Trouble with multiple-current circuit diagram problems.

1. Jan 30, 2010

QuietQuasar

I am having trouble with circuit diagram involving multiple currents. We were taught the method of using junction/loop rules and systems of equations (or matrices) to find the current or voltage in a given spot. I think I understand the basic concept but am setting up my equations incorrectly.

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

Find the current in the 10.0$$\Omega$$ resistor in the drawing (V1 = 15.0 V and R1 = 27.0$$\Omega$$)
http://dl.dropbox.com/u/307214/20_82alt.gif [Broken]

2. Relevant equations
Kirchoff's rules...I don't think any actual equations?

3. The attempt at a solution
First I decided to pick a common start point for the problem as the lower right corner.
Then I labeled the currents in the circuit: I$$_{A}$$ as going to the right in the left branch, I$$_{B}$$ going to the right in the right branch, and I$$_{C}$$ going up across the middle.
I then setup 2 of my equations by going in a loop from my start point and adding or subtracting the voltages, so:
+15 + 10 - 27I$$_{A}$$ - 5I$$_{B}$$ =0 for going around the left branch and
15 + 10 - 10I$$_{C}$$ - 5I$$_{B}$$ =0 for going up the middle.
I then determined (perhaps incorrectly) that I$$_{A}$$+I$$_{B}$$ =I$$_{C}$$
From here I setup matrices on my calculator as such:
Matrix [A]
-27 -5 0
0 -5 -10
1 1 -1
and
Matrix
-25
-25
0
I then did [A]$$^{-1}$$* for the results:
.7042
1.1972
1.9014
None of these answers are correct :/
Any and all help is greatly appreciated!
-QQ

Last edited by a moderator: May 4, 2017
2. Jan 30, 2010

w3390

Try this approach. Start a current from the battery in the center of the circuit diagram. The current moves towards the top where it splits into two other currents, I2 going to the left and I3 going to the right. You can then write an equation for each loop. This will give you two equations but three variables. To solve this problem, you can use the fact that the current I1 is composed of both I2 and I3 (i.e. I1 = I2 + I3). This will allow you to eliminate a variable and solve for I1.

In this sort of diagram, I find it useful to start at the battery in the center of the circuit so that the current splits and rejoins at the bottom. I find it much easier to visualize what currents are adding together.

Last edited: Jan 30, 2010