝐒𝐮𝐩𝐩𝐥𝐲𝐢𝐧𝐠

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 Problem 9
• Use nodal analysis to analyze the
circuit. Find 𝐼𝑥 .
• Determine the current 𝐼𝑦 .

Ans: 𝑰𝒙 = −𝟏. 𝟓 𝒎𝑨; 𝑰𝒚 = 𝟎 𝒎𝑨 Spring'23

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 Nodal Analysis with voltage source between
nodes: (Case 4)
■ Scenario 1 If a voltage source is connected between the reference node and a
nonreference node, we simply set the voltage at the nonreference node equal to
the voltage of the voltage source. For example, 𝑣1 = 10 𝑉.

■ Scenario 2 If a voltage source (dependent

or independent) is connected between two
nonreference nodes, the two nonreference
nodes form a generalized node or supernode.

A supernode is formed when a voltage source

(dependent or independent) is connected
between two nonreference nodes and any
elements connected in parallel with it.

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 Example 2: General Approach (steps 1 & 2)
Use nodal analysis to determine the
current 𝑖. Step 1: Select a node as the reference node and
place a ground to that node.
Step 2: Assign node variables to the remaining
nodes.
V2
V1 V3 Check for supernodes. Check if a voltage source
(dependent or independent) is connected
between two nonreference nodes under
consideration. There can be multiple
supernodes in a circuit.
In this circuit, the 6 V voltage source forms a
supernode between nodes 2 and 3.
0V

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 Example 2: General Approach (step 3)
We need to handle such conditions differently
Use nodal analysis to determine the
because there is no way to know the current
current 𝑖. through a voltage source in advance.
Consider the supernode as a "Whole" node and
apply KCL to the node ignoring the source forming
supernode and anything in parallel with it. There
V2 are 4 wires connected to the supernode,
V1 V3
therefore, the KCL equation for the supernode
should contain 4 terms.
Applying KCL to the supernode,
𝑉2 − 𝑉1 𝑉2 − 8 − 0 𝑉3 − 0 𝑉3 − 𝑉1
+ + + =0
2 1 5 10
After simplification,
6𝑉1 − 15𝑉2 − 3𝑉3 = − 80 (𝑖)
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 Example 2: step 3 (continued … 2)
Use nodal analysis to determine the The next step is to apply KCL to the other
current 𝑖. remaining nonreference nodes except for the
nodes forming the Supernode.
Applying KCL to the node 1,
𝑉1 − 0 𝑉1 − 𝑉2 𝑉1 − 𝑉3
+ + =0
V2 4 2 10
V1 V3 After simplification,
17𝑉1 − 10𝑉2 − 2𝑉3 = 0 (𝑖𝑖)
We have 2 equations, 3 variables, and no
remaining nodes for KCL.
The 3rd equation required, can be found by
applying KVL to the Supernode.

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 Example 2: step 3 (continued … 3)
Use nodal analysis to determine the Applying KVL to the supernode,
current 𝑖. 𝑉2 − 𝑉3 = 6 (𝑖𝑖𝑖)
We got the three equations,
6𝑉1 − 15𝑉2 − 3𝑉3 = − 80
V2 17𝑉1 − 10𝑉2 − 2𝑉3 = 0
V1 V3
𝑉2 − 𝑉3 = 6
Solving … …
𝑉1 = 4.1 𝑉; 𝑉2 = 6.8 𝑉; 𝑉3 = 0.8 𝑉;
The current i can be written as,
0 − −8 − 𝑉2
𝑖= = 1.2 𝐴
1

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 Example 2: Format Approach
Use nodal analysis to determine the Step 1: Identify all the nodes and label them
current 𝑖. (with ground being the 0th node).
Step 2: Write the component equations for all
the voltage sources (voltage difference =
labeled variable).
V2
V1 V3 𝑉4 = 8 𝑉 −−−− −(𝑖)

Check for supernodes. Check if a voltage source

(dependent or independent) is connected
V4 between two nonreference nodes. There can
be multiple supernodes in a circuit.
In this circuit, the 6 𝑉 voltage source forms a
0V supernode between nodes 2 and 3.
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Example 2: Format Approach … (2)
Use nodal analysis to determine the
current 𝑖.
Step 3: Node equation formation.

Node 1, 𝑽𝟏 : 4 , 2 , and 10  resistors are

V2 connected between 𝑉1 and 𝑔𝑟𝑜𝑢𝑛𝑑, 𝑉1 and 𝑉2 ,
V1 V3 and 𝑉1 and 𝑉3 respectively. We write,
1 1 1 0 𝑉2 𝑉3
𝑉1 + + − − − =0
4 2 10 4 2 10
V4 17𝑉1 𝑉2 𝑉3
 − − = 0 −−−−− −(𝑖)
20 2 10

0V

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Example 2: Format Approach … (3)
Use nodal analysis to determine the
current 𝑖. Node 𝟐 & 𝟑 (Supernode): Now we will apply the
same but together in both the nodes 2 & 3.
The 1  and 2  resistors are connected between
𝑉2 and 𝑉4 , and 𝑉2 and 𝑉1 respectively. Again, the 10
V2  and 5  resistors are connected between 𝑉3 and
V1 V3
𝑔𝑟𝑜𝑢𝑛𝑑, and 𝑉3 and 𝑉1 respectively. So,
1 1 𝑉4 𝑉1 1 1 𝑉1 0
𝑉2 + − − + 𝑉3 + − − =0
1 2 1 2 10 5 10 5
V4
With 𝑉4 = 8 𝑉,
3𝑉1 3𝑉2 3𝑉3
0V  − − + 8 = 0 −−−− −(𝑖𝑖)
5 2 10
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Example 2: Format Approach … (4)
Use nodal analysis to determine the Finally, applying KVL to the supernode yields.
current 𝑖.
𝑉2 − 𝑉3 = 6 −−−− −(𝑖𝑖𝑖)
We get the three equations,

