Find Current Drawn from Independent Voltage Source: Node Analysis

In summary: If you are correct, then v1=-6.375 (since the current entering node 1 must be zero), and likewise v1=-6.375 for node 2 (since the current leaving node 2 must be equal to the current entering node 1). However, if you are not correct, then you will get an incorrect answer for node 1 (since the current entering node 1 will be nonzero), and an incorrect answer for node 2 (since the current leaving node 2 will be nonzero). node 2: (v1-v2)/R3 + (v3-v2)/R4 = I('node 3: (v3-v2)/R4 + v3/R5 = 0
  • #1
srh
7
0

Homework Statement


(I have attached a picture of the circuit)
Using node voltage analysis in the circuit, find the current i drawn from the independent voltage source. Let V=3V, R1=1/2, R2=1/2, R3=1/4, R4=1/2, R5=1/4, I=0.5.


Homework Equations


I=V/R


The Attempt at a Solution


I haven't seen one like this w/ the independent voltage source, is it solved the same way? I tried to perform KCL at each node and got:
node 1: (V-v1)/R1 + v1/R2 + (v1-v2)/R3 = 0
node 2: (v1-v2)/R3 + (v3-v2)/R4 = I
node 3: (v3-v2)/R4 + v3/R5 = 0

When I substituted for V, I, R1, R2, R3, R4 and R5 and entered in the matrix I got:
v1=-6.375
Then I related i=(V-v1)/R1. I got an answer of 18.75 A. I thought the answer maybe sounded high and was wondering if I had messed up somewhere.
 

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  • #2
srh said:

The Attempt at a Solution


I haven't seen one like this w/ the independent voltage source, is it solved the same way?

Yes, but it's even better because an independent voltage source puts a constraint on the node voltages. The independent source is the reason that you have 3 node equations instead of 4.

v1=-6.375

Taking the current through R1, R2, and R3 to be going to the left, and taking the current through R2 and R5 to be going down, I wrote down the node equations and solved the model to find that v1=1.375 V (I did this all by hand mind you, so let the buyer beware), which will give a more reasonable current. Check your equations to make sure that you were consistent with the directions of your currents.
 
  • #3
Keep in mind that the voltage source can be treated this way because it's connected to ground. If it wasn't, as in it is inbetween two nodes, you'd just have to use a supernode with a supplemental.
 
  • #4
You only need to perform KCL at the nodes with more than 2 branches. And in the figure, there are only 2 such nodes (excepting the ground), i.e., the one common to R1, R2 and R3; and another common to R3, R4 and the current source.

srh said:
node 1: (V-v1)/R1 + v1/R2 + (v1-v2)/R3 = 0

There is a mistake here. KCL requires either (1) the sum of all currents leaving the node equals zero, or (2) the sum of all currents entering the node equals zero. What you wrote above (for node 1) is not consistent with (1) or (2).
 
  • #5
Indeed, recheck your current directions for node 1 (R1, R2, and R3).
 

Related to Find Current Drawn from Independent Voltage Source: Node Analysis

1. What is node analysis?

Node analysis is a method used to analyze electrical circuits by examining the currents and voltages at different points, or nodes, in the circuit. It is based on Kirchhoff's Current Law, which states that the sum of all currents entering and leaving a node must equal zero.

2. How do you find the current drawn from an independent voltage source using node analysis?

To find the current drawn from an independent voltage source using node analysis, you must first identify the nodes in the circuit and assign variables for the unknown currents. Then, you can write a set of equations based on Kirchhoff's Current Law for each node. Finally, you can solve the equations simultaneously to find the value of the current drawn from the voltage source.

3. What is an independent voltage source?

An independent voltage source is a type of voltage source that maintains a constant voltage regardless of the current passing through it. It is often represented by a battery symbol in circuit diagrams.

4. Are there any limitations to using node analysis?

Node analysis can become more complex for circuits with a large number of nodes. In these cases, it may be more efficient to use other analysis methods such as mesh analysis. Additionally, node analysis assumes that all components in the circuit are linear, meaning their behavior can be described by a straight line on a graph.

5. Can node analysis be used for both DC and AC circuits?

Yes, node analysis can be used for both DC and AC circuits as long as the components in the circuit are linear. However, for AC circuits, the equations used in node analysis must also take into account the phase difference between currents and voltages at each node.

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