How is KVL used in the node method KCL equations?

  • #1
zenterix
480
70
TL;DR Summary
The book I am reading, "Foundations of Analog and Digital Electronic Circuits" by Agarwal has a section on the node method of circuit analysis that says that the KCL equations used in this method contain within them all the information from all the independent KVL equations. I can't seem to see this use of the KVL equations, however.
Consider the following electric circuit in which we have node voltages labeled

1696835576621.png

I have a question about the reasoning present in the book I am reading about the node method of circuit analysis.

If we write KVL equations around the loops we get

$$-V+(V-e)+e=0$$

$$-e+e=0$$

...this choice of voltage variables automatically satisfies KVL. So to solve the circuit it is not necessary to write KVL. Instead, we will directly proceed with writing KCL equations. Furthermore, to save time the KCL equations can be written directly in terms of the node voltages and the resistors' values. Since we have only one unknown, e, we need only one equation. Hence, at node 2,

$$\frac{e-V}{R_1}+\frac{e}{R_2}-I=0\tag{3.5}$$

Notice that the preceding step is actually two substeps bundled into one: (1) writing KCL in terms of currents and (2) substituting immediately node voltages and element parameters for the currents by using KVL and element laws.

Then

Note that in one step we have one unknown and one equation, whereas by the KVL and KCL method of Chapter 2 we would have written eight equations in eight unknowns. Further, note that both the device law for every resistor and all independent statements of KVL for the circuit have been used in writing Equation 3.5.

I don't understand in what way the independent statements of KVL are used when writing out a KCL equation in the node method.

As far as I can see, we are using KCL and then subbing in the expressions for the currents, which come from the element laws.

Where exactly is KVL?
 
Engineering news on Phys.org
  • #2
Ohm Law
## E _ { R _ 1 } = I _ { R _ 1 } \cdot R _ 1 \Rightarrow I _ { R _ 1 } = \frac { E _ { R _ 1 } } { R _ 1 } ##
Kirchhoff Voltage Law
## - V - E _ { R _ 1 } + E _ { R _ 2 } = 0 \Rightarrow E _ { R _ 1 } = - V + E _ { R _ 2 } = - V + e = e - V ##
Ohm Law and Kirchhoff Voltage Law
## I _ { R _ 1 } = \frac { e – V } { R _ 1 } ##
 
  • #4
zenterix said:
I don't understand in what way the independent statements of KVL are used when writing out a KCL equation in the node method.
alan123hk said:
I'm not sure how KVL is used in the nodal method KCL equation
Yeah, for me it is one or the other, not some combination. I prefer KCL usually, since that is a more intuitive method for me.
 

1. What is KVL and how does it relate to the node method using KCL?

Kirchhoff's Voltage Law (KVL) states that the sum of the electrical voltages around any closed network loop is zero. While KVL is typically used in loop analysis, it indirectly relates to the node method (or nodal analysis) which primarily uses Kirchhoff's Current Law (KCL). KCL states that the total current entering a node (or junction) must equal the total current leaving the node. In nodal analysis, KVL is used to express the voltages in terms of the node voltages, which are then used to set up KCL equations at each node.

2. How do you use KVL to set up the node-voltage equations in nodal analysis?

In nodal analysis, each node voltage is defined with respect to a common reference point, typically the ground. KVL is used to express the voltage across each component (like resistors) in the circuit as the difference in node voltages at either end of the component. These voltage expressions are substituted into KCL equations at each node (except the reference node), allowing us to solve for the unknown node voltages.

3. Can you give an example of using KVL in the node method to solve a circuit?

Consider a simple circuit with two nodes and three components: two resistors and a voltage source. Let's say R1 connects the ground (node 0) and node 1, R2 connects node 1 and node 2, and a voltage source is connected between node 2 and ground. Using KVL, the voltage at node 1 (V1) would be V1 - 0 (since the other side of R1 is connected to ground), and the voltage at node 2 (V2) would be V2 - V1 across R2 and V2 - 0 across the voltage source. These voltage drops are then used in KCL equations: at node 1, the current through R1 (from the ground to node 1) equals the current through R2 (from node 1 to node 2); at node 2, the current through R2 equals the current supplied by the voltage source.

4. What are the limitations of using KVL in nodal analysis?

While KVL is essential for defining voltages in nodal analysis, it assumes that there is no electromagnetic interference or fluctuating magnetic fields within the circuit loops. In real-world scenarios, especially in high-frequency circuits or circuits with large inductors, these assumptions might not hold due to induced voltages not accounted for by KVL. Additionally, KVL cannot be applied directly in circuits where components like current sources exist between nodes, as their voltage cannot be defined without additional information.

5. How does KVL help in simplifying complex circuits when using nodal analysis?

KVL helps reduce the complexity of solving circuits by minimizing the number of equations needed. By expressing all voltages in terms of node voltages and using KVL to relate these node voltages, nodal analysis allows for the systematic and straightforward application of KCL at each node. This method is particularly efficient for circuits with multiple nodes and components, as it systematically reduces the problem to a set of linear equations based on the node voltages, which can then be solved using matrix methods or other algebraic techniques.

Similar threads

  • Electrical Engineering
Replies
4
Views
1K
  • Electrical Engineering
2
Replies
56
Views
4K
Replies
12
Views
1K
  • Electrical Engineering
Replies
5
Views
1K
Replies
2
Views
713
Replies
5
Views
950
  • Electrical Engineering
Replies
1
Views
1K
  • Electrical Engineering
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
546
  • Electrical Engineering
Replies
27
Views
1K
Back
Top