How Do Kirchhoff's Rules Determine Currents in a Dual Battery Circuit?

[roam]

Homework Statement

Use Kirchhoff's rules to determine the currents flowing through the two batteries in the following circuit:

[PLAIN]http://img217.imageshack.us/img217/2779/circt.jpg

Homework Equations

Kirchhoff's junction rule and loop rule.

The Attempt at a Solution

The correct answers must be: I_12V=2A, I_3V=0.4A (right to left).

Here's how I labled the currents in the circuit:

[PLAIN]http://img84.imageshack.us/img84/6831/circtcurrents.jpg

I think the current through the 12V battery is I₁, and current through the 3V battery is I₄. So here's four equations in four unknowns:

I₁-1.379-I₃=0
I₄0.5-I₆=0
3-4I₆+4I₃-12+3.723=0
11.274-4I₆+4I₁=0

I put the following equations in the calculator but got an error:

[itex]\left\{\begin{matrix}I_1-I_3=1.379 \\ I_4-I_6=0.5 \\ -I_6+I_3=1.32 \\ -I_6+I_1=-2.82\end{matrix}\right.[/itex]

So is there something wrong with my method or I just need to try solving the equations somehow?

 

[gneill]

I think the current through the 12V battery is I₁, and current through the 3V battery is I₄. So here's four equations in four unknowns:

I₁-1.379-I₃=0
I₄0.5-I₆=0
3-4I₆+4I₃-12+3.723=0
11.274-4I₆+4I₁=0

There's something fishy about your equations. Where are the numbers (like 1.379) coming from? I think you may be making some unwarranted assumptions about certain current values.

 

[NascentOxygen]

Why mark in I₆, when you already have labelled the current though that branch I₃? I also wonder at the origin of your magic numerals, such as 1.379?

 

[roam]

There's something fishy about your equations. Where are the numbers (like 1.379) coming from? I think you may be making some unwarranted assumptions about certain current values.

Oops, I forgot to mention that. 1.379 is the current around the left upper loop it can be found using Kirchhof's voltage law, similarly 0.5 is the voltage in the rightmost loop found using KVL:

3-6I₅=0
I₅=0.5

And

12-I₂(6+2.7)=0
I₂=1.379

So, is there anything wrong with that?

Why mark in I₆, when you already have labelled the current though that branch I₃?

Because I had to use the junction rule...

 

[cmb]

How did you arrive at "3-4I₆+4I₃-12+3.723=0" !?


I₆ = I₃

 

[roam]

How did you arrive at "3-4I₆+4I₃-12+3.723=0" !?


I₆ = I₃

In that case that loop's equation will become 3-4I-12-2.7I=0, but how does this help us? How can we distinguish the current through the 3V battery from the one through the 12V battery?

 

[cmb]

So, what are the currents at the 3-way junction right in the middle?

 

[gneill]

Oops, I forgot to mention that. 1.379 is the current around the left upper loop it can be found using Kirchhof's voltage law, similarly 0.5 is the voltage in the rightmost loop found using KVL:

3-6I₅=0
I₅=0.5

And

12-I₂(6+2.7)=0
I₂=1.379

So, is there anything wrong with that?

Yes, there's something wrong with that! The reason that one writes loop equations and solves them as a set of simultaneous equations is because there are inter-dependencies between the currents in the different loops. This is due to the components that straddle different loops and so carry currents from each; the voltage drop across those components depends upon currents in both loops that they belong to!

You can't just pick a loop and write KVL for it alone and solve for the current in it.

 

[roam]

Yes, there's something wrong with that! The reason that one writes loop equations and solves them as a set of simultaneous equations is because there are inter-dependencies between the currents in the different loops. This is due to the components that straddle different loops and so carry currents from each; the voltage drop across those components depends upon currents in both loops that they belong to!

You can't just pick a loop and write KVL for it alone and solve for the current in it.

Okay so I labled the currents and used KVL like this:

[PLAIN]http://img703.imageshack.us/img703/7882/circuitis.png

I get these 3 equations:

3-6I₁=0 => I₁=0.5...(right loop)

3-4I₃-12-2.7(I₃-I₂)=0 ...(left loop)

12-6I₂-2.7(I₂-I₃)=0 ...(lower loop)

When I solved the last two equations for two unkonwns I got

I₂ = -2.052

I₃ = -2.17

And the current through the 12V battery is I₃-I₂=-2.17+2.052=-0.12, and through the 3V battery: I₁-I₃=0.5+2.17=2.67.

So why is my answer still wrong? I'm sure it's not an algebraic error because I checked the solution of the equations on the calculator. What's wrong with my approach now?

 

[cmb]

...At the risk of being boring, I'll ask the question again: So, what are the currents at the 3-way junction right in the middle?

 

[gneill]

You've drawn I1 and I3 so that they both flow in the same direction through the 3V battery, so check your calculation for the current there.

I solved your equations as given and I did not reach the same values that you did.

 

[roam]

You've drawn I1 and I3 so that they both flow in the same direction through the 3V battery, so check your calculation for the current there.

I solved your equations as given and I did not reach the same values that you did.

Right. So they add up I₁+I₃=0.5-2.17=-1.67, but i'ts still not correct... the answer has to be 0.4A.

 

[gneill]

Right. So they add up I₁+I₃=0.5-2.17=-1.67, but i'ts still not correct... the answer has to be 0.4A.

That's because, as I stated, the values that you arrived at for I2 and I3 from your equations are not correct. Take another look at solving the two equations.

 

[cmb]

C'mon people, this is embarrasing. These aren't the equations to be solved.

You should be solving 6 simultaneous equations in this example. Let's get those right first, eh?

Write out the 6 equations you need to start off with.

 

[gneill]

C'mon people, this is embarrasing. These aren't the equations to be solved.

You should be solving 6 simultaneous equations in this example. Let's get those right first, eh?

Write out the 6 equations you need to start off with.

@cmb: There are only three meshes, one of which is "orphaned" by being separated from the others by constrained voltages (a battery on one leg and a short circuit on another). So that's two mesh equations.

A suitable choice of common reference point would leave one independent node for nodal analysis. So only one equation to find that node voltage. With the node voltages, the branch currents become trivial.

So how do you figure that six equations are required?

 

[cmb]

Well, OK, you are right but the level of the op at the moment suggests he might be better off picking up each circuit leg separately, than each mesh. You still have to deal with 'I2+I1' as a 'variable' to solve. It'd be an equation no more complicated than that. There are 6 legs with different currents in. Therefore you need 6 equations. The first post suggests that solving as 3 meshes with 3 unknowns was not the way the op was directed to solve it.

 

[gneill]

Well, OK, you are right but the level of the op at the moment suggests he might be better off picking up each circuit leg separately, than each mesh. You still have to deal with 'I2+I1' as a 'variable' to solve. It'd be an equation no more complicated than that. There are 6 legs with different currents in. Therefore you need 6 equations. The first post suggests that solving as 3 meshes with 3 unknowns was not the way the op was directed to solve it.

I might agree if the OP hadn't identified and written correct equations for the meshes. Also, the question statement simply said to use "Kirchhoff's rules" to find the battery currents, which sort of leaves the field wide open as to method -- they're all based upon Kirchhoff's Laws! Mesh analysis is simply KVL applied to particular loops, for example.

At the moment, OP roam's problem seems to lie in the algebra.