Find the sum of all these electric currents

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Homework Statement

The electrical circuit consists of amperemeters. Constant voltage was applied to ´E and F. In the result, one of the amperemeters showed 7 mA and another showed 18 mA. What does the sum of results of all amperemeters equal to?

Relevant Equations

$$I_4 = maximim$$
$$I_1 = I_2$$
$$I_3 = I_5$$
$$I_7 = I_ {10} = I_8 = I_6$$

First, I got rid of amperemeters with 0 values. These are 9. 11 and 12.
Amperemeter 4 will show the maximum value of electric current as it is placed directly between E and F. But how to know its value? Will it be 18 mA? I doubt because 18 mA is not said to be the maximum value.
All other amperemeters will show smaller results. Number 1 and 2 will show equal results. Number 5 and 3. And number 7, 10, 8 and 6 as current will not go to the 12-9-11 line and the resistance of wires is neglected. Is it correct?

 

[jbriggs444]

First, I got rid of amperemeters with 0 values. These are 9. 11 and 12.

What reasoning are you using to conclude that no current flows through 9, 11 and 12?

If 9 carries no current then it follows that the node to its right is at the same potential as F.
If 11 carries no current then it follows that the node down and to its left is also at the same potential as F.
It then would follow that 8 carries no current because its ends are at equal potential.
It would also follow that 6 carries no current because any current would have no place to go. So it then follows that the node to the left of 6 is also at the same potential as F.
Then we conclude that 3 carries no current and that 2 carries no current.
It follows then that E is at the same potential as F. Which is patently false.

So whatever reasoning was used to conclude that no current flows through 9, 11 and 12 is questionable.

My first inclination looking at this network is to simplify by folding bottom onto top. By symmetry it is immediately obvious that all of the nodes on the bottom are at equal potential with all corresponding nodes on the top.

Edit: Looking ahead without actually solving the problem, it looks like the network can be solved from that point by further collapsing it one piece at a time.

 

[BvU]

Drat, @jbriggs444 was faster. Here's my two cents (one being I fully agree with him) the other being: we are braiinwashed to see circuits as one line (E) at the top and one line (F) at the bottom. Very enlightening in this case. (even though you are rapidly foirced to draw two lines at the top (2-6-11 // 1-7-12) )

 

[scottdave]

So if they are identical ammeters, you can represent each one by identical resistors. With the information you provided, I think you'd have to make the assumption that wiring is zero resistance.

Make sure you understand why some of them may have zero current. Think about a bridge circuit.

 

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So whatever reasoning was used to conclude that no current flows through 9, 11 and 12 is questionable.

I thought incorrectly that 9, 11 and 12 were out of the field made by E and F so the current wouldn't go that line.

is to simplify by folding bottom onto top.

I thought about that. Thank you. But it still remains unclear to me how to get what will be the maximum value.

it looks like the network can be solved from that point by further collapsing it one piece at a time

Using symmetry?

 

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you can represent each one by identical resistors

And then use Kirchhoff's circuit laws?

 

[BvU]

Sketch I had in mind in #3

working from right to left should be feasible.

 

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Sketch I had in mind in #3

View attachment 252562working from right to left should be feasible.

Thank you! But why are 2, 3, 6, 8, 11 included? Isn't it possible to just fold the symmetrical scheme and rewrite it with resistors 1, 4, 5, 7, 9, 10, 12?

 

[haruspex]

Thank you! But why are 2, 3, 6, 8, 11 included? Isn't it possible to just fold the symmetrical scheme and rewrite it with resistors 1, 4, 5, 7, 9, 10, 12?

Good plan.

 

[BvU]

Is the next step -- to me it dawns only after I've made the sketch (did I mention brainwashing ?)