# Need help to find I in a circuit (mesh analysis?)

1. Feb 8, 2013

### asdf12312

1. The problem statement, all variables and given/known data
http://tinypic.com/r/3478nle/6
need to find the current I

2. Relevant equations
was thinking about using mesh current analysis, but our teacher hasn't gone over it yet.

3. The attempt at a solution
no clue, totally stumped on what to do. was thinking mesh analysis and i attempted that, but i had no idea what i was doing after a while.

2. Feb 8, 2013

### Staff: Mentor

Mesh analysis looks like a fine idea. Why don't you show how you got started so we can see how to help you when you get stuck?

3. Feb 8, 2013

### asdf12312

well i assigned currents I1, I2, and I3 to the right, left, and top loops respectively. moving in clockwise direction this what i got:

right loop:
3V+21(I1)-6(I2)-6(13)=0

left loop:
3V-6(I1)+21(I2)-6(I3)=0

top loop:
-6(I1)-6(I2)+21(I3)=0

now the trouble i'm having is the part where i'm supposed to solve all 3 equations simultaneously. my method is getting I1=I2=I3 in the bottom two equations and then plugging them into eq#1 so i can solve for I1 (only current i need to find).

cancelling out I3:
3.5(3V-6(I1)+21(I2)-6(I3)=0)
-6(I1)-6(I2)+21(I3)=0
--
10.5V-21(I1)+73.5(I2)-21(I3)=0
-6(I1)-6(I2)+21(13)=0
--
-27(I1)+67.5(I2)=-10.5
67.5(I2)=27(I1)-10.5
I2=0.4(I1)-0.16

cancelling out 12:
3V-6(I1)+21(I2)-6(I3)=0
3.5(-6(I1)-6(I2)+21(I3)=0)
--
3V-6(I1)+21(I2)-6(I3)=0
-21(I1)-21(12)+73.5(I3)=0
--
-27(I1)+67.5(I3)=-3V
67.5(I3)=27(I1)-3
I3=0.4(I1)-0.044

substituting into eq#1:
21(I1)-6(I2)-6(13)=-3
21(I1)-6(0.4(I1)-0.16)-6(0.4(I1)-0.044)=-3
21(I1)-2.4(I1)+0.96-2.4(I1)+0.26=-3
16.2(I1)=-4.22
I1=-0.26A

but i know i did this wrong...cause i got a negative current.

4. Feb 8, 2013

### Staff: Mentor

Nothing wrong with a negative current, it just means that your assumed current direction for the loop is the opposite of the actual current direction. Your assumed direction was a clockwise mesh current, right? So the actual current is counterclockwise.

Note the direction that was assigned to I in the diagram; It's the opposite of your assumed mesh current. So I = -I1 ...