How Does Kirchhoff's Law Apply to Calculating Currents and Voltage Differences?

[flyingpig]

Homework Statement

I so got this, but I just need your word to boost my ego

http://img263.imageshack.us/img263/7302/problemq.th.png

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Find the current in each resistor and the potential difference between a and b

The Attempt at a Solution

Before I even go find the current in each resistor, I need to find the branch, and let's just start and end there

http://img819.imageshack.us/img819/4913/partsol.th.png

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So as you can see the direction and the labelled I_a and I_b in the circuit.

Then I get

(1) $$12 - 2I_a - 4I_a + \frac{4}{3}I_b = 0$$

(2) $$8 - 6I_b - \frac{4}{3}I_b + \frac{4}{3}I_a = 0$$Let me explain, for equation one, I did the loop counterclockwise and the (4/3) accounts for the resistor equivalent in that circuit FOR the bigger loop

The second equation follows the same argument. My concern is that for the top smaller loop (I_a), should I have used $$12 - 2I_a - 4I_a + 4I_b - 2I_b= 0$$ instead? I am pretty sure I got equation 2 two.

 

[gneill]

You don't need to do anything with the resistances but use them as they're given. The 4/3 values you've written should just be 2.00's.

 

[flyingpig]

Which loop are you talking about?

 

[gneill]

You've indicated two loops; one with current Ia and one with current Ib. The equations for neither of them require a "4/3" resistance value. They should be 2.00's.

 

[flyingpig]

Why just 2.00s?

 

[cupid.callin]

Hi flyingpig
(Hope this thread is not just for Sammy, that will be awkward )

It doesn't matter if you calculate the net resistance and then apply KIRCHHOFF laws or apply them directly. even if you split the 4Ω resistance into 4 1Ω resistance in series, yo'll get the same result.

But you can't calculate net resistance if two resistances just because they are in series in the loop you have chosen. That will give wrong answer.

ONLY FIND NET RESISTANCE IN SERIES OR PARALLEL IF THEY ARE IN SERIES OR PARALLEL IN ORIGINAL CIRCUIT AND NOT IN THE LOOP CHOSEN

 

[flyingpig]

But how did gneil get just 2.00Ω?

 

[gneill]

Why just 2.00s?

Because it's a 2.00 Ohm resistor in the loops. The fact that it happens to "belong" to two loops doesn't matter; it is still just a 2.00 Ohm resister as far as the loop equations are concerned. The fact that it appears in both equations takes care (mathematically) of the relationship between the loop currents that pass through it.

 

[flyingpig]

But the 4.00Ω also belongs to both loops, so why only include 2.00Ω?

 

[cupid.callin]

If you are talking about the loop with both batteries then yes they both do share the 4Ω resistance.

But why are you so concerned about the resistances shared?

Just do it the old way, write eqn for both loops and solve simply, and here you can't calculate net resistance, that would be wrong

 

[flyingpig]

(1) $$12 - 2I_a - 4I_a +4I_b + 2I_b= 0$$

(2) $$8 - 6I_b + 4I_a + 2I_a = 0$$

 

[gneill]

But the 4.00Ω also belongs to both loops, so why only include 2.00Ω?

Really? The currents that you drew did not both pass through the 4 Ohm resistor. Perhaps you'd better explain precisely how you've chosen your loops.

I see two very obvious loops, namely the one beginning with the 8V supply and passing through the 6.00 Ohm resistor, then the 2.00 Ohm resistor, and returning to the 8V supply, and the other commencing with the 12.0V supply and passing through the 2.00 Ohm resistor and then the 4.00 Ohm resistor and back to the 12V supply. And it appeared that you drew the currents accordingly...

 

[cupid.callin]

(1) $$12 - 2I_a - 4I_a +4I_b + 2I_b= 0$$

(2) $$8 - 6I_b + 4I_a + 2I_a = 0$$

I really can't understand what loops you have taken ... can you show them?

 

[flyingpig]

Wrong loop, if we were to just take the 2Ω resistor then

(1) $$12 - 2I_a - 4I_a + 2I_b= 0$$

(2) $$8 - 6I_b + 2I_a = 0$$

 

[gneill]

Wrong loop, if we were to just take the 2Ω resistor then

(1) $$12 - 2I_a - 4I_a + 2I_b= 0$$

(2) $$8 - 6I_b + 2I_a = 0$$

You missed another 2Ib in the second equation. It's a loop! Ib must flow through each component in the loop, even those components that Ia also flows through!

 

[flyingpig]

(1) $$12 - 2I_a - 4I_a + 2I_b= 0$$

(2) $$8 - 6I_b + 2I_a - 2I_b= 0$$

I still find it unintuitive that we only need the 2 ohm resistor

 

[gneill]

Why? The loops are separate. The loop currents are confined to their loops. They may pass through shared components, but they are separate currents nonetheless.

