How Does Kirchoff’s Voltage Law Apply to Supernodes in Circuit Analysis?

[pekpek]

Hello All,

I am trying to teach myself basic EE. The textbook I am using gave the following example:

1. Find I_S2. Diagram is attached.
Given:
I_S1 = 2A
I₂ = 3A
I₃ = 2A
I₄ = 1A
I₅ = 0A

The answer in the textbook is 1A.

My Method using KCF:
I_S1 + I_S2 - I₂- I₃ - I₄ -I₅ = 0

Substituting the numbers I get: 2 + I_S2 - 3 - 2 - 1 = 0; I_S2 = 4A

I am not sure where I am going wrong here.

Thanks for any insight!

 

[NascentOxygen]

At the junction of R1 and R2 applying KCL you see:
Is1 + Is2 = I2

Voila!

 

[NascentOxygen]

If you look at the equation for the supernode, it won't involve I_S2
current in = current out

I_s1 = I₃ + I₅

Good luck with your studies.

 

[pekpek]

Ahhh. Thanks NascentOxygen!

 

[rezeew]

Hello,

It appears that you are using Kirchoff's Current Law (KCL) to solve this problem. However, in this situation, it would be more appropriate to use Kirchoff's Voltage Law (KVL) and the concept of supernodes.

First, let's define a supernode. A supernode is a combination of two or more nodes that are connected by a voltage source. In this problem, node 1 and node 2 are connected by a voltage source (IS1), so they can be considered as a supernode.

To apply KVL, we need to create a loop that includes all the elements and the supernode. In this case, the loop would be: IS1 - I2 - I3 - IS2 = 0. This is because the voltage across the supernode is equal to the voltage source (IS1), and the sum of the voltage drops around the loop must equal zero.

Now, we can solve for IS2: IS2 = IS1 - I2 - I3 = 2A - 3A - 2A = -3A. However, since IS2 is defined as a current going in the opposite direction as IS1, we need to take the negative value of -3A, which gives us a final answer of 3A.

In summary, KVL and the concept of supernodes can be used to solve this problem and get the correct answer. I hope this helps clarify any confusion you had. Keep up the great work in teaching yourself basic EE!