Solving Kirchhoff's Laws: I1, I2, I3

[IronHide]

Homework Statement

Relavent information is provided in the picture. Trying to solve the currents, I1, I2, I3.

Homework Equations

I(1)+I(2)+I(3)=0

The Attempt at a Solution

Too messy to even show.

Basically I'm having a massive trouble understanding kirchhoffs laws and this problem in particular. It's not that I don't understand the fundamentals of it, just how to approach and solve problems such as these.

EDIT: Direct link for image: http://tinypic.com/r/1znwkzt/5

http://tinypic.com/r/1znwkzt/5

 

[UltrafastPED]

Now use KVL to generate an equation for the two labeled loops ... recall that the sum of the voltage drops around each loop is zero, and use Ohm's law to write the voltage drops for each resistor.

 

[IronHide]

So when I do loop 1 per say; I have to take into account the 15V battery as well and -/+ it into my equation as well?

Cheers for the response too

 

[NascentOxygen]

So when I do loop 1 per say; I have to take into account the 15V battery as well

http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Yes, every element in the loop; the battery is just another element.

 

[IronHide]

Thanks will give it a try now, I can see myself ending up around a lot.. so many questions.. seems physics problems are infinite. ;)

 

[IronHide]

Feel so stupid that I can't even write the equations, either I am really having a huge blank, or I am really dumb. To be honest, need a step by step on how to write them even remotely properly.

 

[tiny-tim]

Hi IronHide! Welcome to PF!

Try KVL for loop 1.

You must have some idea, so show us which bits you can do. ​

 

[IronHide]

I did, I am just not confident.

Obviously the +12V but then I am confused as you go down the middle section, -15V, and the 50 ohm and 20 ohm resistors, use what current? And then I've got other stuff to worry about (Subject wise). Just the kind of the guy who will find out the method and see how stupid I've been and get all embarrassed cause it was so simple!

 

[tiny-tim]

Obviously the +12V but then I am confused as you go down the middle section, -15V, and the 50 ohm and 20 ohm resistors, use what current?

There's only three currents, I₁ I₂ and I₃.

Mark clearly on your diagram, with an arrow, which current is going through which resistor, and in which direction: that's the current you use for that resistor (and it will be + or - depending on whether the current is going the same way as that circular arrow).

So what do you think the KVL contribution should be for the 50Ω and 20Ω ? ​

 

[CWatters]

-15V is wrong.

Consider.. when "you" went through the 12V battery you went from the -ve to the + terminal and gained +12V.

When you go down through the 15V battery you are also going from the -ve to +ve terminal so it should be +15V.

Then for the resistor use ohms law. V = IR or in this case I₃*50. However if you look at the direction of I₃ you will see that there must be a voltage drop as you go through the resistor so the voltage across the resistor will be -I₃*50.

 

[IronHide]

So Loop One: 12+15-I₃*50-I₁*20?

Well I get the right answer for that part of the question.

Just like to say, a thanks to all who helped. I'm so grateful for the swift replies, and no one made me feel that dumb, and now I feel like how did I not realize this earlier. Thanks also for not just plainly writing the answer, and really making me work for it.

Cheers to all,

 

[IronHide]

Also I have another question like this, so can I send my working out to you Tiny-Tim, and make sure you think its all okay?

 

[tiny-tim]

Hi IronHide!

Also I have another question like this, so can I send my working out to you Tiny-Tim, and make sure you think its all okay?

No, private advice is against PF policy …

just start as many new threads as you like!

(a separate one for each question)

 

[tiny-tim]

So Loop One: 12+15-I₃*50-I₁*20?

Looks good.

Perhaps you'd better show us your loop 2 equation also, just to check.

 

[IronHide]

http://i40.tinypic.com/34ytcvq.jpg

Heres the question, so basically you treat it as a normal question? not even a junction really, and just work through it. One of the questions is the current through the 5k ohm resistor, and another is the potential diff across the 8k ohm resistor. So just using V=IR, you can work it out? or just have to use the junction rules etc.?

 

[tiny-tim]

One of the questions is the current through the 5k ohm resistor, and another is the potential diff across the 8k ohm resistor. So just using V=IR, you can work it out? or just have to use the junction rules etc.?

