# How solve for current I?

Discussion in 'Homework Help' started by Techspec2, Oct 15, 2015.

1. ### Techspec2 Thread Starter New Member

Oct 15, 2015
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0
I uploaded the file. I knows there's gotta be an easier way then making alot of mesh loops or using nodal analysis. Any help would be great .

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2. ### WBahn Moderator

Mar 31, 2012
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4,918
Have you considered going through and simplifying the network by combining series/parallel combinations of resistors?

3. ### WBahn Moderator

Mar 31, 2012
18,093
4,918
You might also consider identifying the voltages on as many nodes as you can before doing anything else.

4. ### WBahn Moderator

Mar 31, 2012
18,093
4,918
Hint: By inspection, you can determine that the current is the same as you would get if the batteries were 2V each and the resistances were 5 Ω each.

Nov 13, 2010
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6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,516
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Yeah, what W said. There is a bunch of parallel and series resistors. You should be able to simplify the circuit.

7. ### WBahn Moderator

Mar 31, 2012
18,093
4,918
Ideally yes. But there are lots of schematics and schematic packages that take the position that if either of the two top approaches are taken to show non-connected crossing wires that that means that the bottom can be used to show connected crossing wires.

In this case there are not instances of lines that cross without making a junction that are shown in an unambiguous way, so we have to guess. Assuming that they are connected makes the analysis of the circuit possible by inspection, which I suspect was the intent.

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,301
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At first glance, one might think that so many batteries make it more complicated, but in fact the large number of batteries make it possible to work across the network one mesh at a time, with the batteries providing constraints allowing for solution by inspection.