Nodal analysis, setting up the equations

In summary, the conversation is about setting up KCL equations to solve a circuit with a 30V voltage source and a 15 ohm resistor. The equations involve the voltages at nodes A, B, and C and account for the 4A current source. After solving for the voltages at the nodes, the power at the independent voltage and current sources can be found using voltage drops.
  • #1
gomezfx
20
0

Homework Statement



[PLAIN]http://img844.imageshack.us/img844/8508/circuit.png [Broken]

Homework Equations


KCL set ups


The Attempt at a Solution


I am having trouble setting up the KCL equations

The 30V voltage source on the left with the resistor is messing me up, how do I incorporate that?

Do I just do 30/15 and assume a 2A current source is going into node A?

Here are my equations:

VD = reference node = 0 V

KCL @ Node B
(VB-VA)/25 + (VB-VC)/50 - 1 + 4 = 0

KCL @ Node C
(VC-VD)/50 + (VC-VB)/50 + 1 = 0

KCL @ Node A?
(VA-VB)/25 + (VA-VD)/31.25 - 4 + ?? = 0
 
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  • #2
You have 30V at the other end of the 15 ohm resistor. So the current through it is (VA-30)/15.

ehild
 
  • #3
ehild said:
You have 30V at the other end of the 15 ohm resistor. So the current through it is (VA-30)/15.

ehild

Ok, then the entire equation for voltage at node A is:

(VA-30)/15 + (VA-VB)/25 + (VA-VD)/31.25 - 4 = 0?

After I solve for all the voltages at the nodes, how do I find power at the independent voltage a current sources?

Do I just use voltage drops?
IE, Power at 4A = (VA-VB)*4
 
  • #4
gomezfx said:
Ok, then the entire equation for voltage at node A is:

(VA-30)/15 + (VA-VB)/25 + (VA-VD)/31.25 - 4 = 0?

This is correct.

gomezfx said:
After I solve for all the voltages at the nodes, how do I find power at the independent voltage a current sources?

Do I just use voltage drops?
IE, Power at 4A = (VA-VB)*4

Yes, knowing the potential difference U across the terminals of a source and the current I flowing through it, the power is U*I.ehild
 
  • #5


To incorporate the voltage source on the left, we can use the concept of superposition. This means that we can consider the voltage source as a separate circuit and analyze its effects on the overall circuit. In this case, we can treat the voltage source as a separate current source of 30/15 = 2A going into node A. Therefore, the KCL equation for node A would be (VA-VB)/25 + (VA-VD)/31.25 - 4 + 2 = 0. This takes into account the current entering the node from the voltage source.
 

What is nodal analysis?

Nodal analysis is a method used in electrical circuit analysis to determine the voltage and current at each node or connection point in a circuit. It is based on Kirchhoff's current law, which states that the sum of currents entering a node must equal the sum of currents leaving the node.

How do you set up the equations for nodal analysis?

To set up the equations for nodal analysis, you need to follow these steps:1. Identify all the nodes in the circuit.2. Choose a reference node and assign it a voltage of 0.3. Write Kirchhoff's current law for each node, using Ohm's law to express the currents in terms of the node voltages.4. Solve the resulting system of equations to find the voltages at each node.

What are the advantages of using nodal analysis?

Nodal analysis allows for a systematic and efficient way to analyze complex circuits. It can handle circuits with multiple voltage sources and is particularly useful for circuits with many parallel branches. It also provides a clear understanding of the voltage and current at each node, making it easier to identify and troubleshoot problems in a circuit.

Are there any limitations to nodal analysis?

While nodal analysis is a powerful tool, it does have some limitations. It can only be used for circuits with a single reference node and does not take into account the internal resistance of voltage sources. It also requires solving a system of equations, which can become time-consuming for larger circuits.

Can nodal analysis be applied to any type of circuit?

Nodal analysis can be applied to any linear circuit, meaning that the relationships between voltage and current in the circuit must follow Ohm's law. It can also be used for circuits with nonlinear elements, but only as an approximation. For circuits with nonlinear elements, more advanced analysis methods may be necessary.

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