How solve for current I?

WBahn

Joined Mar 31, 2012
26,148
Have you considered going through and simplifying the network by combining series/parallel combinations of resistors?
 

WBahn

Joined Mar 31, 2012
26,148
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.
 

WBahn

Joined Mar 31, 2012
26,148
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.
 

The Electrician

Joined Oct 9, 2007
2,786
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.
 
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