V2 17𝑉1 𝑉2 𝑉3
V1 V3 − − =0
20 2 10
3𝑉1 3𝑉2 3𝑉3
− − +8=0
5 2 10
V4 𝑉2 − 𝑉3 = 6
Solving …

0V 𝑉1 = 4.1 𝑉; 𝑉2 = 6.8 𝑉; 𝑉3 = 0.8 𝑉

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 Problem 10
• Use nodal analysis to find the node voltages. Use the node voltages to find the
voltage across the 8 Ω resistor. Don't use Source Transformation.

Ans: With the ground placed at the bottom-most

node, 𝟎 𝐕 (𝐆𝐍𝐃), 𝟐𝟐. 𝟓 𝑽, 𝟐. 𝟓 𝑽; 𝑽𝟖 = 𝟐 𝑽

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 Problem 11
• Use nodal analysis, determine the current through the 2  resistance in the
right. Determine the current supplied by the 30 𝑉 source.

Ans: 𝟎 𝑨; 𝟕. 𝟓 𝑨

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 Problem 12
• Use nodal analysis to determine the current through the dependent voltage
source.

Ans: Node voltages = 𝟎 𝑽; 𝟏𝟎 𝑽; 𝟒. 𝟗𝟕 𝑽; 𝟒. 𝟖𝟓 𝑽; − 𝟎. 𝟏𝟐 𝑽;

Current through the 4𝐼0 source = ± 𝟓. 𝟑𝟑 𝑨

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 Problem 13
• Determine voltages 𝑣1 through 𝑣3 in the circuit using nodal analysis.

2Ω

8Ω
4Ω
1Ω

Ans: 𝒗𝟏 = 𝟔. 𝟐𝟑 𝑽; 𝒗𝟐 = 𝟐. 𝟎𝟖 𝑽; 𝒗𝟑 = 𝟏𝟑 𝑽

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 Problem 14
• Use nodal analysis to find 𝑖𝑥 . What is the voltage across the dependent current
source? Find the current through the 10 𝑉 source.

Fall'23 Ans: 𝒊𝒙 = 𝟏. 𝟎𝟓𝟗 𝑨; 𝒗𝟐𝒊𝒙 = 𝟎. 𝟒𝟕 𝑽; 𝒊𝟏𝟎 𝑽 = ± 𝟓. 𝟏𝟐 𝑨

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 Problem 15
• Determine current through the 15 𝑉 source using nodal analysis.

Ans: Node voltages = 𝟎 𝑽 (𝑮𝑵𝑫), −𝟏𝟓 𝑽, 𝟏𝟎 𝑽, −𝟕. 𝟗𝟖𝟐𝟓 𝑽, −𝟐. 𝟏𝟎𝟓𝟑 𝑽, 𝟕. 𝟎𝟏𝟕𝟓 𝑽; 𝑰𝟏𝟓𝑽 = ± 𝟎. 𝟏𝟐 𝑨

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 Problem 16
• Find 𝑣1, 𝑣2, and 𝑣3 using nodal analysis. Determine the currents supplied by the
12 𝑉 and the 2𝑖 source?

Ans: 𝒗𝟏 = − 𝟑 𝑽; 𝒗𝟐 = 𝟒. 𝟓 𝑽; 𝒗𝟑 = − 𝟏𝟓 𝑽; 𝑰𝟏𝟐 𝑽 = 𝟐𝟑. 𝟐𝟓 𝑨; 𝑰𝟐𝒊, = −𝟏𝟒. 𝟐𝟓 𝑨

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 Problem 17
• Determine all the node voltages.

Ans: 𝟎 𝑽 𝑮𝑵𝑫 ; 𝟖 𝑽, 𝟎 𝑽, −𝟖 𝑽, 𝟏𝟐 𝑽, 𝟐𝟒 𝑽

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 Problem 18
• Use nodal analysis and determine 𝐼𝑥 .

Ans: Node voltages = 𝟎 𝑽 𝑮𝑵𝑫 , 𝟏𝟔 𝑽, 𝟐𝟒 𝑽, 𝟒𝟖 𝑽, 𝟏𝟔 𝑽, 𝟑𝟐 𝑽; 𝑰𝒙 = 𝟒 𝑨

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 Problem 19
• Use nodal analysis and determine all the node voltages in the following circuit.

Ans: Node voltages = 𝟎 𝑽 𝑮𝑵𝑫 , 𝟐𝟒 𝑽, 𝟒𝟒 𝑽, −𝟕𝟐 𝑽, 𝟓𝟒. 𝟒 𝑽

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 Problem 20
• Use nodal analysis and determine all the node voltages in the following circuit.

Ans: Node voltages = 𝟎 𝑽 𝑮𝑵𝑫 , −𝟖 𝑽, −𝟏𝟔 𝑽, 𝟒 𝑽, −𝟒 𝑽, −𝟐𝟒 𝑽, 𝟒 𝑽

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 Practice Problems
• Additional recommended practice problems: here
• Other suggested problems from the textbook: here

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 Thank you for your attention

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