Electrical Circuit Analysis: Understanding Current Flow and Direction

[James98765]

I am currently working on a physics homework problem that is giving me some trouble. The problem involves circuit analysis. Here it is:

How much current flows through the bottom wire shown below, and in which direction?

(http://www.cramster.com/answers-jul-08/physics/confused-current-flows-bottom-wire-inthe-figure_286422.aspx)

The answer in the book says "2.1 A, Left to Right". I cannot see how the book came up with that solution.

I made several attempts to solve the problem but the biggest issue I am struggling with is conceptual. If I can understand which way the current moves through the circuit then I think I can determine how much current moves through the bottom wire. I think that the current moves through two loops as follows:

Upward from the 15V battery through the 10 Ohm resistor
Downward through the 12 Ohm and 24 Ohm resistors
From the 24 Ohm resistor back to the 15V battery and from the 12 Ohm resistor to the 9V battery

Upward from the 9V battery through the 6 Ohm resistor
Downward through the 12 Ohm and 24 Ohm resistor
From the 24 Ohm resistor back to the 15V battery and from the 12 Ohm resistor to the 9V battery

Can somebody somebody let me know if I am making any major mistakes in my assumption of the current movements? Thanks for your help!

(I apologize for posting this question in the Classical Physics Forum instead of the Homework Questions Forum. I am not quite sure how to move it.)

 

[Petr Mugver]

Decompose the circuit in 3 loops:

the first is 9V-6R-12R clockwise
the second is 15V-10R-24R anticlockwise
the third is 12R-24R anticlockwise

call i1, i2, i3 the 3 currents in each loop.

Write Ampère's law:

6i1 + 12(i1 + i3) = 9
10i2 + 24(i2 - i3) = 15
12(i3 + i1) + 24(i3 - i2) = 0

I found:

i1 = 0.2234 A
i2 = 0.7340 A
i3 = 0.4149 A

i3 is the current in the bottom wire, from left to right. It doesn't agree with your book's solution. I made the calculations twice, I can't find the mistake, but the procedure is correct.

 

[Curl]

Book is wrong (it happens). The answer is .41A from left to right

 

[James98765]

Thanks for all of your help. I have seen another solution on the internet that disagrees with the book result. You are both correct to the best of my knowledge.

 

[dbell5]

I was lazy and modeled it (B2SPICE).
0.415A, L to R.

Dave

 

[James98765]

Thanks for your help!

 

[get_physical]

Decompose the circuit in 3 loops:

the first is 9V-6R-12R clockwise
the second is 15V-10R-24R anticlockwise
the third is 12R-24R anticlockwise

call i1, i2, i3 the 3 currents in each loop.

Write Ampère's law:

6i1 + 12(i1 + i3) = 9
10i2 + 24(i2 - i3) = 15
12(i3 + i1) + 24(i3 - i2) = 0

Why is it 12R-24R anticlockwise and not -12R+24R since you are going anti clockwise, hence against the current when crossing 24R?

Also, why is it i2-i3 for the 24 term in the second equation?

Thanks