Current Through Two Batteries

  • Thread starter roam
  • Start date
  • #1
1,266
11

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 [Broken]


Homework Equations



Kirchhoff's junction rule and loop rule.

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:

[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? :confused:
 
Last edited by a moderator:

Answers and Replies

  • #2
gneill
Mentor
20,913
2,862
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.
 
  • #3
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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
1,266
11
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...
 
  • #5
cmb
868
43
How did you arrive at "3-4I6+4I3-12+3.723=0" !!?!


I6 = I3
 
  • #6
1,266
11
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?
 
  • #7
cmb
868
43
So, what are the currents at the 3-way junction right in the middle?
 
  • #8
gneill
Mentor
20,913
2,862
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.
 
  • #9
1,266
11
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 [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:
  • #10
cmb
868
43
...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
gneill
Mentor
20,913
2,862
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
1,266
11
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.
 
  • #13
gneill
Mentor
20,913
2,862
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.
 
  • #14
cmb
868
43
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
gneill
Mentor
20,913
2,862
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: :confused: 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
cmb
868
43
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
gneill
Mentor
20,913
2,862
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.
 

Related Threads on Current Through Two Batteries

Replies
14
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
8
Views
20K
  • Last Post
Replies
7
Views
2K
Replies
20
Views
8K
Replies
17
Views
625
Replies
8
Views
3K
Replies
1
Views
8K
Replies
3
Views
11K
  • Last Post
Replies
2
Views
2K
Top