Nodal Analysis - Circuit with Dependent Voltage Source


Determine the power of each source after solving the circuit by the nodal analysis.
Nodal Analysis - Supernode - Dependent Voltage Source 1

Answers:  P_{I_x}=0.497W, P_{1A}=-1.806W, P_{2A}=4.254W, P_{3V}=-3.87W, and  P_{5V}=-3.552W


Solution

I. Identify all nodes in the circuit.
The circuit has 6 nodes as highlighted below.

Nodal Analysis - Supernode - Dependent Voltage Source  - All Nodes
II. Select a reference node. Label it with the reference (ground) symbol.

The right top node is connected to two voltage sources and has three elements. All other nodes also have three elements. Hence, we select the right top node because by this selection, we already know the node voltages of two other nodes, i.e. the ones that the reference node is connected to them by voltage sources.

Nodal Analysis - Supernode - Dependent Voltage Source  - The reference node and node voltages

III. Assign a variable for each node whose voltage is unknown.
We label the remaining nodes as shown above. Nodes of  V_3 and  V_4 are connected to the reference node through voltage sources. Therefore,  V_3 and  V_4 can be found easily by the voltages of the voltage sources. For  V_3 , the negative terminal of the voltage source is connected to the node. Thus,  V_3 is equal to minus the source voltage,  V_3=-5 V . The same argument applies to  V_4 and  V_4=-3V .

IV. If there are dependent sources in the circuit, write down equations that express their values in terms of node voltages.

The voltage of the dependent voltage source is  I_x . We should find this value in terms of the node voltages.  I_x is the current of the  3\Omega - resistor. The voltage across the resistor is  V_2-V_4 . We prefer to define  V_{3\Omega} as  V_2-V_4 instead of  V_4-V_2 to comply with passive sign convention. By defining  V_{3\Omega} as mentioned,  I_x is entering from the positive terminal of  V_{3\Omega} and we have  V_{3\Omega}= 3 \Omega \times I_x . Therefore,  I_x=\frac{V_2-V_4}{3\Omega} .  \to I_x=\frac{V_2}{3} +1

V. Write down a KCL equation for each node.

Nodes of  V_1 and  V_2 :
These two nodes are connected through a voltage source. Therefore, they form a supernode and we can write the voltage of one in terms of the voltage of the other one. Please note that the voltage of the dependent voltage source is  I_x and we have  V_2=V_1+I_x
 \to V_2=V_1+\frac{V_2}{3} +1 \to \frac{2}{3} V_2=V_1+1
 \to V_1=\frac{2}{3} V_2-1

Nodal Analysis - Supernode - Dependent Voltage Source  - The supernode

KCL for the supernode:
 \frac{V_1-V_3}{5\Omega}+\frac{V_1-V_5}{2\Omega}+\frac{V_2-V_4}{3\Omega}-2A=0
 \to \frac{V_1+5}{5}+\frac{V_1-V_5}{2}+\frac{V_2+3}{3}-2=0
 \to \frac{V_1}{5}+\frac{V_1-V_5}{2}+\frac{V_2}{3}=0
 \to 21V_1-15V_5+10V_2=0
Substituting   V_1=\frac{2}{3} V_2-1 ,
 \to 8V_2-5V_5=7

Node of  V_5 :
 \frac{V_5-V_1}{2 \Omega} +\frac{V_5-V_3}{1 \Omega}+1=0
 \to -V_1 +3V_5-2V_3+2=0
Substituting   V_1=\frac{2}{3} V_2-1 and  V_3=-5 V ,
 \to -2 V_2 +9V_5=-39

Here is the system of equations that we need to solve and obtain  V_2 nd  V_5 :

 \left\{ \begin{array}{l} I: 8V_2-5V_5=7 \\ II: -2 V_2 +9V_5=-39 \end{array} \right.

We use elimination method to solve this system of equation:
 (II)\times 4 +(I): 31V_5=-149 \to
 V_5=-4.806 V
 V_2=\frac{9V_5+39}{2} \to
 V_2=-2.127 V
Using   V_1=\frac{2}{3} V_2-1 ,
 V_1=-2.418 V

All node voltages are determined. Now, the power of voltage sources can be calculated from the node voltages. For each source, we need to find the voltage across the source as well as the current flowing through it to compute the power.

 2A current source:
The voltage across the  2A current source is equal to  V_2 . However, the comply with the passive voltage convention, the current should be entering from the positive terminal of the defined voltage as shown below. Therefore,  V_{2A}=-V_2=2.127 V .

 P_{2A}=2A \times V_{2A}=4.254W absorbing power

Nodal Analysis - Supernode - Dependent Voltage Source - Current directions and voltage polarities for sources

 1A current source:

To compliant with the passive sign convention, the voltage  V_{1A} should be defined with polarity as indicated above. We have  V_{1A}= V_5-V_4=-1.806V . Hence,

 P_{1A}=1A \times (-1.806 V)=-1.806W supplying power.

 5V voltage source:

 I_{5V} should be defined such that it enters from the positive terminal of the source in order to use the voltage of the source in power calculation. Another option is to use  V_3 and define the current as entering from the voltage source terminal connected to the node of  V_3 . We use the first approach here. KCL should be applied in the node of  V_3 to determine  I_{5V} .

KCL @ Node of  V_3 :

 -I_{5V}+I_{1\Omega}+I_{5\Omega}=0
 \to I_{5V}=\frac{V_3-V_5}{1\Omega}+\frac{V_3-V_1}{5\Omega}=-0.7104A

 P_{5V}=5V \times (-0.7104 A)= -3.552 W supplying power.

 3V voltage source:

Likewise,  I_{3V} should be defined as shown above to comply with the passive sign convention. We apply KCL to the reference node to find  I_{3V} .

KCL @ the reference node:

 I_{5V}+I_{3V}+2A=0
 \to I_{3V}=-1.29A

 P_{3V}=3V \times (-1.29 A)= -3.87 W supplying power.

The dependent source:
The voltage of the dependent source is  I_x and we define its current  I_{I_x} with the direction illustrated above.  I_{I_x} can be calculated by applying KCL at the node of  V_2 . The current of the  3\Omega resistor is  I_x which is equal to  \frac{V_2}{3} +1= 0.291A .

KCL @ Node of  V_2 :

 -2A+I_x+I_{I_x}=0 \to I_{I_x}=1.709A

 P_{I_x}=I_x \times I_{I_x}=0.291 V \times 1.709 A= 0.497W absorbing power.

The PSpice simulation result is shown below. The PSpice schematics can be downloaded from http://www.solved-problems.com/download/1-23.zip.

Nodal Analysis - Supernode - Dependent Voltage Source - PSpice Simulation Results

7 thoughts on “Nodal Analysis - Circuit with Dependent Voltage Source

    1. Mesh analysis is easier because there are only 4 meshes. However, it is required to be solved with the nodal analysis. I would appreciate it if you share your solution with us. Thanks.

    1. I upgrade your membership. You should be able to create a new post from here (after login). Please write your solution there. I will somehow attach it here.
      To see how to write a post , watch this.
      Thanks.

  1. Hey guys thanx for sharing ur knowledge. the example really helped me. i am currently taking this course in college and i am struggling. this example was more helpful than what the professor has been teaching. what are your tips to find more motivation to help me succeed in this course??

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