I really don't understand Kirchhoff's Law

[flyingpig]

Homework Statement

http://img827.imageshack.us/img827/6426/81545181.th.png Find all the currents I₁, I₂, I₃

2. The attempt at a solution

I find that solving the numbers is trivial, all I got to do is isolate them and put them in an augmented matrix and solve. But setting them is really difficult for me.

I really don't understand what my book is doing.

For the Loop_abcd, in the path d → a the book subtracted -2.0Ω(I₃), WHY?

Isn't this what is actually happening?

http://img156.imageshack.us/img156/4176/98865057.th.png

Because I am assuming that is what's happening to I₁ for the 6.0Ω resistor in Loop_befcb

Now my other question is, why is it 6.0Ω(I₁) and not 6.0Ω(1₂)?3. Solutions given by book

Loop_abcd = 10.0V - 6.0Ω(I₁) - 2.0Ω(I₃)

Loop_befcb = -4.0Ω(I₂) - 14.0V + 6.0Ω(I₁) = 0

 

[futurepocket]

Consider each loop separately. Assign a current to each branch (doesn't matter the direction). Set up a junction rule equation based on the sum of currents and set up two or more loop rules. Isolate and solve for unknown variables. In your specific case, I1 + I2 = I3

 

[sona1177]

Homework Statement

http://img827.imageshack.us/img827/6426/81545181.th.png


Find all the currents I₁, I₂, I₃

2. The attempt at a solution

I find that solving the numbers is trivial, all I got to do is isolate them and put them in an augmented matrix and solve. But setting them is really difficult for me.

I really don't understand what my book is doing.


For the Loop_abcd, in the path d → a the book subtracted -2.0Ω(I₃), WHY?

Isn't this what is actually happening?

http://img156.imageshack.us/img156/4176/98865057.th.png

Because I am assuming that is what's happening to I₁ for the 6.0Ω resistor in Loop_befcb

Now my other question is, why is it 6.0Ω(I₁) and not 6.0Ω(1₂)?


3. Solutions given by book

Loop_abcd = 10.0V - 6.0Ω(I₁) - 2.0Ω(I₃)

Loop_befcb = -4.0Ω(I₂) - 14.0V + 6.0Ω(I₁) = 0

Try reducing the circuits down to simplified ones by looking at each one separately. Follow a path around the loop to find ones that are in series or in parallel and then combine the resistors appropriately. Once you get to a simplified circuit, work backwards to the more complicated ones.

 

[flyingpig]

I don't think you guys are understanding properly...

THIS IS WHERE I AM HAVING PROBLEMS WITH. My path d to a is wrong or the book is

 

[SammyS]

Hello again old friend, flyingpig.

As you go from d to a, you are going in the same direction as the assumed current, I₃. Do you see why?

As the current passes through the 2.0 Ω resistor, it loses energy. In terms of electric potential there is a voltage drop equal to the current times the resistance.

A page or two prior to this example, Serway gives instructions on what sign to use when traversing a resistor or a voltage source as you make your way around a loop.

 

[flyingpig]

Hello again old friend, flyingpig.

As you go from d to a, you are going in the same direction as the assumed current, I₃. Do you see why?

Yes and it therefore it should be -IR

As the current passes through the 2.0 Ω resistor, it loses energy. In terms of electric potential there is a voltage drop equal to the current times the resistance.

A page or two prior to this example, Serway gives instructions on what sign to use when traversing a resistor or a voltage source as you make your way around a loop.

Yes yes, I am using the dumb book, and it says it is -IR

 

[SammyS]

Looks to me like it says: -(2.0Ω)(I₃), which is -IR for the quantities involved.

 

[flyingpig]

What about the 6.0Ω resistor? As you go from c → b, the current flows in the same direction and hence it should be -IR right? They have +IR and it really confuses me

 

[SammyS]

No. They have have - 6.0Ω(I₁) as they go from b → c.

