# Homework Help: Current Through Two Batteries

1. Oct 25, 2011

### roam

1. The problem statement, all variables and given/known data

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 [Broken]

2. Relevant equations

Kirchhoff's junction rule and loop rule.

3. The attempt at a solution

The correct answers must be: I12V=2A, I3V=0.4A (right to left).

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

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

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:

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

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

Last edited by a moderator: May 5, 2017
2. Oct 25, 2011

### Staff: Mentor

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.

3. Oct 25, 2011

### Staff: Mentor

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?

4. Oct 26, 2011

### roam

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?

Because I had to use the junction rule...

5. Oct 26, 2011

### cmb

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

I6 = I3

6. Oct 26, 2011

### roam

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?

7. Oct 26, 2011

### cmb

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

8. Oct 26, 2011

### Staff: Mentor

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.

9. Oct 27, 2011

### roam

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

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

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?

Last edited by a moderator: May 5, 2017
10. Oct 27, 2011

### 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?

11. Oct 27, 2011

### Staff: Mentor

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.

12. Oct 27, 2011

### roam

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.

13. Oct 27, 2011

### Staff: Mentor

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.

14. Oct 28, 2011

### 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.

15. Oct 28, 2011

### Staff: Mentor

@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?

16. Oct 28, 2011

### 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.

17. Oct 28, 2011

### Staff: Mentor

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.