Nodal Analysis and Power Balance in AC Circuits

In summary, the conversation discusses the use of nodal analysis in an ac circuit. The goal is to show a power balance between the sources and the power dissipated in the resistors, as well as a reactive power balance between the sources and the inductors and capacitors. The formula for the sources is given, but there is confusion about whether to use the absolute value of the powers or not. An example is provided, but there is an error in the first node equation. The conversation ends with a request for help in identifying the mistake.
  • #1
sandy.bridge
798
1
I'm working on an ac circuit, and we were told to use nodal analysis. Furthermore, we are asked to show a power balance between the sources and the power dissipated in the resistors, along with a reactive power balance between the sources and the inductors and capacitors.

I have this formula for the sources:
[tex]P=I_{rms}V_{rms}cos(\theta{_v}-\theta{_i})[/tex]

However, when I am using this forumla, some of power in regards to the sources is negative, and some is positive. Therefore, upon summing together each of their contributions, the power is really low in comparison to the power dissipated in the resistors. Am I supposed to be taking the absolute value of these powers or what?
 
Physics news on Phys.org
  • #2
I decided to post an example:
Untitled-11.jpg

At nodes 1, 2 and 3 respectively I have
[tex](0.6-j0.2)V_1+0.2V_2-0.1V_3=0, 0.2V_1+(0.2+j0.05)V_2-0.2V_3=8.75+j2.1651, -0.1V_1-0.2V_2+(0.3+j0.25)V_3=-12.5[/tex]
Upon solving the system of equations, I find that
[tex](V_1, V_2, V_3)=(-18.2797-9.09145j, 61.7064+8.49923j, -2.44807+5.00906j)=(20e^{-j156}, 62.3e^{j7.84}, 5.575e^{j116})[/tex]
The current through the j10V source is
[tex](V_3-j10)/(-j4)=1.38975e^{-j26.1}[/tex]
and the power generated by this source is
[tex]P=(1.38976)(10)cos(90-(-26.1))=-6.1W[/tex]
Current through the 50V source is
[tex](V_3+50-V_2)/5=2.916e^{-j166}[/tex]
and the power is
[tex]P=(2.916)(50)cos(0-(-166))=-141.5[/tex]
The current in the final voltage source is
[tex](V_2-10cos(30)-j10sin(30))/(-j4)=13.29e^{j94}[/tex]
and the power is
[tex]P=(13.29)(10)cos(30-94)=58[/tex]
Total power supplied is 205W.
The power dissipated in the 10-ohm resistor is
[tex]|V_3-V_1|^2/10=42W[/tex]
In the 5-ohm resistor,
[tex]|V_3+50-V_2|^2/5=42W[/tex]
and in the 2-ohm we have
[tex]|V_1|^2/2=200W[/tex]
Combined we have total power dissipated as 284W.
Can anyone see where I have gone wrong here? Obviously this is not adding up, but I've been staring at this problem for a very long time and cannot seem to pin point it.
 
Last edited:
  • #3
I believe that the second term of your first node equation (for node 1) should be imaginary.
That is, j0.2V2.
 

1. What is nodal analysis in AC circuits?

Nodal analysis is a method used in circuit analysis to determine the voltage at each node (or connection point) in a circuit. It is based on the principle that the sum of currents entering a node is equal to the sum of currents leaving the node. This method is particularly useful for analyzing complex AC circuits with multiple voltage sources and resistors.

2. How is nodal analysis different from other circuit analysis techniques?

Nodal analysis is different from other techniques, such as mesh analysis or Kirchhoff's laws, in that it focuses on the voltages at individual nodes in a circuit rather than on the current flow. This makes it particularly useful for circuits with many parallel branches and complex voltage sources.

3. What is the power balance equation in AC circuits?

The power balance equation in AC circuits is used to determine the power dissipated by each component in a circuit. It states that the sum of power consumed by all components in a circuit must be equal to the total power supplied by the voltage sources. This equation is important for understanding how energy is distributed and used in a circuit.

4. How is power balance calculated in AC circuits?

To calculate power balance in AC circuits, you can use the formula P=VI, where P is power, V is voltage, and I is current. The power consumed by a component is equal to the voltage across that component multiplied by the current flowing through it. By calculating the power for each component in the circuit and summing them, you can determine the total power supplied by the voltage sources.

5. What are some practical applications of nodal analysis and power balance in AC circuits?

Nodal analysis and power balance are used in a wide range of practical applications, such as designing electrical circuits for devices, troubleshooting circuit problems, and optimizing energy usage in power grids. They are essential tools for understanding the behavior of AC circuits and ensuring the efficient and safe operation of electronic devices and systems.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
7
Views
816
  • Engineering and Comp Sci Homework Help
Replies
1
Views
203
  • Engineering and Comp Sci Homework Help
Replies
7
Views
859
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
20
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
Replies
3
Views
409
Back
Top