What Are Kirchhoff's Laws and Why Is Zero Current Important?

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Kirchhoff's laws are essential for analyzing electrical circuits, particularly in determining current and voltage relationships. The discussion highlights confusion around calculating currents, specifically the assertion that certain currents are zero despite having defined node voltages. Participants emphasize the importance of using independent equations, suggesting a combination of Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) for accurate analysis. The correct approach involves writing three independent loop equations and two KCL equations to solve for node voltages and subsequently calculate currents. Overall, the conversation underscores the necessity of understanding the relationship between node voltages and branch currents in circuit analysis.
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Homework Statement


5e0985c78f2e078a199f56af2cc6c368.png


Homework Equations


Kirchhoff's laws and I = V/R

The Attempt at a Solution


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658df6808ebf558dc236a71b72964684.png

I asked a question recently about a Kirchhoff's law problem and I received a lot of great help and I feel I better understand it. Or so I thought. This is my attempt at the problem. I don't know if this is right or wrong but it doesn't make sense that the current is 0 for I2, I3, and I4. Maybe someone could shed some light one what I've done wrong.
 
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Your node equations and node voltage solutions look reasonable (I didn't check the solution in detail).

I'm not following what you are trying to do to calculate the currents. What do you get when you just calculate them by hand. You have all the node voltages and resistances, so it should be easy.

BTW, it's just a style thing, but I prefer to write KCL equations as the sum of all currents into a node = 0. That helps me to avoid sign errors, especially in more complicated circuits.
 
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berkeman said:
Your node equations and node voltage solutions look reasonable (I didn't check the solution in detail).

I'm not following what you are trying to do to calculate the currents. What do you get when you just calculate them by hand. You have all the node voltages and resistances, so it should be easy.

BTW, it's just a style thing, but I prefer to write KCL equations as the sum of all currents into a node = 0. That helps me to avoid sign errors, especially in more complicated circuits.
I haven't tried calculating them by hand.. Didn't I write the KCL equations as a sum equal to 0? I feel like I have a sign error somewhere..
 
justin15501 said:
I haven't tried calculating them by hand.. Didn't I write the KCL equations as a sum equal to 0? I feel like I have a sign error somewhere..
Your node voltage equations are correct. If you have Va=-0.928V and Vb=-5.19V, how can I2, I3 and I4 be zero? You have written equations for them in terms of Va and Vb.
 
There are really only three independent loops and you're trying to construct five independent equations using only loop equations. Instead, write three KVL loop equations and two KCL equations at the independent nodes (where Va and Vb are located).

upload_2016-7-21_20-27-32.png
 
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gneill said:
There are really only three independent loops and you're trying to construct five independent equations using only loop equations. Instead, write three KVL loop equations and two KCL equations at the independent nodes (where Va and Vb are located).

View attachment 103627
Doesn't the whole big loop around count? Making it 4 loops? And I don't know how to write the KCL equations? Thanks for the reply!
 
justin15501 said:
Doesn't the whole big loop around count? Making it 4 loops? And I don't know how to write the KCL equations? Thanks for the reply!
Isn't I2 equal to Va/680?
 
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So my equations would be
cnh1995 said:
Isn't I2 equal to Va/680?
How does the Va1 and Va2 play into the equations? Do they count in as batteries? And I'm confused why the loops I picked won't work?
 
justin15501 said:
Didn't I write the KCL equations as a sum equal to 0?
No, you wrote your two KCL equations as "current from the left" = "sum of other currents out of the node". Again, that's fine if you can keep track of it, but for me I make fewer errors when I write "sum of all currents into a node" = 0. No big deal, and as has been pointed out, you solved those equations fine. The issue is in your calculating the currents given the voltages across the resistors... :smile:
 
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  • #10
berkeman said:
No, you wrote your two KCL equations as "current from the left" = "sum of other currents out of the node". Again, that's fine if you can keep track of it, but for me I make fewer errors when I write "sum of all currents into a node" = 0. No big deal, and as has been pointed out, you solved those equations fine. The issue is in your calculating the currents given the voltages across the resistors... :smile:
If my 5 equations are fine as you said, then shouldn't I1-I5 be correct values? I think my problem is over the confusion of Va1 and Va2. Should I have factored these into my equations? I though Va1 and Va2 was just specified so I can't find the potential difference at those points in part 1, and had nothing to do with part 2?
 
  • #11
justin15501 said:
If my 5 equations are fine as you said,
I did not say that. Your first two equations are fine (the KCL equations). Please answer the question you were asked in post #7... :smile:
 
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  • #12
berkeman said:
I did not say that. Your first two equations are fine (the KCL equations). Please answer the question you were asked in post #7... :smile:
I'm very sorry! Yes, the post in #7 makes sense to me.. I added in the top two equations and now I'm getting an answer. How does this look?
e8c0f2e98f41336218d47d3cdeca45a7.png

When I solve the the equation in post #7, i get the I2 that I got in my system of equations. So I guess my answer is right? If it is, I'm still a little unsure about the process my mind should go through when I need to write these equations. First I would write the first two equations that sets the currents equal to each other, and then pick 3 other paths at random for the KVL?
 
