I really don't understand Kirchhoff's Law

[flyingpig]

Please note that I'm studying Kirchoff's Laws the same as you, but I believe you did something wrong. Namely, you did not solve for three equations in three unknowns, which is what the problem called for.

From what I understand -- and PF, please correct me -- Kirchoff's Laws are arbitrary. You assume current flows in such-and-such direction. If you're wrong, you get a negative value, in which case you can just flip some signs around until you get it all correct. A given complex circuit has many more possible equations than you could ever need, from the junction and loop rules. This is really a shotgun approach, not a sniper approach. Just keep spewing out equations until you get an answer that works. From what I understand of Kirchoff's Laws, the math will work itself out.

No the method that I just learned and used doesn't even concern about current splitting. I mean once I get the currents, I can work backwards.

 

[flyingpig]

I should have included this comment in one of my previous posts.

In the above figure, you show the direction of the current, I, to be in the opposite direction of what I₁ is in the figure from the textbook. Therefore, I = -I₁, where I is the current in the above figure, and I₁ is the current in the figure from your textbook.

I would write more: about being consistent with labeling ... , but you do have a tendency to be unclear about what particular point you are addressing when replying. I'll be patient & wait for another opportunity.

Let's try another problem...this one was just plain ambiguous with all the arrows they already labelled it for us

 

[SammyS]

Let's try another problem...this one was just plain ambiguous with all the arrows they already labelled it for us

Actually, it's NOT ambiguous at all.

 

[cepheid]

Then why did you say

Vda = I3*R

Vad = -I3*R

I said that because it was true! There is no contradiction here. V_da is positive, meaning that V_d is higher than V_a, so if you go from d to a, the voltage drops, and hence the change in potential is negative. The change in potential is the final minus the inital, which is actually V_a - V_d. That's what I was trying to explain better with my post #22:

I just thought of a better way of explaining it. If you go from point d to point a, then V_a is your final potential, and V_d is your initial potential. In all cases, the change in potential on a journey between two points is given by:

V_final - V_initial

In this case that is given by:

V_a - V_d

Since our assumed direction of current leads us to believe that V_d > V_a, the change in potential will be negative. That's what I meant when I said, "you're going from high to low, hence you subtract."

 

[flyingpig]

I am just going to use that Maxwell's method from now on, I am not even going to bother with Kirchoff's method, it's too dumb (because I can't do it). It only creates more variables

 

[Dadface]

Don't give up on it.In a nutshell this is what you do.
1.Look at your circuit.You will see that it is made of loop(s).
2.Apply kirchhoffs first law to mark in the currents eg at a junction a current I may split into I1 and I2.It doesn't matter if you mark the currents in the wrong directions because this will show up when you do the maths.
3.Apply the second law to as many loops as are necessary.You will now have some simultaneous equations to solve.Its the maths bit at the end that takes most of the time.

Practise some questions and you will get the hang of it.Good luck.

 

[sona1177]

Don't give up on it.In a nutshell this is what you do.
1.Look at your circuit.You will see that it is made of loop(s).
2.Apply kirchhoffs first law to mark in the currents eg at a junction a current I may split into I1 and I2.It doesn't matter if you mark the currents in the wrong directions because this will show up when you do the maths.
3.Apply the second law to as many loops as are necessary.You will now have some simultaneous equations to solve.Its the maths bit at the end that takes most of the time.

Practise some questions and you will get the hang of it.Good luck.

Exactly, this was always my problem. And then I was in your position, flyingpig. I had use other means to get the answer because when I tried to do a problem using Kirchhoff's loop rule I'd make a silly error with the mathematics. And as you know, one mistake with a +/- will mess everything up. What I can suggest is (and what helped me), try doing a much simpler problem using Kirchhoff's loop rule (if you have time) and then move to these more complex ones. This way you'll know if you made a mistake with the mathematics or if you are having a hard time understanding the actual rule.

Best of luck.

 

[flyingpig]

Now the problem I have is the individual junction when it crosses with another one.

 

[sona1177]

Now the problem I have is the individual junction when it crosses with another one.

You only have to focus on the path you have chosen.

 

[Jokerhelper]

http://en.wikipedia.org/wiki/Mesh_analysis
http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

Not sure why I bumped it, but mesh analysis can be quite useful and quick to solve for currents in these type of circuits.

 

[SammyS]

http://en.wikipedia.org/wiki/Mesh_analysis
http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

Not sure why I bumped it, but mesh analysis can be quite useful and quick to solve for currents in these type of circuits.

Hello Jokerhelper.

I'm not sure why you bumped it either. flyingpig does have conceptual problems with Kirchhof's Laws.

I doubt that Mesh Analysis will help him. He is having problems applying Kirchhoff's Voltage Laws, and as the Wikipedia link says, "Mesh analysis uses Kirchhoff’s voltage law to solve these planar circuits".

 

[Jokerhelper]

Hello Jokerhelper.

I'm not sure why you bumped it either. flyingpig does have conceptual problems with Kirchhof's Laws.

I doubt that Mesh Analysis will help him. He is having problems applying Kirchhoff's Voltage Laws, and as the Wikipedia link says, "Mesh analysis uses Kirchhoff’s voltage law to solve these planar circuits".

From what I can tell, the OP's problem seemed more to deal with understanding things like passive reference configuration and why certain voltages are +IR and others are -IR based on the direction of the loop and the assigned current, rather than KVL itself. The nice thing about mesh analysis is that one doesn't have to worry about the assigned current directions right until the end, since you're just solving for the mesh currents really.

 

[Metaleer]

The problem with mesh analysis is that it can't be used for planar circuits (perhaps this isn't much of a problem as beginner circuit analysis usually features all planar circuits) and that much of the physics is lost, as writing down the mesh equations is mechanical; summing up the values of the resistors each mesh, and the EMF's, etc...

It is better to start off with Kirchhoff's laws, and when you get the hang of this, go to the more systematic methods to solve circuit analysis problems, such as nodal or mesh analysis.