You'll always need to use both KVL and KCL (for KVL, you'll need one less equatioin than there are loops, eg if there were 3 loops, you'd need 2 KVL equations).

And yes, for the potential difference across the 8 kΩ resistor, you just use IR.

 

[UltrafastPED]

Yes - you use Ohm's law to get voltage drops for the resistors in terms of the currents, or currents in terms of the known voltage drops. You use KCL and KVL to generate enough equations to solve for the unknowns.

As your course progresses you will learn additional techniques which speed you through this process - including transformations between equivalent circuits. Most of the techniques are for linear systems.

 

[CWatters]

There are different ways to solve circuits but if you have just been taught kirchhoffs laws then that will probably be what they expect you to use. The general idea is to use KCL and KVL to write simultaneous equations which are then solved. If you have done it correctly you can solve for any unknown they ask for, eg the current through or voltage on any particular branch or component.

In the two cases you have posted there are three loops you could apply KVL to and two nodes (with more then two branches) that you can apply KCL to. So 5 equations.

In the latest example I would probably simplify it first by replacing resistors in series with one.

 

[IronHide]

Thankyou for all quick responses. I don't have answers, (not looking for them), but just wanting to check mine against them when I do it all. And I've been taught to use cramer's rule to solve, but that just seems like a massive waste of time sometimes.

 

[IronHide]

Loop 1 (4v one): 4-I₁*6000-I₃*8000=0

Loop 2 (2v one): 2-I₂*4000-I₃*8000=0

Look about right? Then just use cramers rule to solve?

 

[Enigman]

Can't check the equations, the univ. has blocked the site hosting the pic. sorry.
As for solving the equations: you need the equation you mentioned in first post, otherwise you have got three variables and two equations. Cramer's rule may be used, but I would do it with substitution- Well, stick to what you are comfortable with.

 

[UltrafastPED]

The KVL loops look OK; those plus the KCL for the main node will be sufficient for the solution.

 

[IronHide]

The KVL loops look OK; those plus the KCL for the main node will be sufficient for the solution.

So just make the equation of I₁+_I2+I₃=0, sounds good good to me.

 

[IronHide]

4-i₁*6000-i₃*8000 (1)

2-i₂*4000-i₃*8000 (2)

As I₃=I₁+I₂

Then sub to get rid of I_₂, and then again to sub into each other to get rid of I₂ to get: 2+4000*I₁=6-9000I*₁

Therefore I₁ = 0.000308 A

(Taking I₁ = 4v branch, I₂ = 2v branch, I₃ = 8 k ohm bit.)
Then I did using the answer from above into 4-i₁*6000-i₃*8000 , and found I₃=0.000269 A..

Does this answer seem right? as for this to work, I₂, would have to be negative? and I am not sure whether than works cause of the direction?

 

[UltrafastPED]

KCL says that the current entering a junction is equal to the current leaving a junction ... so unless you have been given directions for the currents, the initial directions are arbitrary - but you have to label them to agree with your equation. Thus if they all flow out of a junction, or all flow into a junction you just set their sum to zero.

In your case you have chose to set your currents to flow in the same directions as your loops ... hence I3 is outbound, and I1 and I2 are inbound at the junction. This doesn't make any difference as long as you are consistent: if your initial choices are "correct" the currents will all be positive; if you picked a wrong direction then the solution for that current will be negative.

You can make them all positive by changing the directions of the current flows that are going "the wrong way" ... but this isn't necessary unless your problem requires you to label all of the currents as they actually flow.

To check for errors it is always good to verify that the currents do obey KCL, and that the voltages do obey KVL for each loop. This self-consistency check should be done for every problem ... it will help you catch errors of logic and algebra.

 

[IronHide]

Yes I₂, came out negative for me, which is weird because of the direction it is in relative to I₂, but I guess at the junction, it's negative because of which way loop 1 flows. Thanks for the help, calculated all voltage drops across loop 1, which came to 0, so works well.

 

[tiny-tim]

… This doesn't make any difference as long as you are consistent: if your initial choices are "correct" the currents will all be positive; if you picked a wrong direction then the solution for that current will be negative.

yup!

when applying Kirchhoff's laws, the arrows you draw are just an intelligent guess …

It does not matter if any arrow is the wrong way round: this will simply result in a negative value for the current when the equations are solved.