  • #13
justin15501 said:
First I would write the first two equations that sets the currents equal to each other, and then pick 3 other paths at random for the KVL?
No. When solving circuits like this choose either a KCL or KVL approach, and write those equations. For this circuit there were 2 KCL equations. Solve those for the node voltages, and then solve for the currents based on those node voltages.
 
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  • #14
berkeman said:
No. When solving circuits like this choose either a KCL or KVL approach, and write those equations. For this circuit there were 2 KCL equations. Solve those for the node voltages, and then solve for the currents based on those node voltages.
So I didn't get the answer right or I did? Because the 5 equations I used, 2 of them are KCL and 3 are KVL?

Assuming I'm wrong, the first thing I would do is write the 2 KCL equations which are:
e3f7a6467c1d61bb31d8a0244c2565de.png

The next step you say is to solve for the node voltages? This is where I start to get confused, sorry.. Can you explain a little more please?
 
  • #15
justin15501 said:
The next step you say is to solve for the node voltages? This is where I start to get confused, sorry.. Can you explain a little more please?
justin15501 said:
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There you are. Those are the node voltages. Then you just do I = V/R for each current you want to calculate.
 
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  • #16
berkeman said:
There you are. Those are the node voltages. Then you just do I = V/R for each current you want to calculate.
What if I wasn't given those two equations? Would I just solve it with the 5 equations I made up?
 
  • #17
justin15501 said:
Doesn't the whole big loop around count? Making it 4 loops? And I don't know how to write the KCL equations? Thanks for the reply!
When writing equations to solve for a set of variables it is essential that the equations be independent, otherwise you're not adding information to the set as you add non-independent (duplicate) information.

The circuit in this problem has three independent loops. Three equations will encapsulate all the information that you can glean by KVL alone. You can draw more than three loops on the figure, but a minimal set of three that incorporates all the components at least once is the best you can do for independent information. More loop equations will be only be duplicating information already obtained. This is why you need another source of information to supplement the information available from KVL. That's where the KCL equations come in.

Once you've chosen a reference node (indicated by the ground symbol in your figure), there are two "essential nodes" remaining. They happen to be labelled Va and Vb in the figure. You can write KCL current sums for those nodes to bring the total number of independent equations to 5, matching the number of unknowns.

It is possible to "solve" this circuit using KVL only using a method called "mesh analysis". There, three "pseudo currents" called "mesh currents" are defined, which are taken to circulate strictly within the three independent loops (see the circular arrows that I indicated in the image in post #5 -- they could correspond to the three mesh currents).

Once the three mesh currents are obtained, the branch currents are obtained by summing the mesh currents that flow through the given branches.
 
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  • #18
gneill said:
When writing equations to solve for a set of variables it is essential that the equations be independent, otherwise you're not adding information to the set as you add non-independent (duplicate) information.

The circuit in this problem has three independent loops. Three equations will encapsulate all the information that you can glean by KVL alone. You can draw more than three loops on the figure, but a minimal set of three that incorporates all the components at least once is the best you can do for independent information. More loop equations will be only be duplicating information already obtained. This is why you need another source of information to supplement the information available from KVL. That's where the KCL equations come in.

Once you've chosen a reference node (indicated by the ground symbol in your figure), there are two "essential nodes" remaining. They happen to be labelled Va and Vb in the figure. You can write KCL current sums for those nodes to bring the total number of independent equations to 5, matching the number of unknowns.

It is possible to "solve" this circuit using KVL only using a method called "mesh analysis". There, three "pseudo currents" called "mesh currents" are defined, which are taken to circulate strictly within the three independent loops (see the circular arrows that I indicated in the image in post #5 -- they could correspond to the three mesh currents).

Once the three mesh currents are obtained, the branch currents are obtained by summing the mesh currents that flow through the given branches.
e5f8afdb38cd420af4f3ee457d68001e.png

So I wrote 3 independent equations for each loop, and the summed up the currents. Would this be correct? You explanation made a tone of sense.
 
  • #19
justin15501 said:
What if I wasn't given those two equations? Would I just solve it with the 5 equations I made up?
If these equations are not given, you can always form them using node voltage method. Your 5 equation method is fine but it is unnecessarily increasing the number of variables. You can do it using only three variables as gneill demonstrated in #5.
Edit: I see gneill has already said that in detail in #17:-p...
 
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  • #20
justin15501 said:
e5f8afdb38cd420af4f3ee457d68001e.png

So I wrote 3 independent equations for each loop, and the summed up the currents. Would this be correct? You explanation made a tone of sense.
Yup. That looks good!
 
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  • #21
Thanks everyone for the help! I learned a lot!
 

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