 

[flyingpig]

Why? The loops are separate. The loop currents are confined to their loops. They may pass through shared components, but they are separate currents nonetheless.

That's an answer I can accept, but it isn't enough to convince me

 

[gneill]

I can only imagine that you must be fuzzy on the concept of what constitutes a loop and how KVL works around a closed path.

 

[SammyS]

OK.

(I've never been inserted into a title before!)

I'll read over the posts & get back to you soon, I hope.

 

[gneill]

Here's you circuit redrawn in a straightforward manner (without the stylistic "jogs" in the wiring), with two possible choices of loops shown. Both choices are fine, and will lead to the same answers if you write the loop equations properly. The only real restrictions on the set of loops you choose is that they don't cross themselves and they must as a set pass through every component in the circuit.

This means that when you "walk" around a loop (one of the closed paths indicated), you include every component that the loop as drawn passes through, and NEVER include components that it does not pass through -- that's the job for other loops that do include those components. It's okay for loops to share components.

When writing a term for a given component in a loop equation, you must include every loop current that passes through it. This is what "links" the simultaneous equations together.

 

[SammyS]

(1) $$12 - 2I_a - 4I_a + 2I_b= 0$$

(2) $$8 - 6I_b + 2I_a - 2I_b= 0$$

I still find it unintuitive that we only need the 2 ohm resistor

For the two currents you are using and the loops that naturally follow from them these are the correct equations.

I see that gneill and his considerable patience have gotten you to the right result.

The 2Ω resistor is the only component common to both loops and which has both currents passing through it. You could rewrite your two equations as:

(1) $$12 - 2(I_a-I_b) - 4I_a = 0$$  As you go around the upper little loop counter-clockwise, the current through the 2Ω resistor is I_a - I_b.

(2) $$8 - 6I_b - 2(I_b-I_a)= 0$$  As you go around the lower little loop counter-clockwise, the current through the 2Ω resistor is I_b - I_a.

My guess is that initially, you were mixing two methods of circuit analysis: Kirchhoff's & the superposition method.

 

[cupid.callin]

Here's how i drew the circuit with currents ...

And maybe you should look at some example of using Kirchhoff laws

Edit:

Here they are:

 

[flyingpig]

Here's how i drew the circuit with currents ...

And maybe you should look at some example of using Kirchhoff laws

Edit:

Here they are:

What do you guys use to draw your diagrams...? I've seen that guy's videos, they are a replacement for my professor basically

 

[flyingpig]

Here is my redrawn path, is this what you guys are implying?

http://img848.imageshack.us/img848/7302/problemq.th.png

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[flyingpig]

Here's you circuit redrawn in a straightforward manner

If only all of them look like that

This means that when you "walk" around a loop (one of the closed paths indicated), you include every component that the loop as drawn passes through, and NEVER include components that it does not pass through

But it is kinda tempting to think it does...

 

[cupid.callin]

What do you guys use to draw your diagrams...? I've seen that guy's videos, they are a replacement for my professor basically

i used paint (win 7)

 

[gneill]

Here is my redrawn path, is this what you guys are implying?

http://img848.imageshack.us/img848/7302/problemq.th.png

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Yes.

 

[gneill]

What do you guys use to draw your diagrams...?

I use Visio, then paste the result into Paint and save it in the desired format (jpg, png, etc.). I wasn't happy with the way Visio saved "non native" formats.

 

[SammyS]

I use Open Office Draw, then export as jpeg, or png or ..., then use gimp to re-size and/or trim the border if needed.

(Usually I simply avoid drawing if possible! LOL.)

 

[flyingpig]

What about the potential difference? My theory is that it is 12 - 8 = 4V because the terminals are opposite.

 

[SammyS]

After all this time?

The potential difference is equal to the voltage dropped across the 2Ω resistor.

 

[flyingpig]

The current running in the small loop is 2.545A and the bigger loop is 1.64A

So the current running through the 2Ω resistor is 0.908A.

So now you want me to 2Ω * 0.908A = 1.82V because this is the only resistor that intercepts both loops.

But I don't like this answer. Does it matter where a and b are placed?

 

[SammyS]

The current running in the small loop is 2.545A and the bigger loop is 1.64A

So the current running through the 2Ω resistor is 0.908A.

So now you want me to 2Ω * 0.908A = 1.82V because this is the only resistor that intercepts both loops.

But I don't like this answer. Does it matter where a and b are placed?

I thought we called these I_a & I_b.

If you want the answer, then you should want to: 2Ω * 0.908A = 1.82V .

As long as a is to the left of the batteries and the 2Ω resistor & b is to the right of all the resistors, their exact location doesn't matter.

 

[flyingpig]

But why is 12V - 8V wrong?