Nodal Analysis Problem with Dependent Voltage and Current Sources

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Electrical Circuits Problems, Resistive Circuits

Solve the circuit with the nodal analysis and determine $i_x$ and $V_y$ .
nodal analysis problem with dependent voltage and current sources

Solution
1) Identify all nodes in the circuit. Call the number of nodes $N$ .
The circuit has 5 nodes. Therefore, $N=5$ .

Nodes

2) Select a reference node. Label it with reference (ground) symbol.
The node at the bottom is the best candidate. It is the node with largest number of elements connected to it.
The reference node

3) Assign a variable for each node whose voltage is unknown.
Four nodes are remaining:
non-reference nodes
Node I is a regular node. A dependent voltage source is located between node II and node III. Therefore, node II and node III form a supernode. Node IV is connected to a voltage source whose other node is the reference node. We label the voltages of node I and node II with $V_1$ and $V_2$ respectively.
Labelling nodes

For node IV, the $2V$ voltage source provides the voltage of the node. Since the positive terminal of the voltage source is connected to the node, the node voltage is $2 V$ .
A voltage source between a node and the reference node

4) If there are dependent sources in the circuit, write down equations that express their values in terms of other node voltages.
The voltage of node III is $V_2+3i_x$ . On the other hand $i_x$ is the current of the $2 \Omega$ resistor in the right hand side. Using the Ohm’s law:
voltage of the $2 \Omega$ resistor = $V_2+3i_x= -2 i_x$ . (Eq. 1)
Please note that the direction of $i_x$ is from the reference node to node III. Since it is always assumed that the reference node is the negative terminal for the defined node voltages, the voltage of node III is $-2 \times i_x$ (instead of $2 \times i_x$ ).
By solving Eq. 1:
$V_2= -5 i_x$ .

Consequently, the voltage of node III is $V_2+3i_x=V_2-\frac{3}{5}V_2=0.4 V_2$ .
Note that we always write unknowns in terms of node voltages in nodal analysis.
Labelled node voltages
Note that $V_2 \neq 2V_y$ . $2V_y$ is the current of the dependent current source. Now, we need to find the current of the dependent current source in terms of the node voltages. $V_y$ is the voltage across node IV and Node I. The positive terminal is connected to node IV and the negative one is connected to node II. Therefore, $V_y=2-V_1$ . You can verify this by applying KVL around the loop consisting of the reference node, node I and node IV.

5) Write down a KCL equation for each node by setting the total current flowing out of the node to zero.
For node I:
$9A + \frac{V_1 - V_2}{1 \Omega}+\frac{V_1 - 2 V}{2 \Omega}=0$ .
$\to - \frac{3}{2} V_1 +V_2=8$ (Eq. 2)

For supernode II&III:

$-2V_y+\frac{V_2-V_1}{1 \Omega}+\frac{0.4V_2}{2 \Omega}+\frac{0.4V_2-2V}{1 \Omega}=0$ . Substituting $V_y=2-V_1$ and simplifying:
$- V_1 -1.6 V_2=-6$ . (Eq. 3)

Solving the set of equations Eq. 2 and Eq. 3 results in $V_1=-2 V$ and $V_2=5V$ . Therefore $V_y=2-V_1=4 V$ and $i_x=-0.2 V_2=-1 A$ .

Comments

17 responses to “Nodal Analysis Problem with Dependent Voltage and Current Sources”

September 3, 2011

René Navarro

Good Job!! You should be a teacher in my University.

Reply
October 8, 2011

Dis Guy

The nodes you labelled in step 3 don’t correspond to what you’re writing later, unless I’ve misunderstood the labeling. Don’t node II and node IV form the supernode, and not node II and III? Also, I don’t see how node IV is connected to the 2V source? That would be node III wouldn’t it?

Reply
1. October 8, 2011
  
  Dr. Yaz Z. Li
  
  You are right.II III and IV must be swapped in the step 3 picture. Thank you.
  
  Reply
  1. October 8, 2011
    
    Dis Guy
    
    I think you would swap III and IV? Since you say that there is a dependent source located between nodes II and IV.
    
    Reply
    1. October 8, 2011
      
      Dis Guy
      
      I mean dependent source between nodes II and III*.
      
      Reply
October 8, 2011

Dis Guy

I also don’t understand how you got V2+3ix=V2-(3/5)V2=0.4V2 for the node voltage in step 4.

Reply
1. October 9, 2011
  
  Dr. Yaz Z. Li
  
  $V_2=-5i_x \to i_x=-\frac{1}{5}V_2$
  Substituting this in $V_2+3i_x$ gives $V_2+3i_x=\frac{2}{5}V_2=0.4V_2$
  
  Reply
November 29, 2011

Rai Sudin

i think u had accidentally labelled wrongly the 3rd and 4th node…then , why u did not consider the 1 ohm resistor at th 3rd node ?

Reply
1. November 29, 2011
  
  Dr. Yaz Z. Li
  
  Well, it is not used in equations. This does not mean that it is missed somewhere. So, the solution and answers are correct. If you check you will see that the current of that resistor is zero so its value is not important, the answers would not change if its replaced with short circuit or open circuit, ….
  
  Reply
May 5, 2013

thipesh

Good job…i understood easily by ur description thnx ya

Reply
October 5, 2013

ishraq

hi,can some one told me about the equition that how we will write equition for a circuit which have a supernode

Reply
1. October 6, 2013
  
  Yaz
  
  Hi Ishraq,
  
  Please download my ebook on Nodal Analysis from . Supernode is also covered.
  
  Reply
October 7, 2013

ankita

plz solve norton equivalent circuit calculating I-norton and R-noton having
1) no independent source and dependent voltage source
2) no independent source and dependent curent source
3)1 independent source and 1 dependent souce

Reply
1. October 8, 2013
  
  admin
  
  Sure. I will add some solved problems for Norton equivalent soon.
  
  Reply
March 4, 2015

retarded jay

i’m stupid and you omitted too much information in your derivations for someone with my brain to understand how you arrived at certain equations.

Reply
April 8, 2020

George

“A dependent voltage source is located between node II and node III. Therefore, node II and node III form a supernode. Node IV is connected to a voltage source whose other node is the reference node. ”

Hi again. I think that you must interchange the names between nodes 3 and 4. You have to name node 4 as node 3 and node 3 as node 4. Please, check it.

Reply
May 2, 2021

Fred

How about showing us a cccs on the perimeter of a circuit , the cccs having no resistors across it nor any voltage sources across it?

Reply