From you post #1: " Loop_abcd = 10.0V - 6.0Ω(I₁) - 2.0Ω(I₃) " = 0 (I added this.)

 

[flyingpig]

I am talking about Loop_befcb = -4.0Ω(I₂) - 14.0V + 6.0Ω(I₁) = 0

 

[SammyS]

I could have sworn you had asked about b → c , not c → b. (Sorry for the brain fart.)

That's a different loop, which you are traversing in the opposite direction relative to the other loop. You are comparing the energy in charge carriers after they have expended their energy in the resistor with charge carriers before they have expended their energy in the resistor. You go from the lower to higher energy end of the resistor.

 

[flyingpig]

They are really related to each other.

In the path from c to b for the top loop

http://img64.imageshack.us/img64/2377/87365174.th.jpg

Uploaded with ImageShack.us

This is how the current is coming back, that is from c to b this should be -IR, but they have +IR

Now the same situation for the 2.0Ω resistor.

http://img849.imageshack.us/img849/6426/81545181.th.png

Uploaded with ImageShack.us

Notice how they have a -IR like it should be?

 

[flyingpig]

Can we use the word traveling and not traversing?

 

[SammyS]

Sure. ...but nothing wrong with expanding your vocab!

Have a good one!

 

[cepheid]

Can we use the word traveling and not traversing?

Actually no. To traverse means to "travel across" (something). (EDIT: and just to pre-empt your inevitable nitpicking, more generally it could mean to travel "along the entire extent of" something). So "traverse" can take a direct object, which is why he said "traversing the loop."

You can't say "travelling the loop." You'd have to add the word "around" in. But why use two words when you can use one that means the same thing?



Okay, I'm done being obnoxious now.

 

[flyingpig]

Sure. ...but nothing wrong with expanding your vocab!

Have a good one!

WHAT! But you didn't answer my other question!?

Actually no. To traverse means to "travel across" (something). So "traverse" can take a direct object, which is why he said "traversing the loop."

You can't say "travelling the loop." You'd have to add the word "around" in. But why use two words when you can use one that means the same thing?



Okay, I'm done being obnoxious now.

...

 

[cepheid]

WHAT! But you didn't answer my other question!?
...

Heh, what are you freaking out about? Maybe he had to go somewhere?

What is your question anyway? Basically as SammyS already explained, the potential decreases in going across a resistor, due to energy losses incurred while traversing (travelling across) it. So, if the current is assumed to flow across a resistor from right to left, the potential at the right end of the resistor should be assumed to be higher than the potential at the left end (by an amount IR).

If the current is assumed to flow across the resistor from left to right, the potential at the left end should be assumed to higher than the potential at the right end (by an amount IR). If your assumed polarity is wrong, you'll get a negative answer for V and I, which is fine.

In the first picture in your latest post, the current is flowing across the resistor from right to left. So the potential at the end of the resistor that is attached to point c is higher than the potential at the other end, which is attached to the battery (let's call this end point e). Therefore, the voltage drop in going from c to b is the sum of the voltages across the resistor and across the battery:

V_cb = V_ce + V_eb = IR + V_batt

where I've used the notation V_xy is defined as V_x - V_y

Does that clear things up?

 

[flyingpig]

So, if the current is assumed to flow across a resistor from right to left, the potential at the right end of the resistor should be assumed to be higher than the potential at the left end.

Yes and it should be -IR

http://img4.imageshack.us/img4/6605/24419253.th.jpg

Uploaded with ImageShack.us

 

[cepheid]

Yes and it should be -IR

http://img4.imageshack.us/img4/6605/24419253.th.jpg

Uploaded with ImageShack.us

The part of my post that you quoted, and what you wrote underneath it (and on the diagram) are in direct contradiction. Perhaps it would help if I were more explicit:

In this example, the current is flowing from right to left. Therefore, the potential at the right end (point c) is HIGHER than the potential at the left end (point e) by an amount IR.. You've stated that you agree with me on this.

