Find Current Drawn from Independent Voltage Source: Node Analysis

Click For Summary
The discussion focuses on using node voltage analysis to find the current drawn from an independent voltage source in a given circuit. The initial attempt yielded an unrealistic current value of 18.75 A, prompting concerns about potential errors in the calculations. Participants emphasized the importance of correctly applying Kirchhoff's Current Law (KCL) and ensuring consistent current direction in the node equations. It was noted that the independent voltage source simplifies the analysis by reducing the number of necessary node equations. Correcting the node equations and verifying current directions are crucial for obtaining accurate results.
srh
Messages
7
Reaction score
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.
 

Attachments

  • circuits2.JPG
    circuits2.JPG
    4.5 KB · Views: 402
Physics news on Phys.org
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.
 
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.
 
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).
 
Indeed, recheck your current directions for node 1 (R1, R2, and R3).
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K