Mathematically, the red part of what I just said is:

V_c > V_e

→ V_c - V_e > 0

Agreed?

The blue part of what I just said includes the additional information that the difference is given by IR:

V_c - V_e = +IR

Hence:

V_ce ≡ V_c - V_e = +IR

This is now step by step. Is that a better explanation?

 

[flyingpig]

Then what about the 2.00Ω resistor? It follows the same argument, but it is -IR

 

[cepheid]

Then what about the 2.00Ω resistor? It follows the same argument, but it is -IR

Using our logic, and given the assumed direction of I3, we have

Vda = I3*R

Vad = -I3*R

So, since the potential is higher at d than at a, you LOSE I3*R volts in going from d to a. So if you're tallying voltages around a loop, you'd subtract I3*R for the portion from d to a, since you went from high to low.

The same true for the journey from c to b.

 

[cepheid]

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]

Sorry for not reading your post right now because I am half asleep (been at it since this afternoon)

But I used something called Maxwell's Loop Method and I got (all going counterclockwise)

(1) 10 - 6I_1 + 14 - 4I_1 + 6I_2= 0

(2) -10 - 2I_2 - 6I_2 + 6I_1 = 0

Solving I get I₁ = 3A and I₂ = 1A

I can't help but feel there is something wrong. How do I match my I₁ and I₂ to the question?

Also, when we apply Maxwell's method, can we ever apply it to the ENTIRE loop?

Thanks! I am going to slee

 

[flyingpig]

Also, since I₁ + I₂ isn't 0 (it is 4A), does that mean I screwed it up?

 

[SammyS]

They are really related to each other.

In the path from c to b for the top loop

http://img64.imageshack.us/img64/2377/87365174.th.jpg

Uploaded with ImageShack.us

This is how the current is coming back, that is from c to b this should be -IR, but they have +IR

"They" show the current, I₁, to be going in the direction b → c. (See your first http://img827.imageshack.us/i/81545181.png/" .) As you travel around the Loop_befcb, you go c → b, opposite the current I₁. Therefore, it's +(6.0Ω)I₁.

Now the same situation for the 2.0Ω resistor.

http://img849.imageshack.us/img849/6426/81545181.th.png

Uploaded with ImageShack.us

Notice how they have a -IR like it should be?

Both are the way they "should be" !

 

[Dadface]

Hi flying pig.In your opening post you wrote down the "solutions given by the book" but these don't seem to make sense.In one equation you wrote:
loop abcd=10V-6I1-2I3
How can a loop equal the things on the right side of the equation.Did you mean to say that :
0=10V-6I1-2I3?
You need to look at the second equation also.

 

[Angry Citizen]

Sorry for not reading your post right now because I am half asleep (been at it since this afternoon)

But I used something called Maxwell's Loop Method and I got (all going counterclockwise)

(1) 10 - 6I_1 + 14 - 4I_1 + 6I_2= 0

(2) -10 - 2I_2 - 6I_2 + 6I_1 = 0

Solving I get I₁ = 3A and I₂ = 1A

I can't help but feel there is something wrong. How do I match my I₁ and I₂ to the question?

Also, when we apply Maxwell's method, can we ever apply it to the ENTIRE loop?

Thanks! I am going to slee

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.

 

[SammyS]

...

http://img64.imageshack.us/img64/2377/87365174.th.jpg

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.

 

[flyingpig]

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."

Then why did you say

Vda = I3*R

Vad = -I3*R

 

[flyingpig]

Hi flying pig.In your opening post you wrote down the "solutions given by the book" but these don't seem to make sense.In one equation you wrote:
loop abcd=10V-6I1-2I3
How can a loop equal the things on the right side of the equation.Did you mean to say that :
0=10V-6I1-2I3?
You need to look at the second equation also.

Sammy corrected it already, I just forgot to edit it and now I can't edit it. God why does PF delete the edit button after a while?

 

[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.